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Studying for a course in fields, and came across the question: is the field $\mathbb{C}(x)$ of rational functions over $\mathbb{C}$ algebraically closed?

At the moment I think it is, as I can't seem to find a simple counter-example nor a good reason as to why it shouldn't be, but it seems as though I may be missing a trick here, and my initial thought was to be suspicious of how simple it seems. Any hints or help greatly appreciated!

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    $\begingroup$ It doesn’t have an element $\alpha$ such that $\alpha^2=x$ $\endgroup$ – J. W. Tanner Aug 22 at 11:51
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    $\begingroup$ To use J.W. Tanner's hint assume a solution $\alpha=p(x)/q(x)$ exists. Expand, clear the denominators, and look at the degrees of the polynomials on both sides of the equation. $\endgroup$ – Jyrki Lahtonen Aug 22 at 11:54
  • $\begingroup$ Ah okay, of course - that seems so obvious now! Thanks for that. $\endgroup$ – jamesmbcn Aug 22 at 11:55
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    $\begingroup$ Feel free to write it up as an answer! That way you get more feedback on the details. We do need to check whether this question has been asked earlier. $\endgroup$ – Jyrki Lahtonen Aug 22 at 11:57
  • $\begingroup$ For sure - thanks for the help. I've posted my attempt at making the solution rigorous, any feedback would be appreciated! $\endgroup$ – jamesmbcn Aug 22 at 12:13
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Thanks to both J.W. Tanner and Jyrki Lahtonen for their help with this question - here's my attempt at making the solution rigorous.

Suppose the field $\mathbb{C}(x)$ of rational functions over $\mathbb{C}$ is algebraically closed. Take the polynomial $P(y) = y^2 - x$; then since $\mathbb{C}(x)$ is algebraically closed, $P$ must have a root in $\mathbb{C}(x)$. But $P(y)=0$ if and only if $y^2 = x$, for some $y \in \mathbb{C}(x)$. Suppose that $y = p(x)/q(x)$. Then we have that: $$ \frac{p^2(x)}{q^2(x)} = x \implies p^2(x) = xq^2(x) $$ Hence considering degrees, we see that we must have $2\text{deg}(p) = 1+2\text{deg}(q)$, and so $2(\text{deg}(p) - \text{deg}(q))=1 \implies \text{deg}(p) - \text{deg}(q) = 1/2$, a contradiction as $p,q$ are polynomials. Thus no such $y$ exists, i.e. $P$ has no root in $\mathbb{C}(x)$ and hence $\mathbb{C}(x)$ is not algebraically closed.

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    $\begingroup$ This a correct answer, but it could be said a lot more concisely - the point is that $x$ has no square roots because of degrees. The paragraph preceding the equation introduces a contradiction you don't need - better to start by declaring that you'll show $y^2-x$ lacks a root and then using contradiction on that more specific statement. Then, once you reach $2\deg(p)=1+2\deg(q)$, you might as well stop - it's clear enough that this can't happen and the further algebra distracts from the point. Having shown a polynomial with no root, you can conclude $\mathbb C(x)$ is not algebraically closed. $\endgroup$ – Milo Brandt Aug 22 at 14:15

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