This comes from a question about the fundamental theorem of algebra. Is the algebraic closure of $\mathbb{C}$ implied by the fact that $\mathbb{C}$ is the algebraic closure of $\mathbb{R}$?

More generally, if a field $K$ is the algebraic closure of a field $F$, is $K$ algebraically closed? Why or why not? Wikipedia says that $K$ must be closed, but is that a result of definition, or is there a more complex reason?

  • $\begingroup$ How are you defining algebraic closure (if not with wikipedias definition)? $\endgroup$ – Lorenzo Najt Apr 3 '16 at 4:02
  • $\begingroup$ I am using wiki's definition, but I am wondering if the closure of K is a necessity through definition. Could there exist an algebraic closure that itself is not algebraically closed? If not, is it because all algebraic closures are defined to themselves be closed? $\endgroup$ – Demetri Pananos Apr 3 '16 at 4:03

Yes, $K$ is also algebraically closed. Consider a polynomial $f(x) \in K[x]$, then let $\alpha$ be a root of $f(x)$,then $\alpha$ generates an algebraic extension over $K$, call it $K(\alpha)$, then it follows that $K(\alpha)$ is algebraic over $F$, because $K$ is algebraic over $F$. Thus $\alpha$ is algebraic over $F$, but then $\alpha \in K$, because $K$ contains all the roots of $F$. It follows that $K$ is algebraically closed.

And yes, a corollary is that the complex numbers are algebraically closed.

  • $\begingroup$ Are "algebraic closure of F" and "algebraic closure of F in E" different? $\endgroup$ – Sophia Mar 31 at 2:08
  • $\begingroup$ And I can't understand why alpha is algebraic over F... $\endgroup$ – Sophia Mar 31 at 2:14
  • $\begingroup$ @Sophia An algebraic closure of $F$ is an algebraic extension of $F$ that is algebraically closed. Once we confirm that it is a subextension/extension of some given extension $E$ then it is unique (So the difference between the two notions in your first comment is that one is unique up to isomorphism and the other is jjust unique, but if that doesn't matter than they're the same, pretty much). Also, $\alpha$ is algebraic over $F$ because $\alpha$ is algebraic over $K$ which is algebraic over $F$, and therefore $\alpha$ is algebraic over $F$. $\endgroup$ – астон вілла олоф мэллбэрг Apr 1 at 4:56
  • $\begingroup$ I got it. Thanks a lot. In my textbook, there is an example that "algebraic closure of $Q$ in $Q(\sqrt{2})$ is $Q$. But $Q$ is not algebraically closed since $x^2+1$ in $Q[x]$ has no zero in $Q$. So algebraic closure of $F$ in $E$ need not be algebraically closed." The difference between this statement and the theorem you proved comes from my first comment, right? $\endgroup$ – Sophia Apr 2 at 4:25
  • $\begingroup$ Yes, that is right, Sophia. $\endgroup$ – астон вілла олоф мэллбэрг Apr 2 at 6:02

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