# Some algebraic results concerning linear recurrence sequences

M. MOUÇOUF

###### Abstract.

We study the set $\mathcal{L}_{F}$ of all $F$-vector spaces $L(P)$, the set of
linear recurrence sequences over $F$ with characteristic polynomial $P$ monic
and splits over $F$. We show that $\mathcal{L}_{F}$ is a commutative semiring
graded by the multiplicative monoid $F$. This study allows us to find
explicitly the polynomial $P\in F[X]$ such that
$L(P)=\prod_{i=1}^{m}L(P_{i})$, where $P_{1},\ldots,P_{m}$ are any monic
polynomials over $F$.

###### Key words and phrases:

Linear recurrence sequences, fields, semirings, vector spaces, polynomials

Department of Mathematics, Faculty of Science,
Chouaib Doukkali University, Morocco
Email<EMAIL_ADDRESS>
##  1\. Introduction

Let $F$ be a field and let $P$ be a polynomial over $F$. We know that the set
$L(P)$ of all linear recurrence sequences over $F$ with characteristic
polynomial $P$ is a $F$-vector space. In this paper, we define on $\mathbb{N}$
a binary operation $\wedge$, which depends on the field $F$, using the
binomial coefficients. This allows us to prove that the set $\mathcal{L}_{F}$
of all $L(P)$, $P\in F[X]$ monic and splits over $F$, is a commutative
semiring graded by the multiplicative monoid $F$. In addition, we show, with
the aid of this result, that if

$P_{1}=X^{s_{1}}Q_{1},\ldots,P_{m}=X^{s_{m}}Q_{m}$

are monic polynomials over $F$, where $Q_{i}(0)\neq 0,\,1\leq i\leq m$, then

$\prod_{i=1}^{m}L(P_{i})=L(X^{\rho}\Upsilon(Q_{1},\ldots,Q_{m})),$

where

$\displaystyle\rho=\begin{cases}\min\\{s_{i}/i\in\Theta\\}&\quad\text{if}\quad\Theta\neq\emptyset\\\
\max\\{s_{i}/1\leq i\leq m\\}&\quad\text{otherwise}\end{cases}$

and

$\Theta=\\{i/Q_{i}=1\\}.$

$\Upsilon(Q_{1},\ldots,Q_{m})$ is a polynomial over $F$ that we determine in
terms of the roots of the polynomials $Q_{1},\ldots,Q_{m}$ in the splitting
field over $F$ of the polynomial $\prod_{i=1}^{m}Q_{i}$. We note that this
problem has been studied in the case where none of the polynomials $P_{i}$
have multiple roots (see, e.g, [3]) and later on by N. Zierler and W. H. Mills
[4], in the case of any polynomials with constants terms, by constructing
special sequences using an arithmetic method. However in this paper, we
present an algebraic method based on a special family of vector spaces, and
especially on their dimensions.

##  2\. Products of linear recurrence sequences

Recall the following definition

######  Definition 2.1.

let $(G,\star)$ be a monoid. A semiring $R$ is said to be $G$-graded (or
graded by $G$) if there exists a family $\\{R_{\alpha}\\}_{\alpha\in G}$ of
additive subsemigroups of $R$ satisfying the following conditions:
1\. $R=\sum_{\alpha\in G}R_{\alpha}$;
2\. $R_{\alpha}R_{\beta}\subseteq R_{\alpha\star\beta}$ for each
$\alpha,\beta\in G$;
3\. If $a_{\alpha_{1}},\ldots,a_{\alpha_{m}}$ are non-zero elements in $R$
where $\alpha_{i}\neq\alpha_{j}$ if $i\neq j$, and each $a_{\alpha_{i}}\in
R_{\alpha_{i}}$, then $\sum_{i=1}^{m}a_{\alpha_{i}}\neq 0$.

Let $\mathcal{C}_{F}$ be the set of all sequences over $F$ and let
$\widetilde{\mathcal{C}_{F}}$ be the set of all subspaces of
$\mathcal{C}_{F}$. It is easily seen that $(\widetilde{\mathcal{C}_{F}},+,.)$
is a commutative semiring, in which
$\operatorname{span}\\{\boldsymbol{0_{F}}\\}$ and
$\operatorname{span}\\{\boldsymbol{1_{F}}\\}$ are the additive identity and
the multiplicative identity respectively, where $H+H^{\prime}$ is the sum of
the subspaces $H$ and $H^{\prime}$, and $HH^{\prime}$ is the subspace of
$\mathcal{C}_{F}$ spanned by all products $CD$ with $C\in H$ and $D\in
H^{\prime}$.
Let $\lambda\in F^{\ast}=F\backslash\\{0\\}$ and $s\in\mathbb{N}$. Then we
note $\langle\lambda\rangle_{s},\langle 0\rangle_{s}$ the subspaces of
$\mathcal{C}_{F}$ spanned by
$\\{\boldsymbol{\lambda}\Lambda_{0},\ldots,\boldsymbol{\lambda}\Lambda_{s-1}\\}$
and $\\{\boldsymbol{0}_{0},\ldots,\boldsymbol{0}_{s-1}\\}$ if $s\neq 0$, and
$\langle\lambda\rangle_{0}=\langle 0\rangle_{0}=0$. It is clair that
$\langle\lambda\rangle_{s}+\langle\lambda\rangle_{t}=\langle\lambda\rangle_{s\vee
t}$, $\vee$ is the the maximum of two integers. Hence
$I_{\lambda}=\\{\langle\lambda\rangle_{s}/s\in\mathbb{N}\\}$ is a submonoid of
$(\widetilde{\mathcal{C}_{F}},+)$ and so is $\sum_{\lambda\in
F^{\ast}}I_{\lambda}$. In view of Proposition 2.1 of [2], $\sum_{\lambda\in
F^{\ast}}I_{\lambda}$ is the internal direct sum of the monoids $I_{\lambda}$.
We denote it by $\mathcal{L}_{F}^{\ast}$.
We shall now try to prove that $(\mathcal{L}_{F}^{\ast},+,.)$ is a subsemiring
of $(\widetilde{\mathcal{C}_{F}},+,.)$ which is graded by the mutiplicative
monoid $F^{\ast}$. To accomplish this, We begin by showing the following
lemma.

