# A polynomial composites

Łukasz Matysiak
Kazimierz Wielki University
Bydgoszcz, Poland
<EMAIL_ADDRESS>
###### Abstract

Polynomial composites were introduced by Anderson, Anderson, and Zafrullah. In
this paper we study many different algebraic properties of polynomial
composites like ACCP, atomic, SR property. We study relationships between
Noetherian polynomial composites certain field extensions.

Keywords: domain, field, irreducible element, polynomial.
2010 Mathematics Subject Classification: Primary 13B25, Secondary 13F20.

##  1 Introduction

By the ring we mean a commutative ring with unity. Let $R$ be an itegral
domain. We denote by $R^{\ast}$ the group of all invertible elements of $R$.

The main motivation of this paper is description polynomial composites as
algebraic object. The related works were started in paper [19], where basic
algebraic properties have been investigated. Continued in [20], where the
focus was on ACCP properties and atomicity. Next, in [22], where we have
research related to various classes of rings as polynomial composites. The
article [21] shows very strong relationships between polynomial composites and
field extensions, and therefore relationships with the Galois theory.

D.D. Anderson, D.F. Anderson, M. Zafrullah in [2] called object $A+XB[X]$ as a
composite, where $A\subset B$ are fields.

In 1972 [15] and in 1976 [6] authors considered the structures in the form
$D+M$, where $D$ is a domain and $M$ is a maximal ideal of ring $R$ with
$D\subset R$. Next, Costa, Mott and Zafrullah ([9], 1978) considered
composites in the form $D+XD_{S}[X]$, where $D$ be a domain and $D_{S}$ be a
localization of $D$ relative to the multiplicative subset $S$. In 1988 [4]
Anderson and Ryckaert studied classes groups $D+M$. Zafrullah in [28]
continued research on structure $D+XD_{S}[X]$ but he showed that if $D$ be a
GCD-domain, then the behaviour of $D^{(S)}=\\{a_{0}+\sum a_{i}X^{i}\mid
a_{0}\in D,a_{i}\in D_{S}\\}=D+XD_{S}[X]$ depends upon the relationship
between $S$ and the prime ideals $P$ of $D$ such that $D_{P}$ be a valuation
domain (Theorem 1, [28]). Fontana and Kabbaj in 1990 ([14]) studied the Krull
and valuative dimensions of composite $D+XD_{S}[X]$. In 1991 there was an
article ([2]) that collected all previous results about composites and authors
began to create a further theory about composites creating results. Of course,
let $K\subseteq L$, where $K$, $L$ are fields, polynomial composites can be
applied to algebraic geometry by the fact that $K+XL[X]/XL[X]=K$, i.e $XL[X]$
is the maximal ideal of $K+XL[X]$. Also in the basics of economics ([5], [17])
we can find the use of polynomials, in particular polynomial composites. In
2009 autors considered a construction of $A+XB[X]$, where $A\subset B$ be an
extension of commutative rings, as particular case of the general construction
of amalgamated algebras introduced in [11], Example 2.5.

In this paper, the considered structures were officially called composites.

In the section 2 we present basic properties of polynomial composites with
possible different configurations of rings, domains, fields, etc. We will show
some simple properties from commutative algebra for these structures. In the
Theorem 2.2 we will show that every nonzero prime ideal in $A+XB[X]$ ($A$, $B$
are fields with $A\subset B$), every prime ideal different from $XB[X]$ is
principal and $A+XB[X]$ is atomic ring. In Corollary 2.3 and Theorem 2.4 we
have a characterization of irreducible polynomials in composites. By
motivation from [23] and [24] in Theorem 2.5 and Corollary 2.6 we show a
characterization of square-free polynomials in composites. Moreover, in [25]
can find a characterization of radical polynomials in composites.

In the section 3 we present many conditions in polynomial composites. Recall,
we say that $R$ satysfies ACCP condition (has ACCP) if every ascending chain
of principal ideals of $R$ stabilizes. Following P.M. Cohn [8] we say that $R$
is atomic if every nonzero nonunit of $R$ can be factored into irreducibles
(atoms). If an integral domain satisfies ACCP, then it is atomic. However,
there are atomic domains that do not satisfy ACCP (the first example was
constructed by A. Grams in [16]). In a Proposition 3.1 we will show when a
ring $R$ such that $A[X]\subseteq R\subseteq B[X]$ ($A$, $B$ are fields) is
ACCP-ring. A complement of this Proposition we have in the Theorem 3.2. In the
Theorem 3.4 we have a condition, when $A+X[B]$ has ACCP ($A$ is domain, $B$ is
quotient field of $A$). The domain $R$ is a bounded factorization domain (BFD)
if it is atomic and for every nonzero nonunit $x\in R$ there is a positive
integer $N$ such that $x=a_{1}\dots a_{n}$ for irreducibles
$a_{1},\dots,a_{n}\in R$ implies that $n\leqslant N$ (Propositions 3.6, 3.7).
We say that $R$ is a half-factorial domain (HFD) if is atomic and any two
factorizations of the same nonzero nonunit of $R$ have the same number of
irreducibles (counting repetitions) (Propositions 3.8, 3.9). Clearly a UFD is
a HFD, a HFD is also a BFD, and hence satisfies ACCP (see [3]). The domain $R$
is an idf-domain (for irreducible-divisor-finite) if each nonzero element of
$R$ has at most a finite number of nonassociate irreducible divisors
(Propositions 3.10, 3.11). A domain is called finite factorization domain
(FFD) if each nonzero nonunit element has only a finite number of nonassociate
divisors (Proposition 3.12). A UFD is a FFD, a FFD is a HFD, a FFD is a BFD
(see [3]). In general,

$\begin{array}[]{ccccccccc}&&HFD\\\
&\mbox{\rotatebox{-135.0}{$\Leftarrow$}}&\Uparrow&\mbox{\rotatebox{135.0}{$\Leftarrow$}}\\\
UFD&\Rightarrow&FFD&\Rightarrow&BFD&\Rightarrow ACCP&\Rightarrow&atomic\\\
&\mbox{\rotatebox{135.0}{$\Leftarrow$}}&\Downarrow\\\ &&idf\end{array}$

Recall that $R$ is an S-domain (S is for Seidenberg) if for each height-one
prime ideal $P$ of $R$, $htP[X]=1$ in $R[X]$ (Proposition 3.15). A commutative
ring $R$ is called a Hilbert ring if every prime ideal of $R$ is an
intersection of maximal ideals of $R$ (Theorem 3.16).

The main motivation of the Section 4 is to answer the following question.

Question:
Is there a relationship between certain field extensions $K\subset L$ and
polynomial composites $K+XL[X]$?

We present a full possible characterization of polynomial composites in the
form $K+XL[X]$, where $K$, $L$ are fields, with respect to a given extension
with appropriate additional assumptions. We also present a full possible
characterization of some extensions of fields $K\subset L$ expressed in the
language of polynomial composites $K+XL[X]$ as Noetherian rings with
appropriate assumptions. From Theorem 4.5 we will know that $K+XL[X]$ is
Noetherian iff $K\subset L$ is finite. In Propositions 4.6 – 4.11 we can find
similar statements with such extensions as algebraic, separable, normal,
Galois. We will remind results of Magid from [18] (Theorems 4.12, 4.13) and we
will supplement these statements with additional conditions (Theorems 4.14 and
4.15).

In the last Section we describe an SR-property in polynomial composites. In
[25] we have general results for any integral domains and monoids. Recall from
[23] that an $SR$-domain (monoid) is a domain (monoid) in which every square-
free element is radical. Naturally, every radical element is square-free, so
in $SR$-domain (monoid) the set of square-free is the same like the set of
radical elements.

##  2 Basic properties

The aim of this chapter will be to examine the simplest structural properties
of the considered structures.

Consider $A$ and $B$ as rings such that $A\subset B$. Put $T=A+XB[X]$. The
structure defined in this way is called a composite. (The definition comes
from [2]). Later results will require more assumptions.

Let’s look at invertible polynomials.

######  Proposition 2.1.

Let $f=a_{0}+a_{1}X+\dots+a_{n}X^{n}\in T$ for any $n\geq 0$. Then $f\in
T^{\ast}$ if and only if $a_{0}\in A^{\ast}$ and $a_{1},a_{2},\dots,a_{n}$ are
nilpotents.

###### Proof.

We know that if $R$ is a ring then $f=a_{0}+a_{1}X+\dots+a_{n}X^{n}\in
R[X]^{\ast}$ if and only if $a_{0}\in R^{\ast}$ and $a_{1},a_{2},\dots,a_{n}$
are nilpotents. In our Proposition we have $a_{1},a_{2},\dots,a_{n}$ are
nilpotents. Of course we get $a_{0}\in A^{\ast}$. ∎

######  Theorem 2.2.

Let $A$, $B$ be fields such that $A\subset B$. Let $X$ be an indeterminate
over $B$ and let $T=A+XB[X]$. Then

  * (1)

every nonzero prime ideal of $T$ is maximal;

  * (2)

every prime ideal $P$ different from $XB[X]$ is principal;

  * (3)

$T$ is atomic.

###### Proof.

(1) First note that $XB[X]$ is maximal since $T/XB[X]\cong K$. Let $P$ be a
nonzero prime ideal of $T$. Now $X\in P$ implies $(XB[X])^{2}\subseteq P$ and
hence $XB[X]\subseteq P$ so $P=XB[X]$. So suppose that $X\notin P$. Then for
$N=\\{1,X,X^{2},\dots\\}$, $P_{N}$ is a prime ideal in the PID
$B[X,X^{-1}]=T_{N}$. (In fact, $T_{P}\supseteq B[X,X^{-1}]$ is a DVR (discrete
valuation ring).) So $P$ is minimal and is also maximal unless $P\subsetneq
XB[X]$. But let $k_{n}X^{n}+\dots+k_{s}X^{s}\in P$ with $k_{n}\neq 0$, where
$k_{n},\dots,k_{s}\in L$ for every $n,s$. Then
$X^{n+1}+k_{n}^{-1}k_{n+1}X^{n+2}+\dots+k_{n}^{-1}k_{s}X^{s}\in P$, so
$X\notin P$ implies that $1+k_{n}^{-1}k_{n+1}X+\dots+k_{n}^{-1}k_{s}X^{s-n}\in
P$, a contradiction. So every nonzero prime ideal is maximal.

(2) If $P$ is different from $XB[X]$, then it contains an element of the form
$1+Xf(X)$, where $f(X)\in B[X]$. Now if $1+Xf(X)$ can be factored in $B[X]$ it
can be written as $(1+Xg(X))(1+Xh(X))$. Hence $1+Xf(X)$ is irreducible in $T$
if and only if it is irreducible in $B[X]$.

Now let $1+Xf(X)$ be irreducible in $T$ and suppose that $1+Xf(X)\mid
h(X)k(X)$ in $T$. Then $1+Xf(X)\mid h(X)k(X)$ in $B[X]$, and so in $B[X]$ we
have, say $1+Xf(X)\mid h(X)$. Then, in $B[X]$, $h(X)=(1+Xf(X))d(X)$. Now
$d(X)$ can be written as $d(X)=aX^{r}(1+Xp(X))$. If $r>0,d(X)\in T$, while if
$r=0,h(X)=(1+Xf(X))(a(1+Xp(X))$ and $a\in A$ because $h(X)\in T$. In either
case, $d(X)\in T$ and so $1+Xf(X)\mid h(X)$ in $T$. Consequently, in $T$ every
irreducible element of the type $1+Xf(X)$ is prime.

Now since every element of the form $1+Xf(X)$ is a product of irreducible
elements of the same form and hence is a product of prime elements, it follows
that every prime ideal of $P$ different from $XB[X]$ contains a principal
prime and hence is actually principal.

