TITLE: Is $\bar{\mathbb{Q}}(x)\cap \mathbb{Q}((x))=\mathbb{Q}(x)$? [unsolved (even though we earlier thought it was)]
QUESTION [15 upvotes]: 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 $\bar{\mathbb{Q}}(x)\cap \mathbb{Q}((x))=\mathbb{Q}(x)$?

REPLY [4 votes]: 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.