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?
Best Answer
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.