Is $\bar{\mathbb{Q}}(x)\cap \mathbb{Q}((x))=\mathbb{Q}(x)$? [unsolved (even though we earlier thought it was)]

Fix the algebraic closure of $\mathbb{Q}((x))$ for this question to make sense. I know that $\mathbb{Q}((x)) \cap \overline{\mathbb{Q}(x)}$ has elements that are not in $\mathbb{Q}(x)$ (in analogy to the "algebraic p-adics"). So I wondered, if we intersect with something even smaller, $\bar{\mathbb{Q}}(x)$, would then the result be $\mathbb{Q}(x)$?

More neatly, does $\bar{\mathbb{Q}}(x)\cap \mathbb{Q}((x))=\mathbb{Q}(x)$?

Solution 1:

For any automorphism $\sigma$ of $\bar{\mathbb{Q}}$, extend it to an automorphism $s$ of $\bar{\mathbb{Q}}((x))$ by acting on coefficients: $$s\left(\sum a_n x^n\right) = \sum \sigma(a_n) x^n$$

I assert the following:

• The subfield of $\bar{\mathbb{Q}}(x)$ fixed by all $s$'s is $\mathbb{Q}(x)$ (note that each $s$ really is an automorphism of $\mathbb{Q}(x)$)
• The subfield of $\bar{\mathbb{Q}}((x))$ fixed by all $s$'s is $\mathbb{Q}((x))$

Therefore, any element of $\bar{\mathbb{Q}}(x)$ that is an element of $\mathbb{Q}((x))$ must also be an element of $\mathbb{Q}(x)$.

One method to see the former assertion is to write elements in a normal form such as a scalar times a ratio (in lowest terms) of monic polynomials. Since the $s$'s act on the coefficients, elements in the fixed field must have all rational coefficients.

Solution 2:

Let $L$ be the field of Puiseux series over $\bar{\mathbb Q}$. This is an algebraic closed field containing $\mathbb Q((x))$ and $\bar{\mathbb Q}(x)$.

Then :

• The elements of $L$ lying in $\mathbb Q((x))$ are the elements with rational coefficients ;
• The elements of $L$ lying in $\bar{\mathbb Q}(x)$ are the elements with no fractional power for which the sequence of coefficient satisfy a linear recurrence with constant coefficients in $\bar{\mathbb Q}$ ;
• The elements of $L$ lying in $\mathbb Q(x)$ are the elements with no fractional power, rational coefficients and for which the sequence of coefficient satisfy a linear recurrence with constant coefficients in $\mathbb Q$.

So it is enough to prove that if a sequence of rational numbers satisfy a linear recurrence with constant coefficients in $\bar{\mathbb Q}$, then it also satisfy a linear recurrence with constant coefficients in ${\mathbb Q}$.

Let $(u_n)\in\mathbb Q^\mathbb N$ such that for all n : $$ u_{n+p} = \sum_{i=0}^{p-1} a_i u_{n+i} $$ for some $(a_i)\in \bar{\mathbb Q}^p$

The coefficients $(a_i)$ lie in a finite extension $K$ of $\mathbb Q$. Consider the trace form $Tr_{K/\mathbb Q}$. Then, for all n : $$ d\cdot u_{n+p} = \sum_{i=0}^{p-1}Tr_{K/\mathbb Q}(a_i) u_{n+i}, $$ with $d$ the degree of the extension. QED.

NOTE

The same argument can be formulated without recurrences the following way :

• The elements of $L$ lying in $\mathbb Q((x))$ are the elements with rational coefficients ;
• The elements of $L$ lying in $\bar{\mathbb Q}(x)$ are the elements $u$ with no fractional power and s.t. there exists $P \in \bar{\mathbb Q}[x]$ s.t. $uP$ is a polynomial ;
• The elements of $L$ lying in $\mathbb Q(x)$ are the elements with no fractional power, rational coefficients and s.t. there exists $P \in {\mathbb Q}[x]$ s.t. $uP$ is a polynomial.

The same argument with traces works as well.

Solution 3:

New answer

The answer is yes.

I think the $\mathbb Q$ and $\overline{\mathbb Q}$ are red herrings.

For instance, if $L/K$ is a field extension and $X$ an indeterminate, one can ask if the inclusion $$ L(X)\cap K((X))\subset K(X) $$ holds. (The answer is yes.) This is clearly equivalent to $$ L(X)\cap K[[X]]\subset K(X). $$ Now let $f$ be in $K[[X]]$, and let $F$ be $K$ or $L$. Then $f$ is in $F(X)$ if and only if the linear equation $$ p-fq=0 $$ has a nonzero solution $(p,q)\in F[X]^2$, and we see that the problem makes sense of $K$ and $L$ are commutative rings.

