0

Question

Let $k\subset E$ be an algebraic extension of fields of characteristic zero. Assume that every non-constant polynomial $f(x)\in k[x]$ has a root in $E$. Show that $E$ is algebraically closed.

Answer

I was following one of the answers given in this link, namely the last one. The answers follows like:

"First take $f\in k[x]$. Then as characteristic is zero $Spl(f)=k(\alpha)$. Then $m_{\alpha,k}(x)$, minimal polynomial of $\alpha$ in $k[x]$, has a root in $E$. Thus, $E$ contains a copy of the splitting field of $f$."

I didn't understand how the last sentence holds. Thanks in advance...

Also related to this question.

confused
  • 489
  • 2
  • 8
  • 1
    The minimal polynomial of $\alpha$ has a root, say $\beta$, in $E$. But then $k(\alpha) $ is isomorphic to $k(\beta) \subseteq E$ and $k(\beta) $ is a splitting field of $f$ as well. – Paramanand Singh Jul 18 '23 at 02:13
  • Ohh that was the fact I was missing. Thanks a lot! – confused Jul 19 '23 at 09:13
  • 1
    The result is called Gilmer's theorem. You should have a look at Keith Conrad's notes here: https://kconrad.math.uconn.edu/blurbs/galoistheory/algclosure.pdf – Paramanand Singh Jul 19 '23 at 10:24

0 Answers0