(3) Thus a general element of $T=A+XB[X]$ can be written as $aX^{r}(1+Xf(X))$,
where $a\in B$ (with $a\in A$ if $r=0$) and $1+Xf(X)$ is a product of primes.
∎

######  Corollary 2.3.

Let $A$, $B$ be fields such that $A\subset B$. Consider $D=A+XB[X]$. Then
$\operatorname{Irr}D=\\{aX,a\in B\\}\cup\\{a(1+Xf(X)),a\in A,f\in
B[X],1+Xf(X)\in\operatorname{Irr}B[X]\\}.$

######  Theorem 2.4.

Consider $T=A+XB[X]$, where $A$, $B$ are fields, $A\subset B$. Then
$f\in\operatorname{Irr}T$ if and only if $f\in\operatorname{Irr}B[X],f(0)\in
A$.

###### Proof.

Suppose that $f\notin\operatorname{Irr}B[X]$ or $f(0)\notin A$. If $f(0)\notin
A$, then $f\notin T$, so $f\notin\operatorname{Irr}B[X]$. Now, assume that
$f\notin\operatorname{Irr}B[X]$. Then $f=gh$, where $g,h\in B[x]\setminus B$.
Let $g=a_{0}+a_{1}X+\dots+a^{n}X^{n},h=b_{0}+b_{1}X+\dots+b_{m}X^{m}$. We have
$f=(a_{0}+a_{1}X+\dots+a^{n}X^{n})(b_{0}+b_{1}X+\dots+b_{m}X^{m})$. Then
$f=\big{(}1+\dfrac{a_{1}}{a_{0}}X+\dots+\dfrac{a_{n}}{a_{0}}X^{n}\big{)}(a_{0}b_{0}+a_{0}b_{1}X+\dots+a_{0}b_{m}X^{m})$,
where $a_{0}b_{0}=f(0)\in A$. Now, suppose that $f\notin\operatorname{Irr}T$.
If $f\notin T$, then $f(0)\notin A$. Now, assume that $f\in T$. Then we have
$f=gh$, where $g,h\in T\setminus A$. This implies $g,h\in B[x]\setminus B$. ∎

######  Theorem 2.5.

Let $K$ and $L$ be fields such that $K\subset L$ and let $T=K+XL[X]$. Then
$\operatorname{Sqf}(T)=(\operatorname{Sqf}(L[X])\cap
T)\cup\\{X^{2}h;h\in\operatorname{Sqf}(L[X]),h(0)\notin\\{a^{2}b;a\in L,b\in
K\\}\\}$.

###### Proof.

Let $f\in\operatorname{Sqf}(T)\setminus\operatorname{Sqf}(L[X])$. There are
some $g\in(L[X])\setminus(L[X]^{\ast})$ and $k\in L[X]$ such that $f=g^{2}k$.
Set $c=g(0)$. If $c\neq 0$ then since $c^{-1}g$, $c^{2}k\in T$ and
$f=(c^{-1}g)^{2}c^{2}k$, we have $c^{-1}g\in T^{\ast}$ and $g\in L[X]^{\ast}$,
a contradiction. Therefore, $c=0$, and thus $f=X^{2}h$ for some $h\in L[X]$.
Since $f\in\operatorname{Sqf}T$, we infer $h(0)=\neq 0$. If $h(0)=a^{2}b$ for
some $a\in L$ and $b\in L$, then $aX$, $a^{-2}h\in T$, and
$f=(aX)^{2}a^{-2}h$, which contradicts the fact that
$f\in\operatorname{Sqf}T$. This implies that $h(0)\notin\\{a^{2}b;a\in L,b\in
K\\}$. Let $r$, $s\in L[X]$ be such that $h=r^{2}s$. Since $h(0)\neq 0$, we
infer $r(0)\neq 0$. Set $d=r(0)$. Then $f=(d^{-1}r)^{2}d^{2}sX^{2}$ and
$d^{-1}r$, $d^{2}sX^{2}\in T$. Consequently, $d^{-1}r\in T^{\ast}$, and hence
$r\in L[X]^{\ast}$. This shows that $h\in\operatorname{Sqf}L[X]$.

Since $(L[X]^{\ast})\cap T=T^{\ast}$, it follows that
$\operatorname{Sqf}(L[X])\cap T\subset\operatorname{Sqf}T$. Now let
$h\in\operatorname{Sqf}L[X]$ be such that $h(0)\notin\\{a^{2}b;a\in L,b\in
K\\}$. It remains to show that $X^{2}h\in\operatorname{Sqf}T$. Clearly,
$X^{2}h\in T$. Let $r$, $s\in T$ be such that $X^{2}h=r^{2}s$. Assume that
$r\in XL[X]$. Then $r=Xt$ for some $t\in L[X]$, so $h=t^{2}s$. Therefore,
$h(0)=t(0)^{2}s(0)\in\\{a^{2}b;a\in L,b\in K\\}$, a contradiction.
Consequently, $r\notin XL[X]$, and thus $s=X^{2}w$ for some $w\in L[X]$. We
infer $h=r^{2}w$, and hence $r\in(L[X]^{\ast})\cap T=T^{\ast}$. ∎

######  Corollary 2.6.

Let $K$ and $L$ be fields such that $K\subset L$ and let $T=K+XL[X]$. Then
$\operatorname{Sqf}T=\operatorname{Sqf}(L[X])\cap T$ iff $L=\\{a^{2}b;a\in
L,b\in K\\}$. In particular, if $L$ is algebraically closed, then
$\operatorname{Sqf}T=\operatorname{Sqf}(L[X])\cap T.$

###### Proof.

It follows from Proposition LABEL:t2.3 that if $L=\\{a^{2}b;a\in L,b\in K\\}$,
then $\operatorname{Sqf}T=\operatorname{Sqf}(L[X])\cap T$. Now let
$L\neq\\{a^{2}b;a\in L,b\in K\\}$. There is some $c\in
L\setminus\\{a^{2}b;a\in L,b\in K\\}$. By Proposition LABEL:t2.3, we have
$X^{2}x\notin\operatorname{Sqf}(L[X])$, and thus
$\operatorname{Sqf}T\neq\operatorname{Sqf}(L[X])\cap T$. Finally, if $L$ is
algebraically closed, then $L=\\{a^{2};a\in L\\}$ and the statement follows. ∎

######  Example 2.7.

Let $L$ be a field with char $(L)=2$ such that $L$ is not perfect, let $K$ be
the prime subfield of $L$ and $T=K+XL[X]$. Then
$\operatorname{Sqf}T\neq\operatorname{Sqf}(L[X])\cap T$.

###### Proof.

Since char $(L)=2$ and $L$ is not perfect, we have $L\neq\\{a^{2};a\in L\\}$.
Since $K=\\{0,1\\}$, this implies that $L\neq\\{a^{2}b;a\in L,b\in K\\}$. It
is an immediate consequence of Corollary LABEL:c2.4 that
$\operatorname{Sqf}T\neq\operatorname{Sqf}(L[X])\cap T$. ∎

In particular, if $T=\mathbb{R}+X\mathbb{C}[X]$, then
$\operatorname{Irr}T=\\{a+bX;a\in\mathbb{R},b\in\mathbb{C}\setminus\\{0\\}\\}$
and $\operatorname{Sqf}T=\\{a\prod_{b\in
B}(1+bX);a\in\mathbb{R}\setminus\\{0\\},B\subset\mathbb{C},B\text{ is
finite}\\}\cup\\{aX\prod_{b\in
B}(1+bX);a\in\mathbb{C}\setminus\\{0\\},B\subset\mathbb{C},B\text{ is
finite}\\}$.

If $L$ and $K$ are finite fields and it is a proper extension, then $K+XL[X]$
is a non-factorial ACCP domain (see , [13]).

Recall that a ring $R$ satisfies ACCP if each chain of principal ideals of $R$
is stabilize.

######  Example 2.8.

In [2] Example 5.1 showed an example of an integral domain $R$ which satisfies
ACCP, but whose integral closure does not satisfy ACCP. It mean
$R=\mathbb{Z}+X\overline{\mathbb{Z}}[X]$, where $\overline{\mathbb{Z}}$ be the
ring of all algebraic integers. $R$ satisfies ACCP. For if not, then there is
an infinite properly ascending chain of pricipal ideals of $R$. Since the
degrees of the polynomials generating these principal ideals are
nonincreasing, the degrees eventually stabilize. The principal ideals in
$\overline{\mathbb{Z}}$ generated by the leading coefficients of these
polynomials gives an infinite ascending chain
$a_{1}\overline{\mathbb{Z}}\subsetneq a_{2}\overline{\mathbb{Z}}\subsetneq...$
where each $a_{n}/a_{n+1}\in\mathbb{Z}$. Thus all $a_{n}\in\mathbb{Q}[a_{1}]$.
Let $A=\overline{\mathbb{Z}}\cap\mathbb{Q}[a_{1}]$. Then $a_{1}A\subsetneq
a_{2}A\subsetneq\dots\subsetneq A$, a contradiction since $A$ is Dedekind.

######  Example 2.9.

Let $R=\mathbb{R}+X\mathbb{C}[X]$. So $R$ is a HFD, so has ACCP, then atomic.

##  3 ACCP, BFD, HFD, idf, FFD, S, Hilbert properties

Note that for $R$ a ring between $A[X]$ and $B[X]$ ($A$, $B$ are fields such
that $A\subset B$), $R$ has ACCP if and only if for every $n\geqslant 0$, any
ascending chain of principal ideals generated by polynomials of degree $n$
terminates. If $B$ be a quotient field of $A$, the Proposition 5.2 [2] may be
used to show that ring $R$ satisfies ACCP. But we can minimalize assumptions
by composites.

######  Proposition 3.1.

Let $A$, $B$ are fields such that $A\subset B$. Let $R$ be a ring with
$A[X]\subseteq R\subseteq B[X]$. Then $R$ has ACCP if and only if $R\cap B$
has ACCP and for each ascending chain of polynomials $f_{1}R\subseteq
f_{2}R\subseteq f_{3}R\subseteq\dots$ where $f_{i}\in R$ all have the same
degree, then there is $d\in(R\cap B)\setminus\\{0\\}$ such that
$df_{i}\in(R\cap B)[X]$.

###### Proof.

($\Rightarrow$) Since $(R\cap B)^{\ast}=R^{\ast}\cap B$, $R$ has ACCP implies
$R\cap B$ has ACCP. The chain $f_{1}R\subseteq f_{2}R\subseteq\dots$ be
stationary, say $f_{n}R=f_{n+1}R=\dots$. So $f_{n+1}=u_{i}f_{i}$, where
$u_{i}$ is a unit of $R\cap B$. Since $f_{n}\in B[X]$, there exists a $0\neq
a\in A\subseteq R\cap B$ with $af_{n}\in A[X]\subseteq R$. But then
coefficients of $af_{n+1}=u_{i}df_{n}$ all lie in $R\cap B$.

($\Leftarrow$) Let $f_{1}R\subseteq f_{2}R\subseteq\dots$ be an ascending
chain in $R$. Since $\deg f_{i+1}\leqslant\deg f_{i}$, eventually $f_{i}$ have
the same degree, so without loss of generality, we can assume that $\deg
f_{1}=\deg f_{2}=\dots$. By hypothesis there exists $0\neq a\in R\cap B$ with
$af_{i}\in(R\cap B)[X]$. Now $f_{i}R\subseteq f_{i+1}R$ implies
$f_{i}=f_{i+1}b$, where $b\in R$ has degree $0$, so $b\in R\cap B$. Hence
$af_{i}(R\cap B)[X]\subseteq af_{i+1}(R\cap B)[X]$. But $R\cap B$ has ACCP and
hence so does $(R\cap B)[X]$. So for large $n$, $f_{n}(R\cap
B)[X]=f_{n+1}(R\cap B)[X]=\dots$, and hence $f_{n}R=f_{n+1}R=\dots$. ∎

A Theorem 3.2 shows that between $A[X]$ and $A+XB[X]$ we can find a structure
which satysfying ACCP condition.