By reasoning like this, we see that the claim boils down to the following statement, which is Corollary $3$ of Proposition II.$3.7$ in Bourbaki's Algèbre.

If $A$ is an associative ring with $1$, if $X$ is a free right $A$-module, and if $(Y_i)_{i\in I}$ is a family of left $A$-modules, then the natural map $$ \phi:X\ \underset{A}{\otimes}\ \prod_{i\in I}\ Y_i\to\prod_{i\in I}\ \left(X\underset{A}{\otimes} Y_i\right) $$ is injective.

Proof. Identifying $X$ to a direct sum $\bigoplus_{j\in J}A$ we can write view $\phi$ as a map $$ \phi:U:=\bigoplus_{j\in J}\ \prod_{i\in I}\ Y_i\ \to\ \prod_{i\in I}\ \bigoplus_{j\in J}\ Y_i=:V $$ Now $U$ and $V$ are subgroups of $$ W:=\prod_{(i,j)\in I\times J}Y_i. $$ More precisely, an element $$ y=(y_{ij})_{(i,j)\in I\times J}\in W $$ is in $U$ if and only if there is a finite subset $J(y)$ of $J$ such that $y_{ij}=0$ if $j$ is not in $J(y)$.

Similarly, $y$ is in $V$ if and only if there is, for each $i$ in $I$, a finite subset $J_i(y)$ of $J$ such that $y_{ij}=0$ if $j$ is not in $J_i(y)$.

So we have $U\subset V$, and it is easy to see that $\phi$ is the inclusion.

Old answer

The answer is yes, as follows from the observation below.

Let $A\subset B$ be commutative rings, $x$ an indeterminate, and $f$ an element of $A[[x]]$. Consider the $A[x]$-linear map $$ \phi:A[x]\times A[x]\to A[[x]],\quad(P,Q)\mapsto P-fQ, $$ and the $B[x]$-linear map $$ \psi:B[x]\times B[x]\to B[[x]],\quad(P,Q)\mapsto P-fQ. $$

If $B$ is $A$-free and $\phi$ injective, then $\psi$ is injective.

Indeed, $\psi$ is the composition of the $B[x]$-linear extension $$ \phi_B:B[x]\times B[x]\to B\otimes_AA[[x]] $$ of $\phi$ and the natural map $$ \theta:B\otimes_AA[[x]]\to B[[x]]. $$ Then $\phi_B$ is injective because $B$ is $A$-flat, and $\theta$ is injective because $B$ is $A$-free.

EDIT A. Assume now that $A$ and $B$ are fields. We claim $$ B(x)\cap A((x))\subset A(x). $$ (The reverse inclusion is obvious.) It suffices to show $$ B(x)\cap A[[x]]\subset A(x). $$ Let $f$ be in $A[[x]]$. Then $f$ is in $A(x)$ if and only if $\phi$ is not injective, and $f$ is in $B(x)$ if and only if $\psi$ is not injective.

EDIT B. Let me prove the injectivity of $\theta$. Let $(b_i)$ be an $A$-basis of $B$, and let $f$ be in $\ker\theta$. Then $f$ can be written in a unique way as $$ f=\sum_i\ b_i\otimes f_i $$ with $f_i\in A[[x]]$ for all $i$, and $f_i=0$ for almost all $i$. Put $$ f_i=\sum_{n\ge0}\ a_{in}\ x^n $$ with $a_{in}\in A$. Then we have $$ 0=\theta(f)=\sum_{n\ge0}\ \sum_i\ b_i\ a_{in}\ x^n. $$ This implies $$ \sum_i\ b_i\ a_{in}=0 $$ for all $n$, and thus $a_{in}=0$ for all $i$ and all $n$.

EDIT C. Let $A\subset B$ be commutative rings and $x$ an indeterminate. The purpose of this edit is to define $A(x),B(x)$ and $A((x))$ in such a way that

(a) the definitions compatible with the usual ones if $A$ and $B$ are fields,

(b) the equality $$ B(x)\cap A((x))=A(x). $$ holds if $B$ free over $A$.

Define $A((x))$ as the ring of Laurent series with coefficients in $A$.

Say that $f\in A((x))$ is in $A(x)$ if there are $P,Q$ in $A[x]$ satisfying $Qf=P$. Then $A(x)$ is a subring of $A((x))$ containing $A[x]$, and the above arguments show that (b) holds.