######  Lemma 2.2.

Let $\lambda\in F$ and let $\wedge_{\lambda}$ be the noncommutative binary
operation on $\mathbb{N}$ defined by

$t\wedge_{\lambda}s=\min(t,s)\delta_{0,\lambda}+t\delta_{0,s}^{c}\delta_{0,\lambda}^{c}$

where $\delta_{e,f}^{c}=1-\delta_{e,f}$ and $\delta_{e,f}$ is the Kronecker
symbol. Then For all $s,t\in\mathbb{N}$ and all $\lambda\in F$ we have

$\langle 0\rangle_{t}\langle\lambda\rangle_{s}=\langle
0\rangle_{t\wedge_{\lambda}s}.$

###### Proof.

We have $\langle 0\rangle_{t}\langle 1\rangle_{s}=\langle 0\rangle_{t}$ for
all $t\in\mathbb{N}$ and $s\in\mathbb{N}^{\ast}$, since
$\boldsymbol{0}_{i}\Lambda_{j}=\binom{i}{j}$. Then, for all $\lambda\in
F^{\ast}$ and for all $(t,s)\in\mathbb{N}\times\mathbb{N}^{\ast}$,

$\displaystyle\langle 0\rangle_{t}\langle\lambda\rangle_{s}$ $\displaystyle=$
$\displaystyle\langle 0\rangle_{t}\langle\lambda\rangle_{1}\langle
1\rangle_{s}$ $\displaystyle=$ $\displaystyle\langle\lambda\rangle_{1}\langle
0\rangle_{t}$ $\displaystyle=$ $\displaystyle\operatorname{span}\\{\langle
0\rangle_{0},\lambda\langle 0\rangle_{1},\ldots,\lambda^{t}\langle
0\rangle_{t}\\}$ $\displaystyle=$ $\displaystyle\langle 0\rangle_{t}$

since $\lambda$ is invertible. Hence, for all $\lambda\in F^{\ast}$ and all
$t,s\in\mathbb{N}$, we have

$\langle 0\rangle_{t}\langle\lambda\rangle_{s}=\langle
0\rangle_{t\delta_{0,s}^{c}}$

But since

$\langle 0\rangle_{t}\langle 0\rangle_{s}=\langle 0\rangle_{\min(t,s)},$

it follows that

$\langle 0\rangle_{t}\langle\lambda\rangle_{s}=\langle
0\rangle_{t\wedge_{\lambda}s}$

for all $\lambda\in F$ and all $t,s\in\mathbb{N}$. ∎

Now let us set $s_{i,j}^{m}=\binom{m}{i\vee j}\binom{i\vee j}{m-\min(i,j)}$,
$s_{i,j}^{m}\in F$, for all $i,j,m\in\mathbb{N}$ such that $i\vee j\leq m\leq
i+j$. Let us consider the sets

$\Delta_{m}(i,j)=\\{(e,t)/e\leq i;t\leq j;e+t=m\\},$
$\Delta(i,j)=\\{m/\exists(e,t)\in\Delta_{m}(i,j);s_{e,t}^{m}\neq 0\\}$

and, for any positive integers $i$ and $j$, let us denote by $i\wedge j$ the
unique integer such that $(i\wedge j)-1$ is the maximum value of the set
$\Delta(i-1,j-1)$. Remark that $i\wedge j\leq i+j-1$ and, since
$s_{0,(i-1)\vee(j-1)}^{{(i-1)\vee(j-1)}}=1$, that $i\vee j\leq i\wedge j$.
Hence

$i\vee j\leq i\wedge j\leq i+j-1.$

With these notations we have

######  Theorem 2.3.

Let $i,j\in\mathbb{N}$ and $\lambda,\mu\in F^{\ast}$. Put $0\wedge s=s\wedge
0=0$ for all $s\in\mathbb{N}$. Then

  1. (i)

$i\wedge j$ is the unique integer such that $\langle 1\rangle_{i}\langle
1\rangle_{j}=\langle 1\rangle_{i\wedge j}$.

  2. (ii)

$\langle\lambda\rangle_{i}\langle\mu\rangle_{j}=\langle\lambda\mu\rangle_{i\wedge
j}$.

  3. (iii)

$(\mathbb{N},\vee,\wedge)$ is a commutative semiring with identity, where
$\vee$ is the the maximum of two integers.

###### Proof.

  1. (i)

It is clair that if $i=0$ or $j=0$, then

$\langle 1\rangle_{i}\langle 1\rangle_{j}=\langle 1\rangle_{0}=\langle
1\rangle_{i\wedge j}.$

Suppose that $1\leq i\leq j$. Since

$\langle
1\rangle_{i}=\operatorname{span}\\{\Lambda_{0},\ldots,\Lambda_{i-1}\\}$

and

$\langle
1\rangle_{j}=\operatorname{span}\\{\Lambda_{0},\ldots,\Lambda_{j-1}\\},$

it follows that

$\langle 1\rangle_{i}\langle
1\rangle_{j}=\operatorname{span}\\{\Lambda_{e}\Lambda_{t}/e\leq i-1;t\leq
j-1\\}.$

Since

$\Lambda_{e}\Lambda_{t}=s_{e,t}^{e\vee t}\Lambda_{e\vee
t}+\cdots+s_{e,t}^{e+t}\Lambda_{e+t}$

for all $e\leq i-1$ and all $t\leq j-1$, it follows that

$\langle 1\rangle_{i}\langle 1\rangle_{j}\subseteq\langle 1\rangle_{i+j-1}.$

Furthermore, Since $\Lambda_{0}\Lambda_{t}=\Lambda_{t}$, it follows that

$\Lambda_{0},\ldots,\Lambda_{j-1}\in\langle 1\rangle_{i}\langle 1\rangle_{j},$

and hence

$\langle 1\rangle_{j}\subseteq\langle 1\rangle_{i}\langle 1\rangle_{j}.$

Thus

$\langle 1\rangle_{j}\subseteq\langle 1\rangle_{i}\langle
1\rangle_{j}\subseteq\langle 1\rangle_{i+j-1}.$

On the other hand, for all $e,t\in\mathbb{N}$ we have

(2.1)
$s_{e,t}^{e+t}=\begin{cases}\binom{2e}{e}=2\binom{2e-1}{e}=2s_{e-1,e}^{2e-1}\quad\text{if}\quad
e=t,\\\
\binom{e+t}{e}=\binom{e+t-1}{e}+\binom{e+t-1}{e-1}=s_{e,t-1}^{e+t-1}+s_{e-1,t}^{e+t-1}\quad\text{if}\quad
e\neq t\end{cases}$