######  Theorem 3.2.

Let $A$, $B$ are fields such that $A\subset B$. Let $C$ be a domain such that
$A[X]\subseteq C\subseteq A+XB[X]$. Suppose that for each $n\geq 0$, there
exists $a_{n}\in A\setminus\\{0\\}$ so that $a_{n}f\in A[X]$ for all $f\in C$
with $\deg f\leq n$. Then $C$ has ACCP if and only if $A$ has ACCP.

###### Proof.

This is a special case of Proposition 2.1. The second part of this proof: let
us recall $R$ a bounded factorization domain (BFD) if for each nonzero nonunit
$a\in R$, there exists a positive integer $N(a)$, so that if $a=a_{1}\dots
a_{s}$ where each $a_{i}$ is nonunit, then $s\leqslant N(a)$. It is very known
fact that a BFD has ACCP but the converse is false. In the proof suppose that
$A$ is a BFD. Let $0\neq f\in C$ have a degree $n$ and leading coefficient
$b$. Write $f=g_{1}\dots g_{s}g_{s+1}\dots g_{m}$, where $g_{1},\dots,g_{m}\in
C$ are nonunits with $g_{1},\dots,g_{s}\in A$ and $g_{s+1},\dots,g_{m}\in C$
have a degree $\geqslant 1$. Now $g_{s+1}\dots g_{m}$ has a degree $n$, so
$m-s\leqslant n$. Also, $r_{n}g_{s+1}\dots g_{m}\in A[X]$, say it has leading
coefficient $c\in A$. Then $r_{n}b=g_{q}\dots g_{s}c$. But $A$ is a BFD, so
there is a bound on the number of factors for $r_{n}b$ and hence on $s$. Thus
the $m$ if $f=g_{1}\dots g_{s}g_{s+1}\dots g_{m}$ has an upper bound.
Conversely, if $C$ be a BFD, easy to see that $A$ be a BFD without any
additional hypothesis on $C$. ∎

######  Proposition 3.3.

Let $A$ be a domain. Then $A$ has ACCP if and only if $A[X]$ has ACCP.

###### Proof.

The equivalene can be easily proved by comparing degrees (resp., orders). ∎

It also turn out that ACCP property moves between $A$ and $A+XB[X]$. This is
important because we do not have to choose a general polynomial, and we can
limit the inclusion to the smallest composite needed. Such a significant
limitation of a polynomial to a composite is important, e.g. in Galois theory
in commutative rings.

######  Theorem 3.4.

Let $A$ be an integral domain with quotient field $B$. An $A$ has ACCP if and
only if $A+XB[X]$ has ACCP.

###### Proof.

From Theorem 3.2 we have $A[X]\subseteq A+XB[X]\subseteq A+XK[X]$, where $K$
be an overfield of $B$. We can to prove that for each $n\geq 0$, there exists
$a_{n}\in A\setminus\\{0\\}$ for all $f\in A+XB[X]$ with $\deg f\leq n$.
Because $A$ has ACCP then from Theorem 3.2 we get $A+XB[X]$ has ACCP.
Conversely, because $A+XB[X]$ has ACCP then $A$ has ACCP. ∎

In papers [19] and [20], polynomial composites with the property of atomicity
and ACCP are presented. The results below are complementary.

######  Proposition 3.5.

Let $T=K+XL[X]$, where $K$, $L$ are fields with $K\subset L$. Let $D$ be a
subring of $K$ and $R=D+XL[X]$. Then:

  * (a)

$R$ is atomic if and only if $T$ is atomic and $D$ is a field.

  * (b)

$R$ satisfies ACCP if and only if $T$ satisfies ACCP and $D$ is a field.

###### Proof.

First suppose that $D$ is not a field. Then $f=d\dfrac{m}{f}$ for each $f\in
XL[X]$ and $d\in D^{\ast}$. Thus no element of $XL[X]$ is irreducible ($XL[X]$
is a maximal ideal of $T$). Hence if $R$ is either atomic or satisfies ACCP,
$D$ must be a field. So let $D$ be a field.

(a) Up to multiplication by a $\alpha\in K^{\ast}$ (resp. $\alpha\in
D^{\ast}$), each element of $T$ (resp. $R$) has the form $f$ or $1+f$ for some
$f\in XL[X]$. Each of these elements is irreducible in $R$ if and only if it
is irreducible in $T$ ([10], Lemma 1.5; 27). If $x$ is a product of
irreducibles, we may assume that each irreducible factor has the form $f$ or
$1+f$ for some $f\in XL[X]$. Thus $x$ is a product of irreducible elements in
$R$ if and only if it is a product of irreducible elements in $T$. Hence $R$
is atomic if and only if $T$ is atomic.

(b) We first observe that a principal ideal of $R$ or $T$ may be generated by
either $f$ or $a+f$ for some $f\in XL[X]$. Let $f$, $g\in XL[X]$. It easily
verified that $(1+f)R\subset(1+g)R$ if and only if $(1+f)T\subset(1+g)T$,
$fR\subset(1+g)R$ if and only if $fT\subset(1+g)T$, and $fR\subset gR$ if and
only if $fT\subset gT$. Also, if $fT\subset gT$, then $fR\subset(\alpha g)R$
for some $\alpha\in K^{\ast}$. Hence, to each chain of principal ideals of
length $s$ in $R$ starting at $fR$ (resp., $(1+f)R$), there corresponds a
chain of principal ideals of length $s$ in $T$ starting at $fT$ (resp.,
$(1+f)T$), and conversely. Thus $R$ satisfies ACCP if only if $T$ satisfies
ACCP. ∎

Propositions 3.6, 3.7 represent the BFD property in polynomial composites.

######  Proposition 3.6.

If $A+XB[X]$ is a Noetherian domain, where $A\subset B$ are domains, then
$A+XB[X]$ is a BFD.

###### Proof.

, Proposition 2.2. ∎

######  Proposition 3.7.

Let $T=K+XL[X]$, where $K\subset L$ are fields. Let $D$ be a subring of $K$
and $R=D+XL[X]$. Then $R$ is a BFD if and only if $T$ is a BFD and $D$ is a
field.

###### Proof.

First suppose that $R$ is BFD. Then $D$ must be a field (, Proposition 1.2).
Again from the proof of (, Proposition 1.2) we get that $R$ is a BFD if and
only if $T$ is a BFD. ∎

Propositions 3.8 and 3.9 represents the HFD property in polynomial composites.

######  Proposition 3.8.

Let $T=K+XL[X]$, where $K\subset L$ are fields. Let $D$ be a subring of $K$
and $R=D+XL[X]$. Then $R$ is a HFD if and only if $D$ is a field and $T$ is a
HFD.

###### Proof.

As in Proposition 1.2 (), $D$ is necessarily a field. The proof of Proposition
1.2 shows that a factorization into irreducibles in $R$ has the same length as
such a factorization in $T$. Hence $R$ is a HFD if and only if $T$ is a HFD. ∎

######  Proposition 3.9.

Let $A$ be a subring of a field $K$. Then $R=A+XK[X]$ is a HFD if and only if
$A$ is a field.

###### Proof.

($\Rightarrow$) Clearly, $R$ a HFD implies that $A$ is a HFD. Suppose that $A$
is not a field, so there is an irreducible element $a\in A$. Then
$X=a^{n}(X/a^{n})$ for all $n\in\mathbb{N}$. Thus $A$ must be a field.

($\Leftarrow$) Suppose that $A$ is a field. By [19] Theorem 2.1, $R=A+XK[X]$
is atomic. The proof of Theorem 2.1 [19] shows that an irreducible element of
$R$ is of the for $aX$, where $a\in K$ or $a(1+Xf[X])$, where $a\in A$,
$f(X)\in K[X]$, and $1+Xf(X)$ is irreducible in $K[X]$. Thus for any $g(X)\in
R$, the number of irreducible factors from $R$ is the same as the number of
irreducible factors in a representation of $g(X)$ as a product of irreducible
factors from the PID $K[X]$. Hence $R$ is a HFD. ∎

Recall that $R$ is an idf-domain if each nonzero element of $R$ has at most a
finite number of nonassociate irreducible divisors.

######  Proposition 3.10.

Let $T=K+XL[X]$, where $K\subset L$ are fields. Let $M$ be a subfield of $K$
and $R=M+XL[X]$. Then:

  * (a)

Suppose that $XL[X]$ contains an irreducible element. Then $R$ is an idf-
domain if and only if $T$ is an idf-domain and the multiplicative group
$K^{\ast}/M^{\ast}$ is finite.

  * (b)

Suppose that $XL[X]$ contains no irreducible elements. Then $R$ is an idf-
domain if and only if $T$ is an idf-domain.

###### Proof.

(a) We first note that an element of $XL[X]$ is irreducible in $R$ if and only
if it is irreducible in $T$. Let $f\in XL[X]$ be irreducible. First suppose
that $R$ is an idf-domain. Then $af\mid f^{2}$ for all $a\in K^{\ast}$. Note
that $af$ and $bf$ are irreducible in both $R$ and $T$, and that they are
associates in $R$ if and only if $a$ and $b$ lie in the same coset in
$K^{\ast}/M^{\ast}$. Hence $K^{\ast}/M^{\ast}$ if finite. Let $y\in T$. By
multiplying by a suitable $a\in K^{\ast}$, we may assume that $y\in R$. Let
$y_{1},y_{2},\dots,y_{n}$ be the distinct nonassociate irreducible divisors of
$y$ in $R$. It is easily verified that any irreducible divisor of $y$ in $T$
is associated to one of the $y_{i}$’s. Thus $T$ is also an idf-domain.
Conversely, suppose that $T$ is an idf-domain and that $K^{\ast}/M^{\ast}$ is
finite. Let $z\in R$. Let $z_{1},z_{2},\dots,z_{r}$ be a complete set of
nonassociate irreducible divisors of $z$ in $T$, which we may assume are all
in $R$, and let $a_{1},a_{2},\dots,a_{s}$ be a set of coset representatives of
$K^{\ast}/M^{\ast}$. Then any irreducible divisor of $z$ in $R$ is an
associate of some $a_{i}z_{j}$. Hence $R$ is an idf-domain.

(b) Since $XL[X]$ has no irreducible elements, an irreducible element in $T$
(resp., in $R$) has the form $a+f$ for some $a\in K^{\ast}$ (resp., $a\in
M^{\ast}$) and $f\in XL[X]$. Hence, up to associates, each has the form $1+f$
for some $f\in XL[X]$. It is then easily verified that
$\\{1+f_{1},1+f_{2},\dots,1+f_{n}\\}$ is a complete set of nonassociate
irreducible divisors of a given element with respect to $R$ if and only if it
is a complete set of nonassociate irreducible divisors with respect to $T$.

∎

######  Proposition 3.11.

Let $T$ be a quasilocal integral domain of the form $K+XL[X]$, where $K\subset
L$ are fields. Let $D$ be a subring of $K$ and $R=D+XL[X]$. If $D$ is not a
field, then $R$ is an idf-domain if and only if $D$ has only a finite number
of nonassociate irreducible elements.

###### Proof.

Let $d$ be a nonzero nonunit of $D$. Then $f=d(f/d)$ shows that no element of
$XL[X]$ is irreducible and $d$ divides each element of $XL[X]$. Also,
$y=d+f=d(1+f/d)$ and $1+f/d\in R^{\ast}$ (since $T$ is quasilocal) shows that
$y$ is irreducible in $R$ if and only if $d$ is irreducible in $D$. Thus $R$
is an idf-domain if and only if $D$ has only a finite number of nonassociate
irreducible elements. ∎

Question If $R$ is an idf-domain, then $R[X]$ be an idf-domain?

The proposition 3.12 represents the FFD property in polynomial composites.

