# Rational functions over $\mathbb{C}$ is algebraically closed

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!

• It doesn’t have an element $\alpha$ such that $\alpha^2=x$ – J. W. Tanner Aug 22 at 11:51
• 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. – Jyrki Lahtonen Aug 22 at 11:54
• Ah okay, of course - that seems so obvious now! Thanks for that. – jamesmbcn Aug 22 at 11:55
• 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. – Jyrki Lahtonen Aug 22 at 11:57
• For sure - thanks for the help. I've posted my attempt at making the solution rigorous, any feedback would be appreciated! – jamesmbcn Aug 22 at 12:13

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.
• 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. – Milo Brandt Aug 22 at 14:15