Thus, by definition of $i\wedge j$, for all $1\leq m\leq i\wedge j-1$ there
exist $e_{m}\leq i-1$ and $t_{m}\leq j-1$ such that $e_{m}+t_{m}=m$ and
$s_{e_{m},t_{m}}^{m}\neq 0$.
Now consider the following equalities

$\Lambda_{e_{m}}\Lambda_{t_{m}}=s_{e_{m},t_{m}}^{e_{m}\vee
t_{m}}\Lambda_{e_{m}\vee
t_{m}}+\cdots+s_{e_{m},t_{m}}^{m}\Lambda_{m},\,\,j\leq m\leq i\wedge j-1.$

From the first equality and the fact that
$\Lambda_{0}\ldots\Lambda_{j-1}\in\langle 1\rangle_{i}\langle 1\rangle_{j}$,
we get that $\Lambda_{j}\in\langle 1\rangle_{i}\langle 1\rangle_{j}$. Using
this result and that $\Lambda_{0}\ldots\Lambda_{j}\in\langle
1\rangle_{i}\langle 1\rangle_{j}$, from the second equality it follows
$\Lambda_{j+1}\in\langle 1\rangle_{i}\langle 1\rangle_{j}$. We proceed in this
way, successively, until we get that $\Lambda_{i\wedge j-1}\in\langle
1\rangle_{i}\langle 1\rangle_{j}$. We conclude that $\langle 1\rangle_{i\wedge
j}\subseteq\langle 1\rangle_{i}\langle 1\rangle_{j}$.
For the reverse inclusion observe that if

$i\wedge j-1<m\leq e+t\leq i+j-2$

for some $e\leq i-1$ and $t\leq j-1$, then we have

$e\vee t\leq j-1\leq i\wedge j-1<m,$

hence

$1\leq d=m-e\vee t\leq\min(e,t)\leq\min(i,j);$

we have then $s_{d,e\vee t}^{m}=0$. But since

$s_{e,t}^{m}=\binom{m}{e\vee t}\binom{e\vee t}{m-\min(e,t)}=s_{d,e\vee
t}^{m}\binom{e\vee t}{m-\min(e,t)},$

we also have $s_{e,t}^{m}=0$. Consequently, $\Lambda_{e}\Lambda_{t}\in\langle
1\rangle_{i\wedge j}$ and hence $\langle 1\rangle_{i}\langle
1\rangle_{j}\subseteq\langle 1\rangle_{i\wedge j}$. The uniqueness of $i\wedge
j$ follows immediately from the fact that $\dim(\langle 1\rangle_{m})=m$ for
all $m\in\mathbb{N}$.

  2. (ii)

Since $\langle\lambda\rangle_{i}=\langle\lambda\rangle_{1}\langle
1\rangle_{i}$, it follows that

$\displaystyle\langle\lambda\rangle_{i}\langle\mu\rangle_{j}$ $\displaystyle=$
$\displaystyle\langle\lambda\rangle_{1}\langle\mu\rangle_{1}\langle
1\rangle_{i}\langle 1\rangle_{j}$ $\displaystyle=$
$\displaystyle\langle\lambda\mu\rangle_{1}\langle 1\rangle_{i\wedge j}.$

That is,

$\langle\lambda\rangle_{i}\langle\mu\rangle_{j}=\langle\lambda\mu\rangle_{i\wedge
j}.$

  3. (iii)

We have

$\langle 1\rangle_{i}\langle 1\rangle_{j}=\langle 1\rangle_{j}\langle
1\rangle_{i}$

and

$(\langle 1\rangle_{i}\langle 1\rangle_{j})\langle 1\rangle_{s}=\langle
1\rangle_{i}(\langle 1\rangle_{j}\langle 1\rangle_{s}),$

then $i\wedge j=j\wedge i$ and $(i\wedge j)\wedge s=i\wedge(j\wedge s)$, that
is, the operation $\wedge$ is commutative and associative. We also have

$\langle 1\rangle_{i}\langle 1\rangle_{1}=\langle 1\rangle_{i}$

then $i\wedge 1=i$ and $(\mathbb{N},\wedge)$ is a monoid with identity element
$1$. It is obvious that $0$ is the absorbing element of $\wedge$. Finally, we
have

$(\langle 1\rangle_{i}+\langle 1\rangle_{j})\langle 1\rangle_{m}=\langle
1\rangle_{i}\langle 1\rangle_{m}+\langle 1\rangle_{j}\langle 1\rangle_{m},$

which gives

$(i\vee j)\wedge m=(i\wedge m)\vee(j\wedge m);$

thus $\wedge$ is distributive with respect to $\vee$.

∎

Let $i$ and $j$ be any positive integers. If $F$ is a field with
characteristic $0$, then Since $s_{i,j}^{i+j}\neq 0$, we have $i\wedge
j=i+j-1$.
Suppose now that $F$ is a field with characteristic $p>0$ and consider the
$p$-expansions of $i-1$ and $j-1$

$i-1=\sum_{m\geq 0}i_{m}p^{m},$ $j-1=\sum_{m\geq 0}j_{m}p^{m}.$

Let $q$ be the smallest non-negative integer such that $i_{m}+j_{m}<p$ for all
$m\geq q$, and set

(2.2) $\displaystyle d=p^{q}+\sum_{m\geq q}(i_{m}+j_{m})p^{m}.$

Then we have the following result which is due to N. Zierler and W. H. Mills
[4]

######  Theorem 2.4.

Let $F$ be a field of characteristic $p>0$. Let $i$ and $j$ be two positive
integers and $d$ be the integer defined by 2.2. Then we have $i\wedge j=d$.