######  Proposition 3.12.

Let $T=K+XL[X]$, where $K\subset L$ are fields. Let $D$ be a subring of $K$
and $R=D+XL[X]$. Then $R$ is a FFD if and only if $T$ is a FFD, $D$ is a
field, and $K^{\ast}/D^{\ast}$ is finite.

###### Proof.

Proof is similar to Proposition 5.2. ∎

Recall an integral domain $D$ is called an S-domain if for each prime ideal
$P$ of $D$ with $htP=1$, $htP[X]=1$.

######  Lemma 3.13.

For an integral domain $D$, the following statements are equivalent.

  * (a)

$D$ is an S-domain.

  * (b)

For each prime ideal $P$ of $D$ with $htP=1$, $D_{P}$ is an S-domain.

  * (c)

For each prime ideal $P$ of $D$ with $htP=1$, $\overline{D_{P}}$ is a Prüfer
domain.

###### Proof.

[2] Lemma 3.1 ∎

######  Lemma 3.14.

For any integral domain $D$, $D[X]$ is an S-domain.

###### Proof.

[2], Theorem 3.2. ∎

######  Proposition 3.15.

Let $D$ be an integral domain and $S$ a multiplicatively closed subset of $D$.
Then $D+XD_{S}[X]$ is an S-domain.

###### Proof.

Let $R=D+XD_{S}[X]$ and let $P$ be a height-one prime ideal of $R$. First
suppose that $P\cap S\neq\emptyset$. Then $P\supseteq XD_{S}[X]P=XD_{S}[X]$.
But since $htP=1$, $P=XD_{S}[X]$. But then $P\cap S=\emptyset$, a
cotradiction. Thus we must have $P\cap S=\emptyset$. Then $P_{S}$ is a height-
one prime ideal in $R_{S}=D_{S}[X]$. By Lemma 3.14, $R_{S}$ is an S-domain.
Hence $R_{P}=R_{S_{P_{S}}}$ is also an S-domain by Lemma 3.13
(a)$\Rightarrow$(b). Thus $R$ is an S-domain by Lemma 3.13
(b)$\Rightarrow$(a). ∎

Recall, a commutative ring $R$ is called a Hilbert ring if every prime ideal
of $R$ is an intersection of maximal ideals of $R$. In [26] it was shown that
if $D\subseteq K$, where $K$ is a field, then $D+XK[X]$ is a Hilbert domain if
and only if $D$ is a Hilbert domain. Thus if $D$ is a PID that is not a field
and $K$ is the quotient field of $D$, then $D+XK[X]$ is a two-dimensional,
non-Noetherian, Bézout-Hilbert domain in which every maximal ideal is
principal.

######  Theorem 3.16.

Let $D$ be an integral domain and $S$ a multiplicatively closed subset od $D$
with the property that for a prime $P$ of $D$ with $P\cap S\neq\emptyset$,
then $Q\cap S\neq\emptyset$ for each prime $0\neq Q\subseteq P$. Then
$R=D+XD_{S}[X]$ is a Hilbert domain if and only if $D$ and $D_{S}$ are Hilbert
domains.

###### Proof.

($\Rightarrow$) Suppose that $R$ is a Hilbert domain. Then $D\cong
R/XD_{S}[X]$ is also a Hilbert domain. Suppose that $D_{S}$ is not a Hilbert
domain. Let $Q$ be a nonzero prime ideal od $D$ with $Q\cap S\emptyset$. Since
$D$ is a Hilbet domain, $Q=\bigcap_{\alpha}M_{\alpha S}$, where
$\\{M_{\alpha}\\}$ is the set of maximal ideals of $D$ containing $Q$. Since
$Q\cap S=emptyset$ by the hypothesis on $S$, each $M_{\alpha}\cap
S=\emptyset$. Hence $Q_{S}=\bigcap M_{\alpha S}$ is an intersection of maximal
ideals of $D_{S}$. So every nonzero prime ideal of $D_{S}$ is an intersection
of maximal ideals. Hence there is a nonzero element $u\in D$ such that $u$ is
in every nonzero prime ideal of $D_{S}$. Consider $u+X\in R$. Let $P$ be prime
ideal of $R$ minimal over ($u+X$) with $P\cap D=0$. (Such a prime $P$ exists
since $(u+X)\cap(D\setminus\\{0\\})=\emptyset$). If $Q$ is a prime ideal of
$R$ with $P\subsetneq Q$, then $Q\cap D\neq 0$. For otherwise in $D_{S}[X]$,
$0\neq P_{S}\subsetneq Q_{S}$ would both contract to $0$. Now if $Q\cap
S\neq\emptyset$, then $X\in XD_{S}[X]\subseteq Q$, while if $Q\cap
S=\emptyset$, then $u\in(Q_{S}\cap D_{S})\cap D\subseteq Q$. So every prime
ideal of $R$ properly containing $P$ contains both $u$ and $X$. Hence $P$ is
not the intersection of the maximal ideals containing it, contradicting the
fact that $R$ is a Hilbert domain. So $D_{S}$ must also be a Hilbert domain.

($\Leftrightarrow$) Let $Q$ be a prime ideal of $R$. Suppose that $Q\cap
S\neq\emptyset$. Then $XD_{S}[X]=XD_{S}[X]Q\subseteq Q$, so $Q=Q\cap
D+XD_{S}[X]$. Since $D$ is a Hilbert domain, $Q\cap D$ is an intersection of
maximal ideals, hence so is $Q$. So we may suppose that $Q\cap S=\emptyset$.
Then since $D_{S}[X]$ is a Hilbert domain, $Q_{S}=\bigcap_{\alpha}M_{\alpha}$,
where $\\{M_{\alpha}\\}$ is the set of maximal ideals of $D_{S}[X]$ containing
$Q_{S}$. Then $Q=\bigcap_{\alpha}(M_{\alpha}\cap R)$. So it suffices to show
that each $M_{\alpha}\cap R$ is a maximal ideal of $R$. So let $M$ be a
maximal ideal of $D_{S}[X]$. Then $M=N_{S}$, where $N$ is a prime ideal of
$D[X]$. Now $M$ maximal implies $M\cap D_{S}$ is maximal since $D_{S}$ is
Hilbert domain. If $M\cap D_{S}=0$, then $D_{S}$ is a field and hence $R$ is a
Hilbert domain ([26], Theorem 5). So we may assume that $M\cap D_{S}\neq 0$.
Then by hypothesis on $S$, $(M\cap D_{S})\cap D=N\cap D$ must also be maximal.
Since $N\supsetneq(N\cap D)[X]$, $N$ must be a maximal ideal of $D[X]$. Hence
$D[X]/N\subseteq R/M\cap R\subseteq D_{S}[X]/M=D_{S}[X].N_{S}=D[X]/N$ since
$D[X]/N$ is a field. Therefore $M\cap R$ is a maximal ideal. ∎

The next Proposition says that every polynomial composite is a one-dimensional
Bézout domain.

######  Proposition 3.17.

Let $K\subset L$ be a pair of fields with $L$ purely inseparable over $K$
(that is, $charK=p>0$ and for each $l\in L$, there exists a natural number
$n=n(l)$ with $l^{p^{n}}\in K$). Then every ring $R$ between $K[X]$ and $L[X]$
is a one-dimensional almost Bézout domain.

###### Proof.

Since $K[X]\subset L[X]$ is an integral extension, $\dim R=\dim K[X]=1$. For
each $f\in L[X]$, $f^{p^{n}}\in K[X]$ for $n$ large enough. Hence for $f,g\in
R$, $f^{p^{n}},g^{p^{n}}\in K[X]$ for some $n\in\mathbb{N}$. But
$(f^{p^{n}},g^{p^{n}})K[X]$ is principal. Hence $(f^{p^{n}},g^{p^{n}})R$ is
principal. ∎

Let $K$ be a field, $D$ a subring of $K$. Every ring $R$ between $D[X]$ and
$K[X]$ has a composite cover ([2] Section II), i.e. the unique minimal
overring of $R$ that is a composite. Recall $I(B,A)=\\{f(X)\in B[X]\mid
f(A)\subseteq A\\}$.

######  Proposition 3.18.

(a) Let $R$ be a domain with qoutient field $K$. Suppose that for each $0\neq
r\in R$, $R/(r)$ is finite. Then the composite cover of $I(K,R)$ is $R+XK[X]$.

(b) Let $A\subseteq B$ be rings where $A$ is finite. Then the composite cover
of $I(B,A)$ is $A+XB[X]$.

###### Proof.

(a) Let $r$ be a nonzero nonunit of $R$ and let
$R/(r)=\\{r_{1}+(r),\dots,r_{n}+(r)\\}$. Set
$f(X)=\dfrac{1}{r}(X-r_{1})\dots(X-r_{n})\in K[X]$. Now for $a\in R$,
$a+(r)=r_{i}+(r)$ for some $i$, so $a-r_{i}=sr$ for some $s\in R$. Hence
$f(a)=\dfrac{1}{r}(sr)\prod_{j\neq 1}(a-r_{j})\in R$. So
$f(X)=\dfrac{1}{r}X^{n}+\dots\in I(K,R)$ and hence $I(K,R)$ has composite
cover $R+XK[X]$.

(b) For each $b\in B$, $f(X)=b(\prod_{a\in A}(X-a))\in I(B,A)$. ∎

##  4 Characterization of field extensions in terms of polynomial composites

Let’s start with an auxiliary lemmas that will help.

######  Lemma 4.1.

Let $\varphi\colon K_{1}\to K_{2}$ be an isomorphism of fields and
$\varPsi\colon K_{1}[X]\to K_{2}[X]$ be an isomorphism of polynomials ring. If
polynomial $f_{1}\in\operatorname{Irr}K_{1}[X]$, $f_{1}$ has a root $a_{1}$ in
an extended $L_{1}$ of $K_{1}$ and polynomial $f_{2}=\varPsi(f_{1})$ has a
root $a_{2}$ in an extended $L_{2}$ of $K_{2}$, then there exists
$\varPsi^{\prime}\colon K_{1}(a_{1})\to K_{2}(a_{2})$, which is an extension
of $\varphi$ and $\varPsi^{\prime}(a_{1})=a_{2}$ holds.

###### Proof.

[7], Lemma 2, p. 105. ∎

######  Lemma 4.2.

If $\varphi\colon K\to L$ be an embedding field $K$ to an algebraically closed
field $L$, and $K^{\prime}$ be an algebraic extension of $K$, then there
exists an embedding $\varPsi\colon K^{\prime}\to L$ which is an extension of
$\varphi$.

###### Proof.

[7], Lemma 4, p. 109. ∎

######  Lemma 4.3.

If $L$ be a finite field extension of $K$, then $L$ is a Galois extension of
$K$ if and only if $|G(L|K)|=(L\colon K)$.

###### Proof.

[7], Corollary 1, p. 126. ∎

######  Lemma 4.4.

If there exists a nonzero ideal $I$ of $L[X]$, where $L$ is a field, that is
finitely generated as an $K+XL[X]$-module, then $K$ is a field and $[L\colon
K]<\infty$.

###### Proof.

Clearly, $I$ is finitely generated over $L[X]$, and hence $XL[X]I\neq I$. For
otherwise, $XL[X]L[X]_{XL[X]}\cdot IL[X]_{XL[X]}=IL[X]_{XL[X]}$ and therefore
$IL[X]_{XL[X]}=0$, by Nakayama’s lemma. This is impossible, since $0\neq
I\subseteq IL[X]_{XL[X]}$. It follows that $I/XL[X]I$ is a nonzero
($L[X]/XL[X]=L$)-module that is finitely generated as an
($K+XL[X]/XL[X]=K$)-module. Since $L$ is a field, $I/XL[X]I$ can be written as
a direct sum of copies of $L$. Thus, $L$ is a finitely generated $K$-module.
But then $K$ is a field, since the field $L$ is integral over $K$ and
obviously $[L\colon K]<\infty$. ∎

######  Theorem 4.5.

Let $K\subset L$ be a field extension. Put $T=K+XL[X]$. Then the following
conditions are equivalent:

  * (1)

$T$ is Noetherian.

