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!
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.