###### Proof.

Without loss of generality, we can suppose that $i\leq j$. Let

$i-1=\sum_{m\geq 0}i_{m}p^{m}$

and

$j-1=\sum_{m\geq 0}j_{m}p^{m}$

be the $p$-expansions of $i-1$ and $j-1$, and suppose first that $q=0$. In
this case we have

$\displaystyle d$ $\displaystyle=$ $\displaystyle 1+i-1+j-1$ $\displaystyle=$
$\displaystyle i+j-1,$

and hence $d-1=(i-1)+(j-1)$; that is,

$(i-1,j-1)\in\Delta_{d-1}(i-1,j-1).$

On the other hand, since $q=0$, it follows that

$d-1=\sum_{m\geq 0}(i_{m}+j_{m})p^{m}$

is the $p$-expansion of $d-1$. Then in view of Lucas’s theorem
$\binom{d-1}{i-1}\neq 0$, hence $d-1\in\Delta(i-1,j-1)$. But since

$d-1=i+j-2\geq e+t$

for all $e\leq i-1$ and $t\leq j-1$, it follows that $d-1=(i\wedge j)-1$; that
is $d=i\wedge j$.
Suppose now that $q\neq 0$. In this case we have $i\geq 2$ and $j\geq 2$. In
addition, we also have

$\displaystyle i+j-2$ $\displaystyle=$ $\displaystyle\sum_{m\leq
q-1}(i_{m}+j_{m})p^{m}+\sum_{m\geq q}(i_{m}+j_{m})p^{m}$ $\displaystyle\geq$
$\displaystyle p^{q}+\sum_{m\geq q}(i_{m}+j_{m})p^{m}$ $\displaystyle=$
$\displaystyle d$

since $i_{q-1}+j_{q-1}\geq p$. Hence

$d-i\leq j-2\leq j-1;$

that is,

$(i-1,d-i)\in\Delta_{d-1}(i-1,j-1).$

To prove that $d-1\in\Delta(i-1,j-1)$, it is thus sufficient to show that
$\binom{d-1}{i-1}\neq 0$. To see this, observe that, under our assumption that
$q\neq 0$, the $p$-expansion of $d-1$ is

$p-1+(p-1)p+\cdots+(p-1)p^{q-1}+\sum_{m\geq q}(i_{m}+j_{m})p^{m}.$

Hence $(d-1)_{m}\geq i_{m}$ for all non-negative integer $m$. Thus, using
Lucas’s theorem again, we see that $\binom{d-1}{i-1}\neq 0$. Now let

$e=\sum_{m\geq 0}e_{m}p^{m}\leq i-1$

such that $d-e\leq j-1$. Suppose that $e_{m}=0$ for all $m\leq q-1$, i.e.,

$e=\sum_{m\geq q}e_{m}p^{m}.$

Since $e\leq i-1$, it follows that

$e\leq\sum_{m\geq q}i_{m}p^{m},$

and since $d-e\leq j-1$, we conclude that

$d-\sum_{m\geq q}i_{m}p^{m}\leq j-1.$

Therefore,

$p^{q}\leq\sum_{m\leq q-1}i_{m}p^{m}$

which is a contradiction. Thus there exists $m\leq q-1$ such that $e_{m}\neq
0$. It follows that $\binom{d}{e}=0$. Hence, we have proved that
$s_{e,t}^{d}=0$ whenever $(e,t)\in\Delta_{d}(i-1,j-1)$. Consequently, in view
of formula 2.1, this property must also be true for all integers greater than
or equal to $d$. Therefore we have proved $d=i\wedge j$, as desired. ∎

It is well-known that the definition of $i\wedge j$ given by Zierler and Mills
in [4] is equivalent to the following one proposed by Göttfert and
Niederreiter [1].

######  Definition 2.5.

For two positive integers $s$ and $t$, let $s\wedge t$ be the maximum value of
$i+j+1$ such that $\binom{i+j}{i}\neq 0$ (in $F$) where $0\leq i\leq s-1$ and
$0\leq j\leq t-1$.

The operation $\wedge$ depends only on the characteristic of $F$ and has the
following properties

######  Lemma 2.6.

Let $i,j$ be positive integers and $s,s^{\prime},t,t^{\prime}$ non-negative
integers.

  1. 1.

If $F$ is a field with characteristic $0$, then we have $i\wedge j=i+j-1$.

  2. 2.

If $s^{\prime}\leq s$ and $t^{\prime}\leq t$, then $s^{\prime}\wedge
t^{\prime}\leq s\wedge t$ and $s^{\prime}\wedge_{\lambda}t^{\prime}\leq
s\wedge_{\lambda}t$ for all $\lambda\in F$.

  3. 3.

$i\wedge j$ is the unique integer $d$ such that $\binom{d-1}{i-1}\neq 0$ and
$\binom{d}{e}=0$ for all integer $e$ such that

$i\wedge j\leq e+j-1\leq i+j-2$

  4. 4.

If $F$ is a field with characteristic $p>0$, then

    1. (i)

$i\wedge j=i+j-1$ if and only if $q=0$, where $q$ is the integer defined in
formula 2.2.

    2. (ii)

$ip^{s}\wedge jp^{s}=p^{s}(i\wedge j)$.

    3. (iii)

$ip^{s}\wedge j$ is divisible by $p^{s}$.

###### Proof.

  1. 1.

Follows immediately from the fact that $i+j-2$ is the maximum value of the set
$\Delta(i-1,j-1)$.

  2. 2.

Follows immediately from the fact that $\langle 1\rangle_{s^{\prime}}\langle
1\rangle_{t^{\prime}}$ and $\langle
0\rangle_{s^{\prime}}\langle\lambda\rangle_{t^{\prime}}$ are respectively
subspaces of $\langle 1\rangle_{s}\langle 1\rangle_{t}$ and $\langle
0\rangle_{s}\langle\lambda\rangle_{t}$.

  3. 3.

See the proof of the above theorem.

  4. 4.

    1. (i)

See the proof of the above theorem.
Let

$i-1=\sum_{m\geq 0}i_{m}p^{m}$

and

$j-1=\sum_{m\geq 0}j_{m}p^{m}$

be the $p$-expansions of $i-1$ and $j-1$ respectively.

    2. (ii)

Let $q$ be the smallest non-negative integer such that $i_{m}+j_{m}<p$ for all
$m\geq q$. Since

$ip^{s}-1=(i-1)p^{s}+p^{s},$

it follows that

$\displaystyle ip^{s}-1$ $\displaystyle=$ $\displaystyle\sum_{m\geq
0}(ip^{s})_{m}$ $\displaystyle=$ $\displaystyle
p-1+(p-1)p+\cdots+(p-1)p^{s-1}+\sum_{m\geq 0}i_{m}p^{m+s}$

is the $p$-expansion of $ip^{s}-1$. Similarly,

$\displaystyle jp^{s}-1$ $\displaystyle=$ $\displaystyle\sum_{m\geq
0}(jp^{s})_{m}$ $\displaystyle=$ $\displaystyle
p-1+(p-1)p+\cdots+(p-1)p^{s-1}+\sum_{m\geq 0}j_{m}p^{m+s}$

is the $p$-expansion of $jp^{s}-1$. It is clair that $q+s$ is the smallest
non-negative integer such that