  * (2)

$[L\colon K]<\infty$.

###### Proof.

$(1)\Rightarrow(2)$ Since $XL[X]$ is a finitely generated ideal of $K+XL[X]$,
it follows from Lemma 4.4 that $[L\colon K]<\infty$. Thus, $L[X]$ is module-
finite over the Noetherian ring $K+XL[X]$.

$(2)\Rightarrow(1)$ $L[X]$ is Noetherian ring and module-finite over the
subring $K+XL[X]$. This is the situation covered by P.M. Eakin’s Theorem [12].
∎

All our considerations began with the Theorem 4.5. This Theorem motivated us
to further consider polynomial composites $K+XL[X]$ in a situation where the
extension of fields $K\subset L$ is algebraic, separable, normal and Galois,
respectively.

######  Proposition 4.6.

Let $K\subset L$ be a fields extension such that $L^{G(L\mid K)}=K$. Put
$T=K+XL[X]$. Then the following conditions are equivalent:

  * (1)

$T$ is Noetherian.

  * (2)

$K\subset L$ be an algebraic extension.

###### Proof.

$(1)\Rightarrow(2)$ Since $T=K+XL[X]$ is Noetherian, where $K\subset L$ be
fields extension, then by Theorem 4.5 we get that $K\subset L$ is a finite
extension. And every finite extension is algebraic.

$(2)\Rightarrow(1)$ Assume that $K\subset L$ be an algebraic extension.
Assuming $L^{G(L\mid K)}=K$ we get directly from the definition of the Galois
extension. Since $K\subset L$ be the Galois extension, then $K\subset L$ be a
normal extension. Every normal extension is finite, then by Theorem 4.5 we get
that $K+XL[X]$ be a Noetherian. ∎

######  Proposition 4.7.

Let $K\subset L$ be fields extension such that $K$ be a perfect field and
assume that any $K$-isomorphism $\varphi\colon M\to M$, where $\varphi(L)=L$
holds for every field $M$ such that $L\subset M$. Put $T=K+XL[X]$.

  * (1)

$T$ is Noetherian.

  * (2)

$K\subset L$ be a separable extension.

###### Proof.

$(1)\Rightarrow(2)$ By Theorem 4.5 $K\subset L$ be a finite extension. Every
finite extension be an algebraic extension. Since $K$ be the perfect field,
then $K\subset L$ be a separable extension.

$(2)\Rightarrow(1)$ First we show that if $L$ be a separable extension of the
field $K$, then the smallest normal extension $M$ of the field $K$ containing
$L$ be the Galois extension of the field $K$.

If $L$ be a separable extension of the field $K$, and $N$ be a normal
extension of the field $K$ containing $L$, then let $M$ be the largest
separable extension of $K$ contained in $N$. So we have $L\subset M$ and
therefore it suffices to prove that $M$ be the normal extension of $K$.

Let $g\in\operatorname{Irr}K[X]$ has a root $a$ in the field $M$. Because $N$
be the normal extension of $K$ and $a\in N$, so it follows that all roots of
polynomial $G$ belong to the field $N$. The element $a$ is separable relative
to $K$, and so belong to $M$. Hence polynomial $g$ is the product of linear
polynomials belonging to $M[X]$, which proves that $M$ be the normal extension
of the field $K$.

Since $M$ be the normal extension of $K$ and the Galois extension of $K$, then
$L$ be the normal extension of $K$ by the assumption ([7], Exercise 4, p.
119).

Because $L$ be the normal extension of $K$, then $L$ be the finite extension
of $K$. And by Theorem 4.5 we get that $K+XL[X]$ is Noetherian. ∎

######  Proposition 4.8.

Let $K\subset L$ be fields extension and let $T=K+XL[X]$. Assume that if a map
$\varphi\colon L\to a(K)$ is $K$-embedding, then $\varphi(L)=L$. Then the
following conditions are equivalent:

  * (1)

$T$ is Noetherian.

  * (2)

$K\subset L$ is a normal extension.

###### Proof.

$(1)\Rightarrow(2)$ By Proposition 4.6 $K\subset L$ be the algebraic
extension.

Let $c$ be a root of polynomial $g$ belonging to $L$, and $b$ be the arbitrary
root of $g$ belonging to $a(K)$. Because polynomial
$g\in\operatorname{Irr}K[X]$, so by Corollary from Lemma 4.1 there exists
$K$-isomorphism $\varphi^{\prime}\colon K(c)\to K(d)$. By Lemma 4.2 it can be
extended to embedding $\varphi\colon L\to a(K)$. Hence towards $\varphi(L)=L$
and $\varphi(K(c))=\varphi^{\prime}(K(c))=K(d)$ we get that $K(d)\subset L$,
so $b\in L$. Hence every root of polynomial $g$ belong to $L$, so polynomial
$g$ is the product of linear polynomials belonging to $L[X]$ ($\ast$).

For every $c\in L$ let $g_{c}\in\operatorname{Irr}K[X]$ satisfying
$g_{c}(c)=0$. By ($\ast$) every roots of $g_{c}$ belong to the field $L$.
Hence $L$ is a composition of splitting field of all polynomials $g_{c}$,
where $c\in L$. Hence $L$ be the normal extension of $K$.

$(2)\Rightarrow(1)$ If $L$ be a normal field extension of the field $K$, then
$K\subset L$ be the finite extension. Then by Theorem 4.5 we get that
$K+XL[X]$ be Noetherian. ∎

In the above Proposition we can replace the assumption that ”if a map
$\varphi\colon L\to a(K)$ is $K$-embedding, then $\varphi(L)=L$” to
”$L^{G(L\mid K)}=K$”. Then:

######  Proposition 4.9.

Let $K\subset L$ be fields extension such that $L^{G(L\mid K)}=K$. If
$T=K+XL[X]$ is a Noetherian, then $K\subset L$ be a normal extension.

###### Proof.

By Proposition 4.6 we get that $K\subset L$ is the algebraic field extension.
Assuming $L^{G(L\mid K)}=K$ we get directly from the definition of the Galois
extension, and so normal extension. ∎

######  Proposition 4.10.

Let $T=K+XL[X]$ be Noetherian, where $K\subset L$ be fields. Assume $|G(L\mid
K)|=[L\colon K]$ and any $K$-isomorphism $\varphi\colon M\to M$, where
$\varphi(L)=L$ holds for every field $M$ such that $L\subset M$. Then the
following conditions are equivalent:

  * (1)

$T$ is a Noetherian.

  * (2)

$K\subset L$ be a Galois extension.

###### Proof.

$(1)\Rightarrow(2)$ By Theorem 4.5 we get $K\subset L$ be the finite
extension. By the assumption we can use Lemma 4.3 and we get that $K\subset L$
be Galois extension.

$(2)\Rightarrow(1)$ If $K\subset L$ be Galois fields extension, then is
separable. By Proposition 4.7 we get that $K+XL[X]$ be Noetherian. ∎

In the above Proposition we can swap the assumptions that ”$|G(L\mid
K)|=[L\colon K]$ and any $K$-isomorphism $\varphi\colon M\to M$, where
$\varphi(L)=L$ holds for every field $M$ such that $L\subset M$” to
”$L^{G(L\mid K)}=K$”. Then:

######  Proposition 4.11.

Let $T=K+XL[X]$, where $K\subset L$ be fields such that $K=L^{G(L\mid K)}$.
The following conditions are equivalent:

  * (1)

$T$ is a Noetherian.

  * (2)

$K\subset L$ be a Galois extension.

###### Proof.

$(1)\Rightarrow(2)$ By Proposition 4.6 we get that $K\subset L$ be the
algebraic extension. Assuming $K=L^{G(L\mid K)}$ we get directly from the
definition of the Galois extension.

$(2)\Rightarrow(1)$ If $K\subset L$ be Galois fields extension. Then $K\subset
L$ be a normal extension. Hence by Proposition 4.9 we get that $K+XL[X]$ be
Noetherian. ∎

We present the full possible characterization of field extensions. Combining
the Magid results and from this paper, we get the following two theorems.

######  Theorem 4.12 ([18], Theorem 1.2.).

Let $M$ be an algebraically closed field algebraic over $K$, and let $L$ such
that $K\subseteq L\subseteq M$ be an intermediate field. Then the following
are equivalent:

  * (a)

$L$ is separable over $K$.

  * (b)

$M\otimes_{K}L$ has no nonzero nilpotent elements.

  * (c)

Every element of $M\otimes_{K}L$ is a unit times an idempotent.

  * (d)

As an $M$-algebra $M\otimes_{K}L$ is generated by idempotents.

######  Theorem 4.13 ([18], Theorem 1.3.).

Let $M$ be an algebraically closed field containing $K$, and let $L$ be a
field algebraic over $K$. Then the following are equivalent:

  * (a)

$L$ is separable over $K$.

  * (b)

$M\otimes_{K}L$ has no nonzero nilpotent elements.

  * (c)

Every element of $M\otimes_{K}L$ is a unit times an idempotent.

  * (d)

As an $M$-algebra $M\otimes_{K}L$ is generated by idempotents.

Below we have conclusions from the above results.

######  Theorem 4.14.

In Theorems 4.12 and 4.13 if assume $L^{G(L\mid K)}=K$, then the conditions
(a) – (d) are equivalent to

  * (e)

$K+XL[X]$ be a Noetherian.

  * (f)

$[L\colon K]<\infty$

  * (g)

$K\subset L$ be an algebraic extension.

  * (h)

$K\subset L$ be a Galois extension.

###### Proof.

(h)$\Rightarrow$(a) – Obvious.

(a)$\Rightarrow$(g)$\Rightarrow$(e)$\Rightarrow$(h) If $K\subset L$ be a
separable extension, then be an algebraic extension. By Proposition 4.6
$K+XL[X]$ be a Noetherian. By Proposition 4.11 $K\subset L$ be a Galois
extension.

(e)$\Rightarrow$(f) – Theorem 4.5. ∎

######  Theorem 4.15.

In Theorem 4.14 if assume that $K$ is a perfect field and $L^{G(L\mid K)}=K$,
then the conditions (a) – (h) are equivalent to

(i) $K\subset L$ be a normal extension.

###### Proof.

(i)$\Rightarrow$(a) If $K\subset L$ be a normal extension, then be an
algebraic extension. By definition perfect field $K\subset L$ be a separable
extension.

(h)$\Rightarrow$(i) Obvious. ∎

##  5 An SR property in polynomial composites

Recall from [24] a monoid $H$ is called $SR$-monoid, if
$\operatorname{Sqf}H=\operatorname{Gpr}H$. We present facts based on monoids.
They can be easily transferred to integral domains.

######  Proposition 5.1.

If $H$ is an $AP$-monoid, then $R$ is an $SR$-monoid.

###### Proof.

Let $s\in\operatorname{Sqf}R$. Then $s=q_{1}\dots q_{n}$, where $q_{1}$,
$\dots$, $q_{n}\in\operatorname{Irr}R$, $s_{i}\mbox{\small\rm rpr}s_{j}$.
Since $R$ is $AP$-monoid, then all $q_{i}\in\operatorname{Prime}R$ and since
$s=q_{1}\dots q_{n}$ and $q_{i}\mbox{\small\rm rpr}q_{j}$ then
$s\in\operatorname{Gpr}H$. ∎

The opposite fact generally does not hold.

######  Example 5.2.

Let $H$ be a non-group monoid such that every element of $H$ is a square. For
example $\mathbb{Q}_{\geqslant 0}$. Then
$\operatorname{Sqf}H=\operatorname{Gpr}H=H^{\ast}$, so $H$ is an SR-monoid.
But since $\operatorname{Irr}H\subset\operatorname{Sqf}H$, then
$\operatorname{Irr}H$ must be empty. Similarly $\operatorname{Prime}H$ is also
empty. There is not an $AP$-property. So $H$ is $SR$-monoid, non-$AP$-monoid.

