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I have the feeling that this is a rather silly question, but I couldn't figure it out myself: How can I see that $\bar{\mathbb{Q}}(t)$ is algebraically closed in $\mathbb{C}(t)$?

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    @Michalis, originally the title of the question was asking for algebraic closure of $\mathbb{Q}(t)$ in $\mathbb{C}$, which was mildly confusing. I did miss the part about the closure being relative. Sorry about that.2012-04-19

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Edit: I think that should be OK: Suppose $P/Q\in \mathbb C(t)$ is algebraic over $\overline{\mathbb{Q}}(t)$. By the fundamental theorem of algebra

$ P/Q=c\frac{\prod(t-\lambda_i)}{\prod(t-\delta_j)}, $ with $c,\lambda_i,\delta_i\in\mathbb C$ and $\lambda_i\neq \delta_j$. Suppose that $P/Q\notin\overline{\mathbb{Q}}(t)$.

If all the $\lambda_i,\delta_j$ where in $\overline{Q}$ then $\frac{1}{c}P/Q\in\overline{\mathbb{Q}}(t)$ and hence $c\in\mathbb{C}$ algebraic over $\overline{\mathbb{Q}}(t)\Rightarrow c\in \overline{\mathbb{Q}}\Rightarrow P/Q\in\overline{\mathbb{Q}}(t)$. Contradiction.

Now suppose one of the $\lambda_i$ or $\delta_j$ is not in $\overline{\mathbb Q}$. By inverting $P/Q$ if necessary we can assume that $\lambda_1\notin\mathbb C$. Now since $P/Q$ is algebraic over $\overline{\mathbb{Q}}(t)$ we have $a_i\in\overline{\mathbb{Q}}(t)$ with \begin{eqnarray} a_n(t)(P(t)/Q(t))^n+\ldots+a_0(t)=0\\ a_n(t)P(t)^n+\ldots+a_0(t)Q(t)^n=0\\ (t-\lambda_1)Z(t)=-A(t)Q(t)^n, \end{eqnarray} where $Z(t),A(t)\in \mathbb C[t]$ are polynomials. The last line was obtained by subtracting $a_0(t)Q(t)^n$ and multiplying with the denominators of the $a_i(t)$. Since $\lambda_1\notin \overline{\mathbb{Q}}$ the right hand side is not divisible by $t-\lambda_1$. A contradiction.

Remark: This is actually wrong, but is correctable (see comments): You can use that the algebraic closure of $\overline{\mathbb{Q}}(t)$ is the field $P$ of Puiseux series (Not true!) with coefficients in $\overline{\mathbb Q}$, which embeds into the algebraic closure of $\mathbb C(t)$, the Puiseux series with coefficients in $\mathbb C$. Now the algebraic closure of $\overline{\mathbb{Q}}(t)$ in $\mathbb C(t)$ is $P\cap \mathbb C(t)=\overline{\mathbb{Q}}(t)$.

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    @Hurkyl: You're right, but I think the additional work that has to be done makes the remark essentially useless :) the original problem is easier.2012-04-20