$(ip^{s})_{m}+(jp^{s})_{m}<p\quad\text{for all}\quad m\geq q+s.$

Thus

$\displaystyle ip^{s}\wedge jp^{s}$ $\displaystyle=$ $\displaystyle
p^{s+q}+\sum_{m\geq q}(i_{m}+j_{m})p^{s+m}$ $\displaystyle=$ $\displaystyle
p^{s}(i\wedge j).$

    3. (iii)

Let $q$ be the smallest non-negative integer such that

$(ip^{s})_{m}+j_{m}<p\quad\text{for all}\quad m\geq q.$

If $q\geq s$, then $ip^{s}\wedge j$, which is divisible by $p^{q}$, is
certainly divisible by $p^{s}$. Suppose that $q\leq s-1$. Since

$p-1+(p-1)p+\cdots+(p-1)p^{s-1}+\sum_{m\geq 0}i_{m}p^{m+s}$

is the $p$-expansion of $ip^{s}-1$ and

$j_{0}+j_{1}p+\cdots+j_{s-1}p^{s-1}+\sum_{m\geq 0}j_{m+s}p^{m+s}$

is that of $j-1$, it follows that

$j_{q}=\cdots=j_{s-1}=0$

and then

$\displaystyle ip^{s}\wedge j$ $\displaystyle=$ $\displaystyle
p^{q}+(p-1)p^{q}+\cdots+(p-1)p^{s-1}+\sum_{m\geq 0}(i_{m}+j_{m+s})p^{m+s}$
$\displaystyle=$ $\displaystyle p^{s}+p^{s}\sum_{m\geq 0}(i_{m}+j_{m+s})p^{m}$

is divisible by $p^{s}$. In both cases $ip^{s}\wedge j$ is divisible by
$p^{s}$, as was to be shown.

∎

######  Remark 2.7.

We will see in section $4$ that Property $(iii)$ of the above Lemma is key in
the proof of Theorem 2.14 in the inseparable case.

Let us consider $(\mathcal{L}_{F}^{\ast},+,.)$ where multiplication is given
by

$\langle\lambda\rangle_{i}\langle\mu\rangle_{j}=\langle\lambda\mu\rangle_{i\wedge
j},$

extended via distributivity.
For all monic polynomial $P\in F[X]$, let us denote by $L(P)$ the $F$-vector
space of all linear recurrence sequences over $F$ with characteristic
polynomial $P$. Let $\mathcal{R}_{F}^{\ast}$ be the set of all $L(P)$ where
$P$ is a polynomial with all roots in $F^{\ast}$. Let us partition

${F^{\star}}^{m}=\bigcup_{i\in T}\Omega_{i}$

into equivalence classes under the equivalence relation

$(\lambda_{1},\ldots,\lambda_{m})\mathcal{R}(\mu_{1},\ldots,\mu_{m})\quad\text{if
and only if}\quad\lambda_{1}\cdots\lambda_{m}=\mu_{1}\cdots\mu_{m}.$

Let $\wedge_{m}$ denote the map defined as follows

$\displaystyle\wedge_{m}:\mathbb{N}^{m}$
$\displaystyle\longrightarrow\mathbb{N}$ $\displaystyle(t_{1},\ldots,t_{m})$
$\displaystyle\longmapsto t_{1}\wedge\cdots\wedge t_{m}$

and let $f:{F^{\star}}^{m}\longrightarrow\mathbb{N}^{m}$ be a mapping
satisfying Property $(\mathcal{D})$ below.

$\displaystyle\textbf{Property}(\mathcal{D}):$ $\displaystyle\text{For
all}\,\,(\mu_{1},\ldots,\mu_{m})\in{F^{\star}}^{m},\,\,\text{and}\,\,\sigma\in
S_{m},$ $\displaystyle\text{the symmetric group of degree}\,\,m,$
$\displaystyle
f(\mu_{\sigma(1)},\ldots,\mu_{\sigma(m)})=(t_{\sigma(1)},\ldots,t_{\sigma(m)})$
$\displaystyle\text{whenever}\,\,(t_{1},\ldots,t_{m})=f(\mu_{1},\ldots,\mu_{m}).$

Suppose further that $f$ has finite support, and let us use the following
notations for simplicity

$\displaystyle\langle\mu_{1}\rangle_{f}\cdots\langle\mu_{m}\rangle_{f}$
$\displaystyle=$
$\displaystyle\langle\mu_{1}\rangle_{t_{1}}\cdots\langle\mu_{m}\rangle_{t_{m}}$
$\displaystyle=$
$\displaystyle\langle\mu_{1}\cdots\mu_{m}\rangle_{\overline{f}(\mu_{1},\ldots,\mu_{m})}$
$\displaystyle\overline{f}$ $\displaystyle=$ $\displaystyle\wedge_{m}\circ f$
$\displaystyle\widehat{\Omega}_{i}$ $\displaystyle=$
$\displaystyle\mu_{1}\cdots\mu_{m}\quad\text{if}\quad(\mu_{1},\ldots,\mu_{m})\in\Omega_{i}$
$\displaystyle\Omega_{i}^{f}$ $\displaystyle=$
$\displaystyle\max_{(\mu_{1},\ldots,\mu_{m})\in\Omega_{i}}(\overline{f}(\mu_{1},\ldots,\mu_{m}))$

where $f(\mu_{1},\ldots,\mu_{m})=(t_{1},\ldots,t_{m})$ and $i$ is an element
of $T$.
Under these notations, we have the following lemma:

######  Lemma 2.8.

Let $m$ be a positive integer and
$f:{F^{\star}}^{m}\longrightarrow\mathbb{N}^{m}$ be a map with finite support
satisfying Property $(\mathcal{D})$. Then we have

$\sum_{(\lambda_{1},\ldots,\lambda_{m})\in{F^{\star}}^{m}}\langle\lambda_{1}\rangle_{f}\cdots\langle\lambda_{m}\rangle_{f}=\sum_{i\in
T}\langle\widehat{\Omega}_{i}\rangle_{\Omega_{i}^{f}}.$

###### Proof.

We have

$\displaystyle\sum_{(\lambda_{1},\ldots,\lambda_{m})\in{F^{\star}}^{m}}\langle\lambda_{1}\rangle_{f}\cdots\langle\lambda_{m}\rangle_{f}$
$\displaystyle=\sum_{i\in
T}\sum_{(\lambda_{1},\ldots,\lambda_{m})\in\Omega_{i}}\langle\lambda_{1}\rangle_{f}\cdots\langle\lambda_{m}\rangle_{f}$
$\displaystyle=\sum_{i\in
T}\sum_{(\lambda_{1},\ldots,\lambda_{m})\in\Omega_{i}}\langle\lambda_{1}\cdots\lambda_{m}\rangle_{\overline{f}(\lambda_{1},\ldots,\lambda_{m})}$
$\displaystyle=\sum_{i\in
T}\sum_{(\lambda_{1},\ldots,\lambda_{m})\in\Omega_{i}}\langle\widehat{\Omega}_{i}\rangle_{\overline{f}(\lambda_{1},\ldots,\lambda_{m})}$
$\displaystyle=\sum_{i\in
T}\langle\widehat{\Omega}_{i}\rangle_{\Omega_{i}^{f}}$

∎

######  Remark 2.9.

In Lemma 2.8, we may replace the condition that $f$ is with finite support
with the more general assumption that $\overline{f}$ is bounded on every
$\Omega_{i}$.