######  Proposition 5.3.

Let $R$ be a factorial monoid. Then the following conditions are equivalent:

  * (a)

$H$ is an $AP$-monoid,

  * (b)

$H$ is an $SR$-monoid.

###### Proof.

From Proposition 5.1 we have (a) $\Rightarrow$ (b). So it sufficient to show
that every square-free is radical. Let $s\in\operatorname{Sqf}H$ and $s\mid
x^{m}$ for all $x\in H$ and $m\in\mathbb{N}_{>1}$. Since $H$ is factorial,
then $s=q_{1}\dots q_{n}$, with
$q_{i}\in\operatorname{Irr}H=\operatorname{Prime}H$ and $q_{i}\mbox{\small\rm
rpr}q_{j}$ and $x=p_{1}\dots p_{r}$, with
$p_{i}\in\operatorname{Irr}H=\operatorname{Prime}H$. Then $q_{1}\dots
q_{n}\mid(p_{1}\dots p_{r})^{m}$.

We see that for all $i\in\\{1,\dots,n\\}$ we have $q_{i}\mid(p_{1}\dots
p_{r})^{m}$. Since $q_{i}\in\operatorname{Prime}H\subset\operatorname{Gpr}H$,
then $q_{i}\mid p_{1}\dots p_{r}$, i.e. $q_{i}\mid x$. Thus
$\operatorname{lcm}(q_{1},\dots,q_{n})\mid x$ and $q_{i}\mbox{\small\rm
rpr}q_{j}$, then $s\mid x$. ∎

######  Corollary 5.4.

$H$ is factorial if and only if $H$ is atomic and $H$ is $SR$-monoid.

The $SR$-property holds in pre-Schreier monoids, $GCD$-monoids. So it is
enough to consider square-free dependencies.

######  Proposition 5.5.

Let $K$ be a field and $L$ its a field of fractions. Then $K+XL[X]$ is a pre-
Schreier domain.

###### Proof.

It follows from definition of pre-Schreier domain and it sufficient to check
free coefficients. ∎

######  Proposition 5.6.

Let $H$ be a monoid and $S\subset H$ be an multiplicative subset. Then, if $H$
is an $SR$-monoid, then $H_{S}$ is $SR$-monoid.

###### Proof.

Assume $a\in\operatorname{Sqf}H=\operatorname{Gpr}H$. Suppose
$a\in\operatorname{Sqf}H_{S}\setminus\operatorname{Gpr}H_{S}$. By [24],
Proposition 3.1 if $a\in\operatorname{Sqf}H$, then
$a\in\operatorname{Sqf}H_{S}$. By [27] Lemma 2.1, if
$a\notin\operatorname{Gpr}H_{S}$, then $a\notin\operatorname{Gpr}H$. A
contradiction. ∎

######  Corollary 5.7.

If $T=K+XL[X]$ is atomic and $\operatorname{Irr}T\subset\operatorname{Gpr}T$,
then $T$ is radical factorial.

## References

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# Reusable Self-Attention-based Recommender System for Fashion

Marjan Celikik<EMAIL_ADDRESS>,  Jacek Wasilewski
<EMAIL_ADDRESS>,  Sahar Mbarek<EMAIL_ADDRESS>,  Pablo
Celayes<EMAIL_ADDRESS>,  Pierre Gagliardi
<EMAIL_ADDRESS>,  Duy Pham<EMAIL_ADDRESS>,  Nour Karessli
<EMAIL_ADDRESS>and  Ana Peleteiro Ramallo
<EMAIL_ADDRESS>Zalando SEBerlinGermany

(2022)

###### Abstract.

A large number of empirical studies on applying self-attention models in the
domain of recommender systems are based on offline evaluation and metrics
computed on standardized datasets, without insights on how these models
perform in real life scenarios. Moreover, many of them do not consider
information such as item and customer metadata, although deep-learning
recommenders live up to their full potential only when numerous features of
heterogeneous type are included. Also, typically recommendation models are
designed to serve well only a single use case, which increases modeling
complexity and maintenance costs, and may lead to inconsistent customer
experience. In this work, we present a reusable Attention-based Fashion
Recommendation Algorithm (AFRA), that utilizes various interaction types with
different fashion entities such as items (e.g., shirt), outfits and
influencers, and their heterogeneous features. Moreover, we leverage temporal
and contextual information to address both short and long-term customer
preferences. We show its effectiveness on outfit recommendation use cases, in
particular: 1) personalized ranked feed; 2) outfit recommendations by style;
3) similar item recommendation and 4) in-session recommendations inspired by
most recent customer actions. We present both offline and online experimental
results demonstrating substantial improvements in retention and engagement.

Recommendation Systems, Transformers, Fashion Industry

††copyright: acmcopyright††journalyear: 2022††doi:
XXXXXXX.XXXXXXX††conference: Fourth Workshop on Recommender Systems in
Fashion; September 18–23, 2022; Seattle, WA, USA††booktitle:
FashionXRecSys’22: Workshop on Recommender Systems in Fashion, September 23,
2022, Seattle, WA††price: 15.00††isbn: 978-1-4503-XXXX-X/18/06††ccs: Computing
methodologies Neural networks††ccs: Information systems Recommender systems

##  1\. INTRODUCTION

(a) ”Get the Look” personalized feed (catalog) of ranked outfits.

(b) ”Style preview” carousel with personalized outfits selection in different
styles.

(c) ”Inspired by you” carousel with in-session outfit recommendations based on
customer’s recent interactions.

(d) ”You might also like” carousel with personalized outfit recommendations
(right) related to an anchor outfit (left).

Figure 1. Different outfit recommendation and ranking use cases.

Fashion contributes to our everyday life in reflecting personality, culture,
and style. With the vast choice of items available in e-commerce, it has
become increasingly difficult for customers to find relevant content, combine
it and match with a specific style. Finding inspiration, such as for example
influencers, has become crucial to support customers in discovering
inspirational and fresh outfits tuned to their taste.

¡Org¿ is a leading European online platform for fashion and lifestyle, and one
of its main goals is to provide fashion inspiration and support customers in
their fashion discovery journey. We do this through different experiences,
such as for example creator outfit inspirational content, in touch points such
as: a) ”Get the Look” (GTL), which personalized feed of all available
influencer outfits (1(a); b) ”Style” carousel, providing personalized outfit
recommendations in different styles (e.g. classic, casual, streetwear; 1(b));
c) ”Inspired by you” carousel, providing real-time in-session recommendation
based on the customer’s recent interactions (1(c)); and d) ”You might also
like” carousel, providing personalized recommendations related to the
currently viewed outfit (1(d)).

A complexity that arises when building recommendation models for inspirational
content in the e-commerce domain is the availability of input signals.
Interaction data with non-product fashion entities, such as influencer
outfits, is usually low, which makes the data extremely sparse when compared
to interactions with individual items (e.g., a shirt). In this work, we
investigate how interactions with other ”high-traffic” fashion entities could
be used and improve ”low-traffic” entities such as outfit recommendations.

In e-commerce, often different but related customer experiences are served by
completely different recommender systems, for example either to steer more to
in-session recommendations or to capture longer-term customer preferences.
This can significantly increase not only modeling complexity and maintenance
costs of those different systems, but also lead to inconsistent customer
experience. In contrast to the common belief that different recommenders are
needed for different use cases (netflixdl), we show that a single Transformer-
based recommender system can be trained on diverse types of interactions
coming from various sources and successfully re-used across many related use
cases, such as session-based recommendation for short-term interests,
personalized ranking based on long-term user preferences and even provide
item-related or similar recommendations.

Moreover, in e-commerce, many users are either new or do not login when
navigating, thus it is important to use a model that can provide meaningful
recommendations to new, cold-start users. The model we propose can serve
”partial” cold-start users (users that have not interacted with outfits but
have interacted with the platform) by utilizing other types of interactions in
the session, as well as ”full” cold-start users (without any platform
activity) by utilizing contextual information about the user and the use case,
such as premise (site or product), device, country, language, etc.

In recent years sequential recommendation approaches have gained significant
traction (huifang). These model customers as sequences of interactions to
better infer context and customer preferences that change over time. In
addition, representing customers as sequences of actions at least partially
simplifies the feature engineering and modeling efforts. Sequential
recommendation approaches often employ NLP techniques such as Word2Vec
(mikolov), RNNs (hidasi2015sessionbased; rnns) and attention mechanisms
(sasrec; bert4rec; alibabapog). Transformer-based models are the most
promising sequential and session-based models to date due to their ability to
model long-range sequences and their scalability thanks to efficient parallel
training (trm4rec; sasrec; bert4rec). However, the majority of previous
studies are based on offline evaluation and metrics computed on open datasets,
which may not be directly comparable with industrial datasets and online
results. Moreover, many of them do not consider side information such as
categorical inputs to represent items or customers, although it is well known
that deep learning recommender systems live up to their full potential only
when numerous features of heterogeneous types are used (netflixdl). Due to
these shortcomings, even if relevant, most of the research work on self-
attention models for recommendations does not explore the potential
effectiveness in real-world industry applications. In this work, we contribute
to bridge this gap by presenting online experimental results of using a self-
attention-based recommender system that can utilize contextual information
about the customer and the use case. We show case that our personalized outfit
recommendation model can improve engagement and customer retention in
different use cases throughout the customer journey.

In summary, the main contributions of this work are as follows: (1) a reusable
transformer-based recommender system for fashion recommendations that utilizes
diverse types of interactions on various fashion entities. It is able to
provide session-based recommendations as well as take into consideration long-
term user preferences and contextual information, to both recurring and cold
start users; (2) extensive A/B testing that shows that this approach can be
successfully applied to different recommendation use cases including
personalized ranked feed; outfit recommendations by style; similar item
recommendation and in-session recommendations inspired by most recent customer
activities; (3) extensive offline experiments and evaluation considering
different ranking losses and metrics where we show how our approach
substantially increases customer engagement and retention. The rest of the
paper is structured as follows: in section 2 we discuss related work; section
3 describes the proposed approach; we present offline evaluation in section 4
and online experiment results in section 5; finally, we provide discussion of
conclusions and future work in section 6.

##  2\. RELATED WORK

The main objective of recommender systems is to recommend relevant content
tailored to the customer’s preferences and intent. Fashion e-commerce has
heavily invested in developing recommender systems for different use cases to
aid online shopping experience (Chakraborty2021; Deldjoo2022) including
recommending relevant items (Deng2018; Cardoso2018; Zhou2018), complete
outfits (Lin2019; Celikik2020; Denk2020), and size recommendation
(sembium2018; Lasserre2020).

Transformer-based recommendation systems started with the SASRec algorithm
presented in (sasrec), where, similarly to GPT (gpt), a Transformer encoder
with masked attention for causal language modeling (CLM) was used to predict
the next item in the sequence of user interaction. A similar work in (ssept)
builds on this idea by concatenating user embeddings to each learnt item
embedding from the same sequence to add contextual user information. BERT4Rec
(bert4rec) is another work that employs the Transformer encoder trained with
the masked language modeling (MLM) approach, while masking only the last item
of the sequence during inference to avoid leaks of future information.

There are a few works in the literature that use complex hierarchical
architectures or mix together different network architectures. The SDM
algorithm introduced in (sdm) is fusing LSTMs and self-attention networks to
obtain a final user behavior vector. (seqhierarchicalattention) proposes a
similar approach by using a two-layer hierarchical attention-based network
while (graphselfattention) employs a graph and a self-attention network. In
all of these approaches no evidence is provided as to why the complex
architecture is needed. Moreover, these methods are only compared to baselines
that are either not based on self-attention or not based on deep learning
methods. Hence, it is unclear whether the improvement comes from the choice of
the architecture. In our case, we present a effective and simple approach that
is able to capture both short-term and long-term user interests.

