Is $K(X)$ never algebraically closed?

Question

I saw an exercise like this:

Let $K$ be a algebraically closed field. Show that $K(X)$ is not algebraically closed.

I was trying to prove that $T^2-X$ has not root in $K(X)$. How do I show it? Does it really use the hypothesis about $K$?

Thank you!

@Xoff: thank you, but I wrote "I was trying to prove that T^2-X has not root in K(X)" — Alphonse, Aug 22 '16 at 13:55
So if $T=P(X)/Q(X)$ is a root of $T^2-X$, then $P(X)^2=XQ(X)^2$. We indeed have an even degree that is equal to an odd degree ($Q ≠ 0$). So this does not depend on $K$ being algebraically closed? — Alphonse, Aug 22 '16 at 13:57
As @Bernard said, and no, you don't need to know anything about $K$. — Xoff, Aug 22 '16 at 13:57
Related on MO: http://mathoverflow.net/questions/206362/algebraic-closure-of-the-field-of-rational-functions — Alphonse, Aug 23 '16 at 13:08

score 1 · Accepted Answer · answered Mar 03 '18 at 11:00

As in the comments:

If $T=P(X)/Q(X)$ is a root of $T^2-X$, then $P(X)^2=XQ(X)^2$. We indeed have an even degree that is equal to an odd degree ($Q ≠ 0$). This is a contradiction. Therefore, the field $K(X)$ is not algebraically closed, for any field $K$.

Is $K(X)$ never algebraically closed?

1 Answers1

Linked

Related