######  Remark 2.10.

It is natural to impose the requirement that $f$ satisfies Property
$(\mathcal{D})$, since, if not, $f$ could not be well-defined.

We may now state the following result

######  Theorem 2.11.

Let $F$ be a field. Then

  1. (i)

$(\mathcal{L}_{F}^{\ast},+,.)$ is a commutative semiring graded by the
multiplicative monoid $F^{\ast}$.

  2. (ii)

$\mathcal{L}_{F}^{\ast}=\mathcal{R}_{F}^{\ast}$.

  3. (iii)

Let $P_{i}=\prod_{\lambda\in F^{\ast}}(X-\lambda)^{\lambda(P_{i})}$, $1\leq
i\leq m$, be monic polynomials with all roots in $F^{\ast}$, where
$\lambda(P_{i})$ designates the multiplicity of $\lambda$ in $P_{i}$. Then

$\prod_{i=1}^{m}L(P_{i})=L(\Upsilon(P_{1},\ldots,P_{m})),$

where

$\Upsilon(P_{1},\ldots,P_{m})=\prod_{i\in
T}(X-\widehat{\Omega}_{i})^{\Omega_{i}^{f}}$

and $f$ is the map defined by

$\displaystyle f:{F^{\star}}^{m}$ $\displaystyle\longrightarrow\mathbb{N}^{m}$
$\displaystyle(\lambda_{1},\ldots,\lambda_{m})$
$\displaystyle\longmapsto(\lambda_{1}(P_{1}),\ldots,\lambda_{m}(P_{m})).$

###### Proof.

  1. (i)

Obviously $(\mathcal{L}_{F}^{\ast},+,.)$ is a commutative semiring in which
$\langle 1\rangle_{1}$ is the multiplicative identity. On the other hand, we
have

$\mathcal{L}_{F}^{\ast}=\Sigma_{\lambda\in F^{\ast}}I_{\lambda}$

where $\\{I_{\lambda}\\}_{\lambda\in F^{\ast}}$ is a family of additive
subsemigroups of $\mathcal{L}_{F}^{\ast}$ satisfying
$I_{\lambda}I_{\mu}\subseteq I_{\lambda\mu}$ by Theorem 2.3. Let now
$\lambda_{1},\ldots,\lambda_{m}$ be elements of $F^{\ast}$. Since

$\langle\lambda_{i}\rangle_{t_{i}}\subseteq\langle\lambda_{1}\rangle_{t_{1}}+\cdots+\langle\lambda_{m}\rangle_{t_{m}},$

it follows that

$\langle\lambda_{1}\rangle_{t_{1}}+\cdots+\langle\lambda_{m}\rangle_{t_{m}}\neq\operatorname{span}\\{\boldsymbol{0}\\},$

if there exists at least one $\lambda_{i}$ such that
$\langle\lambda_{i}\rangle_{t_{i}}\neq\operatorname{span}\\{\boldsymbol{0}\\}$.
Thus $(i)$ holds.

  2. (ii)

Let $L(P)\in\mathcal{R}_{F}^{\ast}$, then there exists a unique element

$(\lambda(P))_{\lambda\in F^{\ast}}\in\bigoplus_{F^{\ast}}\mathbb{N}$

such that

$P=\prod_{\lambda\in F^{\ast}}(X-\lambda)^{\lambda(P)}.$

Hence

$L(P)=\sum_{\lambda\in
F^{\ast}}\langle\lambda\rangle_{\lambda(P)}\in\mathcal{L}_{F}^{\ast}.$

Now let

$\sum_{\lambda\in
F^{\ast}}\langle\lambda\rangle_{t_{\lambda}}\in\mathcal{L}_{F}^{\ast}.$

Then

$\sum_{\lambda\in F^{\ast}}\langle\lambda\rangle_{t_{\lambda}}=L(P)$

where $P$ is the unique polynomial that satisfies $\lambda(P)=t_{\lambda}$.
Hence

$\sum_{\lambda\in
F^{\ast}}\langle\lambda\rangle_{t_{\lambda}}\in\mathcal{R}_{F}^{\ast}.$

Thus $\mathcal{L}_{F}^{\ast}=\mathcal{R}_{F}^{\ast}$.

  3. (iii)

Let

$\Upsilon(P_{1},\ldots,P_{m})=\prod_{i\in
T}(X-\widehat{\Omega}_{i})^{\Omega_{i}^{f}}.$

Since the $\widehat{\Omega}_{i}$’s are pairwise distinct elements of
$F^{\ast}$, it follows that

$L(\Upsilon(P_{1},\ldots,P_{m}))=\sum_{i\in
T}\langle\widehat{\Omega}_{i}\rangle_{\Omega_{i}^{f}}.$

On the other hand, by Lemma 2.8, we have

$\displaystyle\prod_{i=1}^{m}L(P_{i})$
$\displaystyle=\prod_{i=1}^{m}(\sum_{\lambda\in
F^{\ast}}\langle\lambda\rangle_{\lambda(P_{i})})$
$\displaystyle=\sum_{(\lambda_{1},\ldots,\lambda_{m})\in{F^{\star}}^{m}}\langle\lambda_{1}\rangle_{f}\cdots\langle\lambda_{m}\rangle_{f}$
$\displaystyle=\sum_{i\in
T}\langle\widehat{\Omega}_{i}\rangle_{\Omega_{i}^{f}}.$

Then

$\prod_{i=1}^{m}L(P_{i})=L(\Upsilon(P_{1},\ldots,P_{m})),$

in other words (iii) holds.