Most of the published studies do not consider side information for
representing items or users and instead work with user-item interaction data
only. The few works that include side information usually consider only a few
item features such as category and brand (sdm). The BST algorithm from Alibaba
(alibabapog) considers richer item features and user profiles as well as
contextual features. The differences to our approach are twofold: first, the
contextual features are not part of the self-attention mechanism but rather
concatenated with the output of the Transformer, and second, the use of a
positional encoding for the entire sequence instead of representing different
sessions separately.

There are only a few lines of work that we are aware of that consider
(re)using models for multiple recommendation use cases. Parallel to our work,
the work in (rnnlong) shows that RNNs can perform well in both, short and
long-term recommendation tasks when certain improvements are applied and even
over-perform more complex hierarchical models. The work (multigraph) presents
a multi-graph structured multi-use case recommendation solution which
encapsulates interaction data across various use cases and demonstrates
increase in CTR and video views per user. Another work in
(knowledgetransferrecsys) surveys the benefits of pre-training and knowledge
transfers in recommender systems in order to alleviate the data sparsity
problem.

##  3\. ALGORITHM

Figure 2. Overview of the model trained on interactions coming from different
fashion entities. Interactions are converted into embeddings of the same
structure in terms of feature representation. Contextual features
corresponding to the use case such as premise (site) and device are added to
input sequences, where temporal session data and interaction metadata is
concatenated directly to the learned embeddings. Masking is used in order to
consider target outputs only of certain entity types (e.g. outfits).

###  3.1. Problem Formulation

Let $\mathcal{U}=\\{u_{1},u_{2},...,u_{|\mathcal{U}|}\\}$ denote the set of
users and $\mathcal{V}=\\{v_{1},v_{2},...,v_{|\mathcal{V}|}\\}$ a set of
items, we represent the interaction sequence as
$\mathcal{S}_{u}=\\{v_{1}^{(u)},...,v_{t}^{(u)},...,v_{n_{u}}^{(u)}\\}$ with
interactions ordered chronologically for the user $u\in\mathcal{U}$, where
$v_{t}^{(u)}\in\mathcal{V}$ is the item that the user $u$ has interacted with
at time $t$, and $n_{u}$ is the length of interaction sequence for user $u$.
Let $\mathcal{C}_{u}$ be the contextual information about user $u$, such as
country, device, premise and date. Given the interaction history
$\mathcal{S}_{u}$ and context $\mathcal{C}_{u}$, sequential recommender system
aims to predict the item that user $u$ is most likely to interact with at time
$n_{u}+1$, i.e. we would like to find an item $i$ so that the probability
$p\left(v_{n_{u}+1}^{(u)}=i|\mathcal{S}_{u},\mathcal{C}_{u}\right)$ is
maximized.

###  3.2. Model Architecture

We use a standard Transformer encoder (sasrec; attention; trm4rec) trained
with the causal language model (CLM) approach as illustrated in Figure 3. A
causal attention mask is provided to the self-attention mechanism so that each
position can attend only to earlier positions in the sequence. Note that our
approach is oblivious of the training logic and it can be trained with the
masked language model (MLM) approach as well (bert4rec; trm4rec). Given an
input sequence of length $t$ and a matrix of learnable input embeddings
$V\in\mathbb{R}^{t\times d}$, a single layer of the Transformer model computes
hidden representations $H^{1}=\text{Trm}(V)\in\mathbb{R}^{t\times d}$. By
stacking multiple layers, we have $H^{l}=\text{Trm}(H^{l-1})$. For details we
refer the reader to (attention) and (sasrec). The final output of the last
position is fed into a softmax layer over the item set. We employ categorical
cross-entropy as a loss function, however, experiment with other losses as
well.

Figure 3. Illustration of a single Transformer encoder (left) and overview of
our model architecture (right). Model inputs consist of contextual information
and a sequence of: item embedding $E_{v_{t}}$, action embedding $A_{v_{t}}$
and session embedding $T_{v_{t}}$. Target mask filters predictions of entities
other than the target ones – in this example target mask is set for outfit
entities.

Given an input sequence, the expected target output is a shifted version of
the same sequence. In our setting, an item can refer to a fashion article, an
outfit or an influencer. We train on every item in a sequence, but predict
only those items that are relevant for the use case. To that end, we assign a
boolean mask to each relevant item. The mask is set to 1 only if the next
position in the sequence is associated with an item that should be predicted
by the model as a valid recommendation. For example, if our model recommends
outfits, then all inputs corresponding to valid outfits will have a mask of 1,
for other entities the mask will be set to 0 (all use cases presented in
Figure 1 consider outfits, so in our experiments we focus on outfit
predictions). The mask is passed to the loss function and the positions
corresponding to zeroes do not contribute to the loss. In addition, items that
are valid but not available (e.g. out of stock), will have an output mask set
to 0 as well.

Contextual information about the use case and the customer are encoded as
embeddings with the same dimensionality as the input item embeddings and are
set as the first positions of the sequence so that every other position can
attend and utilize this information when making predictions. Inputs such as
location, market and language play an important role for cold-start customers
that are new to the platform and do not have any interactions yet. For cold-
start customers that are new to the outfit use cases but not new to the
platform itself, the model can make recommendations based on interactions with
other fashion entities, either historical or from the current session. Inputs
such such as premise and device help the model to hone in on the particular
use case. Figure 2 provides a summary of the modeling choices and the
different sources of data used for training.

###  3.3. Input Embeddings

We represent each item $v_{t}$ (for brevity we omit the superscript) in a user
sequence $\mathcal{S}_{u}$ as a concatenation of learned embeddings that
correspond to its different features. The features are always encoded in the
same order. Since each input position in the sequence must consist of the same
embeddings, we pad the corresponding part of the input tensor with zeroes if
the item does not have a certain feature. For example, an outfit might have an
influencer (creator), however, a single article does not have one. An item
$v_{t}$ will have the following representation:

$E_{v_{t}}=E_{f_{1}}(v_{t})\oplus E_{f_{2}}(v_{t})\oplus...\oplus
E_{f_{n}}(v_{t})$

where $E_{f_{k}}$ is the embedding matrix of a feature $f_{k}$ and $\oplus$ is
the concatenation operator. Depending on the feature type and its cardinality,
we represent features through learned embedding matrices or by using 1-hot
encoded vectors. Note that if $v_{t}$ does not have the feature $f_{k}$,
$E_{f_{k}}(v_{t})$ is the 0-vector. Examples of categorical features that we
represent as embedding matrices are brand, category, color and influencer,
while features such as average price bucket and outfit style are represented
by using 1-hot encoding.

As there are clear hierarchical relationships between fashion entities, e.g.
an outfit consists of articles and influencers can be represented by the set
of outfits they have created, we employ a simple representation of more
complex items by averaging the embeddings of the individual items. For
example, given an outfit $o$, let $E_{B}$ be an embedding matrix for the brand
feature. We represent the brand embedding of outfit $o$ as

$E_{B}(o)=\sum_{a\in o}E_{B}(a)$

where $a$ is an article in $o$. We represent individual articles in a sequence
as outfits with a single article.

For the model to be able to utilize the interaction type of an item and learn
to treat different interactions differently, we concatenate a one-hot encoding
of the interaction type of the item with the item representation, i.e.
$E_{v_{t}}\oplus A_{v_{t}}$, where $A_{v_{t}}$ is one-hot encoding of the
$v_{t}$’s interaction type.

###  3.4. Modeling Sessions for Long and Short-Term Interests

In this section we present a simple but yet effective mechanism that allows
our algorithm to utilize information from the current and past browsing
sessions in order to model short in-session and long-term customer
preferences. A user sequence $\mathcal{S}_{u}$ consists of different browsing
sessions. A session is a list of consecutive interactions within a given time
frame, for example a day, when the user has a clear shopping intent while
their interests can change sharply when they start a new session, say a week
later. Hence, modeling user sequences directly while ignoring this structure
will affect performance negatively (as observed in our experiments).

We model sessions via introducing temporal inputs in the form of interaction
recency defined as number of days passed since the action has taken place
relative to the model training timestamp during training and the serving
timestamp during serving. We discretize recency and consider only the integer
part of the timestamp. Each temporal input is assigned its own interaction
recency that is concatenated with the rest of the item embeddings. For
simplicity, we consider user activity during a single day as a single session.
Hence, interactions within the same session will have the same recency. For
example, if the next prediction is in-session then the recency of the last
action will be 0. However, the model is able to attend to previous actions as
well and use the attention mechanism to select actions that are relevant for
the prediction.

Note that with CLM training, the model does not have bi-directional access to
positions in the sequence, hence it has no information about when the next
action has taken place. However, this information is needed for a correct
prediction. Namely, the next action may happen on the same day or a week after
the previous action. To account for this, in addition to recency, we include
the time gap between two actions and make this information accessible to the
previous action as one of its inputs. At inference time, the time gap is
always 0. Hence, the final input representation of a single interaction to the
model is

$E_{v_{t}}\oplus T_{v_{t}}\oplus A_{v_{t}}$

where $E_{v_{t}}$ is the item embedding, $A_{v_{t}}$ is the action embedding
for item $v_{t}$, and $T_{v_{t}}=r_{v_{t}}\oplus t_{v_{t}}$ is the session
embedding of item $v_{t}$, $r_{v_{t}}$ is the learnt embedding of the
discretized recency and $t_{v_{t}}$ is the time gap between item $v_{t}$ and
item $v_{t+1}$.

##  4\. OFFLINE EVALUATION

###  4.1. Dataset

Our data consists of a sample of 60 days of user interactions on the platform.
We perform a time-based split, where we train on the first 59 days of data and
evaluate on the last day. We aggregate the interaction per user into
sequences. During the evaluation, we feed only those interactions as inputs
that took place before the split timestamp (if any). We filter out sequences
without outfit interactions. Our training data contains roughly 6.6m sequences
on 9.9k distinct outfits created by 677 unique creators. The test data
contains roughly 172k interactions on 9.4k distinct outfits. The average
outfit length was 4.6 articles. The total number of distinct articles in the
dataset was 1.6m. Roughly 23% of the customers were completely new to the
platform, 31% new to outfits but had already interacted with articles on the
platform and 46% of the customers had already interacted with outfits. To
confirm the consistency of our results, we set fixed random seeds, trained and
evaluated on 3 different datasets (collected with respect to 3 different last
days) and reported averages.

###  4.2. Experimental Setting

We evaluate our new Attention-based Fashion Recommendation Algorithm (AFRA)
against a set of existing recommendation algorithms some of which have been
powering outfit recommendation use cases presented in Figure 1. In the
following we present each of the compared algorithms.

  * •

Neural LTR: powers the ”Get the Look” personalized feed and the ”Style”
carousel use cases. It is a neural Learning-To-Rank approach implemented by
using the TensorFlow-Ranking (TensorflowRankingKDD2019) library to rank
outfits based on outfit and user inputs. The outfit inputs consist of learnt
embeddings of categorical features such as brand, influencer, color and the
market of the outfit. The user features are represented by the top-3 favorites
brand, colors, influencers, market as well as historical normalized
frequencies of how often the customer interacted with certain content types. A
feed-forward network is used as the model architecture, with 2 dense layers
with 128 and 64 layers units each. The output is passed through a $tanh$
activation function. A pairwise logistic loss is used for training. For the
”Style” carousel use case, the output is further filtered by style.

  * •

IB-CNN-kNN: is a kNN-based recommendation algorithm powering the ”Inspired by
you” carousel. Outfit recommendations are retrieved based on cosine similarity
between item embeddings of previously interacted fashion articles and item
embeddings of articles in the outfit. The similarity is defined as the average
cosine similarity between the embeddings of the best matching article from the
user’s history, considering the last 10 article interactions in near real-time
and the 200 most recent outfits on the platform. Each item is represented by
latent embeddings computed by using a fine-tuned CNN as in (fdna).

