Show that $\overline {\mathbb Q(x)}$ is isomorphic to a subfield of $\mathbb C$.

Here $\mathbb Q(x)$ is the field of rational functions (the field of fractions of the polynomial ring $\mathbb Q[x]$), and $\overline {\mathbb Q(x)}$ is its algebraic closure.

I'm confused by the definition of $\overline {\mathbb Q(x)}$. Is a general element of $\overline {\mathbb Q(x)}$ of the form $p(x)\over q(x)$, where $p(x), q(x)$ are polynomials whose coefficients are algebraic numbers, and $q(x) \neq 0$, i.e. $\overline {\mathbb Q(x)}= \overline {\mathbb Q}(x)$?

I have tried multiple approaches (factorizing $p(x)$ and $q(x)$ into linear factors; using the fact that $\mathbb Q[x]/(p(x)) \cong \mathbb Q(\alpha)$ for irreducible $p(x)$), but none of these are close to prove that there exists an isomorphism.

Thanks for your help


$\mathbb{Q}(x)$ is isomorphic to $\mathbb{Q}(\pi)$, so $\overline{\mathbb{Q}(x)}$ is isomorphic to $\overline{\mathbb{Q}(\pi)}$, which is a subfield of $\mathbb{C}$.

( $\mathbb{Q}(\pi) \subseteq \mathbb{R} \Rightarrow \overline{\mathbb{Q}(\pi)} \subseteq \overline{\mathbb{R}} = \mathbb{C}$ ).

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$\bar{\mathbb{Q}}(x)$ is not isomorphic to $\overline{\mathbb{Q}(x)}$. To see this, note that $\bar{\mathbb{Q}}(x)$ is not algebraically closed. Consider, for example the polynomial $f(y)\in \bar{\mathbb{Q}}(x)[y]$ defined by $f(y)=y^2-x$. The roots of $f(y)$, namely $\pm\sqrt{x}$ do not lie in $\bar{\mathbb{Q}}(x)$, and thus, $\bar{\mathbb{Q}}(x)$ is not algebraically closed.

However, $\overline{\mathbb{Q}(x)}$ is algebraically closed by definition: it is the algebraic closure of $\mathbb{Q}(x)$.

To show that $\overline{\mathbb{Q}(x)}$ is isomorphic to a subfield of $\mathbb{C}$, try to find an embedding $\overline{\mathbb{Q}(x)}\to \mathbb{C}$. Why should $\overline{\mathbb{Q}(x)}$ be inside $\mathbb{C}$?

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  • $\begingroup$ Can you be more specific? $\mathbb Q(x)$ is countable, so $\overline {\mathbb Q(x)}$ should be countable? But that doesn't imply the existence of an embedding $\overline{\mathbb{Q}(x)}\to \mathbb{C}$ $\endgroup$ – ros2868 Nov 22 '13 at 4:42
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    $\begingroup$ It is a fact that for every uncountable cardinal, algebraically closed fields are unique up to isomorphism. So if you embed $\overline{\mathbb{Q}(x)}$ in any uncountable algebraically closed field (say of same cardinality as $\mathbb{C}$), it must be isomorphic to $\mathbb{C}$. $\endgroup$ – userCaltech Nov 22 '13 at 6:14

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