∎

######  Remark 2.12.

We note that if one of the $P_{i}$ is $1$, then $\Omega_{i}^{f}=0$ for all
$i\in T$; that is $L(\Upsilon(P_{1},\ldots,P_{m}))=0$.

Let

$\mathcal{L}_{F}=I_{0}+\mathcal{L}_{F}^{\ast}=\sum_{\lambda\in F}I_{\lambda}$

be the set of all $L(P)$ where $P$ is a monic polynomial with all roots in
$F$. It is easily seen that every element $L(P)$ of $\mathcal{L}_{F}$ can be
written in just one way in the form

$L(P)=\langle 0\rangle_{s}+L(Q);$

more precisely,

$s=0(P)\,\,\text{and}\,\,Q=\prod_{\lambda\in
F^{\ast}}(X-\lambda)^{\lambda(P)}.$

Furthermore, because $(\mathcal{L}_{F}^{\ast},+,.)$ is a commutative semiring
graded by the multiplicative monoid $F^{\ast}$, and since $\langle
0\rangle_{s}\langle\lambda\rangle_{t}=\langle 0\rangle_{s\vee_{\lambda}t}$, it
follows that $(\mathcal{L}_{F},+,.)$ is a commutative semiring graded by the
multiplicative monoid $F$.
Now let us consider

$L(P_{i})=\langle 0\rangle_{s_{i}}+L(Q_{i})\neq 0\in\mathcal{L}_{F},\,\,1\leq
i\leq m,$

and let $f$ be the mapping defined by

$\displaystyle f:{F^{\star}}^{m}$ $\displaystyle\longrightarrow\mathbb{N}^{m}$
$\displaystyle(\lambda_{1},\ldots,\lambda_{m})$
$\displaystyle\longmapsto(\lambda_{1}(Q_{1}),\ldots,\lambda_{m}(Q_{m})).$

Since

$\langle 0\rangle_{t}L(Q)=\langle 0\rangle_{t}\sum_{\lambda\in
F^{\ast}}\langle\lambda\rangle_{\lambda(Q)}\,\,\text{for
all}\,\,L(Q)\in\mathcal{L}_{F}^{\ast},$

it follows that

$\langle 0\rangle_{t}L(Q)=\langle
0\rangle_{t}\,\,\text{if}\,\,L(Q)\neq\operatorname{span}\\{\boldsymbol{0}\\},$

i.e. $Q\neq 1$. Put

$\Theta=\\{i/Q_{i}=1\\},$

and let

$\displaystyle\rho=\begin{cases}\min\\{s_{i}/i\in\Theta\\}&\quad\text{if}\quad\Theta\neq\emptyset\\\
\max\\{s_{i}/1\leq i\leq m\\}&\quad\text{otherwise}.\end{cases}$

Note that since we have assumed that none of the subspaces $L(P_{i})=\langle
0\rangle_{s_{i}}+L(Q_{i})$ is zero, we have that $s_{i}\neq 0$ whenever
$i\in\Theta$. Using this remark it is straightforward to check that

$\prod_{i=1}^{m}L(P_{i})=\left\\{\begin{array}[]{lll}\langle
0\rangle_{\rho}+L(\Upsilon(Q_{1},\ldots,Q_{m}))&\quad\text{if}&\Theta=\emptyset\\\
\langle 0\rangle_{\rho}&\quad\text{if}&\Theta\neq\emptyset\end{array}\right.$

and since $L(\Upsilon(Q_{1},\ldots,Q_{m}))=0$ whenever one of the $Q_{i}$ is
$1$, it follows that, even if $\Theta\neq\emptyset$, we have
$\prod_{i=1}^{m}L(P_{i})=\langle
0\rangle_{\rho}+L(\Upsilon(Q_{1},\ldots,Q_{m}))$. Then in all cases, we have,

$\displaystyle\prod_{i=1}^{m}L(P_{i})$ $\displaystyle=$ $\displaystyle\langle
0\rangle_{\rho}+L(\Upsilon(Q_{1},\ldots,Q_{m}))$ $\displaystyle=$
$\displaystyle L(X^{\rho}\Upsilon(Q_{1},\ldots,Q_{m})).$

Thus we have proved the following:

######  Theorem 2.13.

Let $P_{1}=X^{s_{1}}Q_{1},\ldots,P_{m}=X^{s_{m}}Q_{m}$ be non-constant monic
polynomials over $F$ with all roots in $F$, where $Q_{i}(0)\neq 0,\,1\leq
i\leq m$, and put $\Theta=\\{i/Q_{i}=1\\}$. Then we have

$\prod_{i=1}^{m}L(P_{i})=L(X^{\rho}\Upsilon(Q_{1},\ldots,Q_{m})),$

where

$\displaystyle\rho=\begin{cases}\min\\{s_{i}/i\in\Theta\\}&\quad\text{if}\quad\Theta\neq\emptyset\\\
\max\\{s_{i}/1\leq i\leq m\\}&\quad\text{otherwise}.\end{cases}$

We close this section by proving that the result in Theorem 2.13 still holds
true for any monic polynomials $P_{1},\ldots,P_{m}$ over $F$.

######  Theorem 2.14.

Let $P_{1}=X^{s_{1}}Q_{1},\ldots,P_{m}=X^{s_{m}}Q_{m}$ be non-constant monic
polynomials over $F$, where $Q_{i}(0)\neq 0,\,1\leq i\leq m$, and put
$\Theta=\\{i/Q_{i}=1\\}$. Then we have
1\. The polynomial $\Upsilon(Q_{1},\ldots,Q_{m})$, which is a polynomial over
the splitting field of the polynomial $\prod_{i=1}^{m}Q_{i}$ over $F$, is also
a polynomial over $F$.
2\. We have

$\prod_{i=1}^{m}L(P_{i})=L(X^{\rho}\Upsilon(Q_{1},\ldots,Q_{m})),$

where

$\displaystyle\rho=\begin{cases}\min\\{s_{i}/i\in\Theta\\}&\quad\text{if}\quad\Theta\neq\emptyset\\\
\max\\{s_{i}/1\leq i\leq m\\}&\quad\text{otherwise}.\end{cases}$

###### Proof.

1\. Let us note first that if there exist

$(\alpha_{1},\ldots,\alpha_{m})\in\Omega_{i}\,\,\text{and}\,\,(\beta_{1},\ldots,\beta_{m})\in\Omega_{j}$

such that $\alpha_{1}\cdots\alpha_{m}$ and $\beta_{1}\cdots\beta_{m}$ are
conjugate over $F$, then