  * •

IB-CF-kNN: is a kNN-based Collaborative-Filtering algorithm powering the ”You
might also like” carousel. We retrieve outfit recommendations based on cosine
similarity between user-outfit vectors.

  * •

AFRA: AFRA has been implemented as described in section 3 by using the CLM
training approach, 2 Transformer layers each with 8 attention heads, dff layer
with 1024 units, dropout of 0.1, $d_{model}$ set to 128, batch size of 64 and
learning rate set to 0.01. The algorithm was trained with 10 epochs. Each
interaction sequence passed to the model has been truncated to the 100 most
recent interactions. The model uses a set of article-based categorical
features such as brand, color, material, fit, pattern; outfit based-features
such as influencer and style, general interaction features such interaction
type and interaction recency; and contextual features such as premise, market
and device type. We employ two strategies to evaluate AFRA: AFRA-RT in which
the customer actions are available to the algorithm in near real-time and
AFRA-Batch in which only customer actions performed up to the previous day are
available to the algorithm.

  * •

SASRec : SASRec (sasrec) is a Transformer-based recommendation algorithm
trained with the CLM approach. We apply the same hyper-parameters that we use
in AFRA. We experiment with the authors’ implementation of the algorithm. The
model uses only embedding representation of user and item IDs as inputs. The
model is trained on both outfit and article interactions, while predictions
are made on outfits only.

It is worth noting that AFRA is able to make use of additional content-based,
contextual or temporal features that other approaches reasonably cannot (CF,
kNN, LTR) or do not (SASRec) without introducing substantial changes in the
algorithms.

###  4.3. Offline Experiments

In this section we evaluate different aspects of our proposed algorithm such
as _relevance_ , _diversity_ and _freshness_. We experiment by using different
variants of AFRA by training on different types of fashion items. In addition,
we evaluate the performance of AFRA by using other ranking loss functions that
are commonly considered in the literature.

####  4.3.1. Relevance

We run offline evaluation using a number of relevance metrics, including
recall@k, precision@k, hitrate@k, nDCG@k and mAP@k according to their standard
definition provided in (offlineeval). For each algorithm we calculated the
metrics for the top $k$ results, where $k\in\\{5,15,30\\}$ which corresponds
to the use cases that we consider, e.g. $k=15$ and $k=30$ corresponds to the
first and the second page of results in a feed, while $k=5$ corresponds to a
single carousel of recommendations. The recommendation task consists of
predicting outfit clicks, regardless whether they are sequential (i.e. in-
session), historical (i.e. when customers return from another day) or cold-
start (when customers perform an action for the first time). Since all of the
relevance metrics were strongly correlated with each other, for brevity we
report only recall@k, which in turn correlates well with click-through-rate
(CTR) (hidasi2015sessionbased).

Customer segment | Algorithm | Recall@5 | Recall@15 | Recall@30
---|---|---|---|---
All customers | AFRA-RT | 0.135 | 0.233 | 0.301
AFRA-Batch | 0.081 | 0.156 | 0.224
AFRA-RT (outfits only) | 0.093 | 0.158 | 0.214
| AFRA-Batch (outfits only) | 0.056 | 0.108 | 0.156
| LTR | 0.025 | 0.104 | 0.153
| IB-CNN-kNN | 0.056 | 0.078 | 0.110
| IB-CF-kNN | 0.052 | 0.073 | 0.102
| SASRec | 0.061 | 0.102 | 0.144
| SASRec (outfits only) | 0.051 | 0.075 | 0.099
New customers (cold-start) | AFRA-RT | 0.082 | 0.164 | 0.232
AFRA-Batch | 0.045 | 0.102 | 0.163
AFRA-RT (outfits only) | 0.069 | 0.131 | 0.183
| AFRA-Batch (outfits only) | 0.045 | 0.093 | 0.134
| LTR (w/ recency) | 0.033 | 0.085 | 0.122
| Popularity | 0.030 | 0.052 | 0.091
| IB-CNN-kNN | 0.034 | 0.042 | 0.055
| IB-CF-kNN | 0.047 | 0.057 | 0.074
| SASRec | 0.041 | 0.074 | 0.101
| SASRec (outfits only) | 0.005 | 0.011 | 0.032
Table 1. Relevance in terms of recall@k on all compared algorithms computed on
new and on all customers. It can be observed that all AFRA variants perform
the best in general for the given use cases. The label ”outfits only” means
that the model has been trained by using outfit interactions only.

Table 1 shows the recall@k for all the compared algorithms. The following main
observations can be made. First, the two versions of AFRA that simulate real-
time and batch scenarios perform better than all other algorithms, including
SASRec, on all and on cold-start customers specifically. Second, training by
using diverse item interaction data substantially improves the relevance
compared to training on outfits only. This can be observed on both algorithms,
AFRA and SASRec. Third, AFRA performs substantially better than SASRec even
when trained on the much sparser outfit interactions only, thanks to using
rich heterogeneous data and session encoding. Fourth, AFRA performs
substantially better on cold-start customers compared to all other algorithms
thanks to its ability to utilize contextual data. This is especially
pronounced in the real-time use case where session data is available to the
algorithm. Fifth, using real-time session data has a large impact on
performance: in certain scenarios AFRA-RT achieves twice as high a recall
compared to AFRA-Batch.

Loss | Recall@5 $\Delta$ % | Recall@30 $\Delta$ %
---|---|---
Sampled Cross Entropy | -19% | -13%
Binary Cross Entropy (sasrec) | -7% | 0%
BPR | -32% | -11%
TOP1 | -73% | -20%
Table 2. Change in AFRA’s recall@k when various loss functions are used
compared to full cross entropy loss. In summary, the gain in training speed
did not justify the loss in the offline metrics.

Algorithm | Freshness@30
---|---
LTR | 53
AFRA | 51
\+ age feature | 40
\+ age decay | 20
Table 3. Average outfit age in the top-30 recommendations when different
freshness strategies are used.

####  4.3.2. Freshness

Recommender systems have bias towards older items since those usually have
more interactions. Users, however, prefer fresh content, although not at the
expense of relevance (youtube). We focus on freshness in the ”Get the Look”
feed use case, which is our main entry point for fashion inspirations. We
measure freshness of recommendations as the average age of the top-30
recommended outfits, given in days. Table 3 shows this metric for the top-30
recommendations. We can observe that the previous approach, LTR, and AFRA
provide similar freshness. Moreover, we have conducted experiments on how this
metric could be improved, and considered two strategies for adjusting for
freshness. The first one is inspired by (youtube), where an item age feature
is added during training which is set to 0 during inference to ”de-bias” old
outfits that have higher chance of being interacted with. The second one
introduces a tuned age exponential decay, simulating ”content aging”, with a
half-life of 3 weeks obtained by parameter tuning. This value is used to
weight the ranking score produced by the model during inference, which can be
seen as a re-ranking of the results. Exponential decay proved to be
particularly effective as it did not harm relevance and substantially
increased fresh content among the top-$k$ recommendations, decreasing the
average age from 51 to 20 days. On the other hand, the age feature decreased
the average age to 40 days.

####  4.3.3. Diversity

Diversity is another aspect of recommender systems important to prevent filter
bubbles that cause the customers to lose interest over time due to
recommendations that are too similar (relevancediversity). We measure two
types of diversity: _inter-list diversity_ (measuring the content diversity
within a list of recommendations) and _temporal-diversity_ (measuring the
difference in recommendations from one to the next visit). As a proxy for
inter-list diversity we use the maximum consecutive sublist in the top-k
created by the same creator (consecutive recommendations from the same creator
are undesirable for our product). AFRA and LTR had the highest diversity among
all algorithms. For both algorithms this metric on average was less than 2.0,
with AFRA outperforming LTR by up to 20%. We define temporal diversity as a
normalized set difference between the recommendation lists from two
consecutive visits. Both algorithms perform similarly with temporal diversity
roughly around 70%. As a future work, we would like to introduce impression
data to AFRA to improve temporal diversity by down-ranking items the user has
already seen but not interacted with.

####  4.3.4. Loss functions

We have employed other standard ranking functions based on negative sampling
such as BPR (bpr), TOP1 (rnns) and binary-cross entropy (sasrec) to improve
relevance and/or training speed. Table 3 shows the relative change in recall
compared to standard categorical cross-entropy (softmax loss). We experimented
by using 30 and 100 negative samples (without replacement). The main
observation is that loss functions based on negative sampling were not very
effective in our setting. The training speed improvements obtained were modest
and always less than 2x. Hence, the decrease in relevance did not justify the
improvements in training speed. One of the possible reasons for decreased
relevance metrics could be the choice of negative samples not matching the
background distribution well. Improving the negative sampling distribution as
well as using hard negatives is one of our future work directions.

##  5\. ONLINE RESULTS

In order to confirm the efficacy of our algorithm in real scenarios, we have
performed A/B tests on three of our use cases: ”Get the Look”, ”Style preview”
and ”Inspired by you” (1(a); 1(b), 1(c) respectively). Each A/B test has been
run for 3 to 4 weeks (until convergence). For both ”Get the Look” and ”Style
preview”, we compare AFRA-Batch against LTR in order to be consistent with the
previous approaches that used daily updates of customer interaction data.
Real-time session data was used only for new customers to address the cold-
start problem. For the ”Inspired by you” use case we use near real-time data
and therefore compare AFRA-RT against the IB-CNN-kNN algorithm that uses near
real-time data as well. Table 4 summarizes the results from the A/B tests on
the _retention_ and _engagement_ KPIs. We define retention as the share of
users with multiple interactions within 7 days, and engagement as outfit
interaction rate per user.

In summary, AFRA performs substantially better on all KPIs on all tested use
cases. We can observe that on the ”Get the Look” and ”Style preview” use cases
the improvements in retention range from 23% up to 42%. The improvement was
even higher on the ”Inspired by you” use case where it ranged from 109% to
130%. We observe a strong improvement on the engagement KPI as well that
ranges from 30% to 47% for the first two use cases, reaching 201% on ”Inspired
by you”. On both KPIs, the improvement was stronger on the cold-start
customers on the first two use cases, thanks to introducing in-session
recommendations and contextual inputs in AFRA.

These strong results are consistent with our offline experiments. We believe
this is because we train on diverse sources of interaction data from all
products and premises which in turn helps to provide more personalized and
relevant content and dampens feedback loops and selection biases (netflixdl;
criticialfashionxrecsys).

Customer segment | Get the Look | Style preview | Inspired by you
---|---|---|---
Retention | Engagement | Retention | Engagement | Retention | Engagement
All customers | +28.5% | +33.1% | +23.9% | +30.2% | +130.5% | +201.3%
New customers
(cold-start) | +42.0% | +47.1% | +39.7% | +39.7% | +109.2% | +137.5%
Existing customers | +27.0% | +32.0% | +23.1% | +29.6% | +130.5% | +201.9%
Table 4. A/B test results showing increase in user retention and engagement
metrics of our transformer-based recommender system, compared against the
existing algorithms for three use cases illustrated in Figure 1.

##  6\. CONCLUSIONS

In this paper we have presented a reusable Transformer-based recommender
system that is able to utilize different types of interactions with various
fashion entities. We have shown our approach is able to model short term
customer interests by providing session-based recommendations as well as take
into consideration long-term user preferences and contextual information about
the customer and the use case. We have demonstrated its effectiveness on
different use cases, with extensive offline and online experiments that show
that our approach substantially improves both customer retention and
engagement.

Future work that we consider worth exploring is introducing impressions to
AFRA to improve temporal diversity by organically down-ranking items the user
has already seen but not interacted with. Furthermore, we would like to
abstract the prediction head of our recommender system to allow the
flexibility of re-ranking and hence the ability to apply AFRA on use cases
with many millions of items.