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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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I'm assuming that $t$ is transcendental over the complex numbers (otherwise, what's the point of adjoining it to $\mathbb{C}$). Why do you think it would be algebraically closed? Where's $\sqrt t$ for one? – Jyrki Lahtonen Apr 19 '12 at 10:48
@JyrkiLahtonen: He said algebraically closed in $\mathbb C(t)$. That means that every element of $\mathbb C(t)$ that is algebraic over $\overline{\mathbb Q}(t)$ is in $\overline{\mathbb Q}(t)$. – Michalis Apr 19 '12 at 10:53
Yes, $t$ is transcendental over $\mathbb{C}$. I meant that $\bar{\mathbb{Q}}(t)$ is the algebraically closed in $\mathbb{C}(t)$. So that should explain where $\sqrt{t}$ is. – 1904 Apr 19 '12 at 10:53
@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. – Jyrki Lahtonen Apr 19 '12 at 12:24
up vote 0 down vote accepted

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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This makes sense to me although I have to agree, it's not a very direct argument. – 1904 Apr 19 '12 at 11:25
@1904: OK I missed my first train, but I hope I added the correct answer :) – Michalis Apr 19 '12 at 11:38
Hmm. Isn't the field of Puiseux series an algebraic closure of the field of Laurent series? I doubt that, say, $e^t$ would be algebraic over $\mathbb{C}(t)$. Therefore the algebraic closure of $\overline{\mathbb{Q}}(t)$ in $P$ is not the field of Puiseux series with coefficients in $\overline{\mathbb{Q}}$. – Jyrki Lahtonen Apr 19 '12 at 12:33
@Jyrki: I was about to say the same thing. I'm pretty sure we can still make use of the idea in the remark, though, by embedding the relevant algebraic closures into the relevant fields of Puiseux series. – Hurkyl Apr 19 '12 at 12:36
You are right. So could I say since the algebraic closure of $\overline{\mathbb{Q}}(t)$ has to be a subset of the algebraic closure of $\overline{\mathbb{Q}}((t))$, i.e. the Pusieux series with coefficients in $\overline{\mathbb{Q}}$, the algebraic closure of $\overline{\mathbb{Q}}(t)$ in $\mathbb{C}(t)$ is a subset of $P \bigcap \mathbb{C} (t)$ and therefore $\overline{\mathbb{Q}}(t)$? – 1904 Apr 19 '12 at 19:27

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