$\Omega_{i}^{f}=\Omega_{j}^{f},$

where $f$ is the mapping defined by

$\displaystyle f:{F^{\star}}^{m}$ $\displaystyle\longrightarrow\mathbb{N}^{m}$
$\displaystyle(\mu_{1},\ldots,\mu_{m})$
$\displaystyle\longmapsto(\mu_{1}(Q_{1}),\ldots,\mu_{m}(Q_{m})),$

$\mu_{i}(Q_{i})$ is the multiplicity of $\mu_{i}$ in the splitting field of
$Q_{i}$. In fact, suppose that there exists an $F$-isomorphism

$\sigma:F(\alpha_{1}\cdots\alpha_{m})\longrightarrow
F(\beta_{1}\cdots\beta_{m})$

such that

$\displaystyle\sigma(\alpha_{1}\cdots\alpha_{m})$ $\displaystyle=$
$\displaystyle\sigma(\alpha_{1})\cdots\sigma(\alpha_{m})$ $\displaystyle=$
$\displaystyle\beta_{1}\cdots\beta_{m}.$

Then $(\sigma(\alpha_{1}),\ldots,\sigma(\alpha_{m}))\in\Omega_{j}$. Since
$\alpha_{i}$ and $\sigma(\alpha_{i})$ are $F$-conjugate roots of $Q_{i}$, they
have the same multiplicity in the splitting field of $Q_{i}$, i.e.,
$\alpha_{i}(Q_{i})=\sigma(\alpha_{i})(Q_{i})$. Hence

$\displaystyle\alpha_{1}(Q_{1})\wedge\cdots\wedge\alpha_{m}(Q_{m})$
$\displaystyle=$
$\displaystyle\sigma(\alpha_{1})(Q_{1})\wedge\cdots\wedge\sigma(\alpha_{m})(Q_{m})$
$\displaystyle\leq$ $\displaystyle\Omega_{j}^{f},$

but since

$\sigma(\mu_{1})\cdots\sigma(\mu_{m})=\beta_{1}\cdots\beta_{m}\,\,\text{for
all}\,\,(\mu_{1},\ldots,\mu_{m})\in\Omega_{i},$

we obtain $\Omega_{i}^{f}\leq\Omega_{j}^{f}$. Similarly,
$\Omega_{j}^{f}\leq\Omega_{i}^{f}$, and thus $\Omega_{i}^{f}=\Omega_{j}^{f}$.
Let us then partition $\bigcup_{i\in T}\Omega_{i}$ into subsets $G_{i},i\in
N$, each of which contains the conjugate of each of its elements. Then

$\Upsilon(Q_{1},\ldots,Q_{m})=\prod_{i\in N}(\prod_{\omega\in
G_{i}}(X-\omega))^{\theta_{i}}$

where $\theta_{i}=\Omega_{j}^{f}$ for some $j\in T$.
To prove that the polynomial $\Upsilon(Q_{1},\ldots,Q_{m})$ is still belongs
to $F[X]$ we treat two cases separately, according to whether
$\prod_{i=1}^{m}Q_{i}$ is separable or not.
Suppose that $\prod_{i=1}^{m}Q_{i}$ is separable. Then so is
$\Upsilon(Q_{1},\ldots,Q_{m})$, as a polynomial with coefficients in the
splitting field of $\prod_{i=1}^{m}P_{i}$. Since all conjugates, over $F$, of
a root of $\Upsilon(Q_{1},\ldots,Q_{m})$ are roots of
$\Upsilon(Q_{1},\ldots,Q_{m})$, we have $\prod_{\omega\in G_{i}}(X-\omega)\in
F$ for all $i\in N$, and so is the polynomial $\Upsilon(Q_{1},\ldots,Q_{m})$.
Now suppose that $\prod_{i=1}^{m}Q_{i}$ is inseparable. Let $p$ be the
characteristic of $F$ and fix $j\in N$. It is clear that there exists
$\varpi=\alpha_{1}\cdots\alpha_{m}$ such that

$\theta_{j}=\alpha_{1}(Q_{1})\wedge\cdots\wedge\alpha_{m}(Q_{m}).$

On the other hand, we have

$Q_{i}(X)=Irr(X,\alpha_{i},F)^{t_{i}}H_{i}(X)$

where $Irr(X,\alpha_{i},F)$ and $H_{i}$ are coprime. Since
$Irr(X,\alpha_{i},F)$ is an irreducible polynomial over $F$, it follows that
over a suitable extension of $F$ we have

$Irr(X,\alpha_{i},F)=D_{i}^{q_{i}},$

where $D_{i}$ is a polynomial without multiple roots, and $q_{i}$ is a power
of $p$. Therefore $\alpha_{i}(Q_{i})=t_{i}q_{i}$, and then

$\theta_{j}=t_{1}q_{1}\wedge\cdots\wedge t_{m}q_{m}.$

Let $F_{j}$ be the extension of $F$ obtained by adjoining the coefficients of
the polynomials $D_{1},\ldots,D_{m}$ and set

$q=q_{1}\vee\cdots\vee q_{m}.$

It is clear that $x^{q}\in F$ for all $x\in F_{j}$, and then $\omega^{q}\in F$
for all $\omega\in G_{j}$ since $G_{j}\subseteq F_{j}$. Now, by using Lemma
2.6(iii), we see $\theta_{j}$ is divisible by $q$ and then the coefficients of

$(\prod_{\omega\in G_{j}}(X-\omega))^{\theta_{j}}$

are in $F$, and so are the coefficients of $\Upsilon(Q_{1},\ldots,Q_{m})$.
2\. Let $E$ be an extension field of $F$. To avoid confusion, let us denote by
$L_{E}(P)$ the set of all linear recurrence sequences over $E$ with
characteristic polynomial $P$. It is well known, and easy to show, that

$L_{F}(P_{1})\cdots L_{F}(P_{m})=\mathcal{C}_{F}\bigcap L_{E}(P_{1})\cdots
L_{E}(P_{m});$

hence if further $Q_{1},\ldots,Q_{m}$ splits over $E$ then we obtain

$L_{E}(P_{1})\cdots L_{E}(P_{m})=L_{E}(X^{\rho}\Upsilon(Q_{1},\ldots,Q_{m}))$

by Theorem 2.13. Therefore

$L_{F}(P_{1})\cdots L_{F}(P_{m})=L_{F}(X^{\rho}\Upsilon(Q_{1},\ldots,Q_{m})).$

Thus the proof is complete. ∎

## References

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