# Why $E$ is the algebraic closure of $K$?

Let $E/K$ be a separable, algebraic extension such that every noncostant polynomial in $K[x]$ has a root in $E$, then $E$ is an algebraic closure of $K$. Could you help me to solve this exercise? (there is this hint: use the primitive element theorem).

EDIT: well it's enough to prove $E$ is algebraically closed. So take $f(x)\in E[x]$ I want to prove that it has a root in $E$. One of the thing that I was trying is to prove that the minimal polynomial of $\alpha$ over $K$ divides $f(x)$, but I couldn't, I think it's not true.

• Where, exactly, is your problem? What have you tried so far? What didn't work out? – Nils Matthes May 9 '12 at 15:04
• @Berry: You must use separability. I don't see what the primitive element theorem has to do with the question. See also this answer. – Zhen Lin May 9 '12 at 15:11
• Separability is not essential for the truth of the result: if $E/K$ is an algebraic extension and every nonconstant in $K[x]$ has a root in $E$ then $E$ is an algebraic closure of $K$. But it's trickier to prove this without assuming $E/K$ is separable. – KCd May 9 '12 at 17:52
• @KCd: Could you please sketch a proof of the statement you mentioned or give a reference? It sounds interesting that we don't have to assume that $E/K$ be separable. – Nils Matthes May 10 '12 at 11:16
• @Nils Why don't you ask it as a new question? – Makoto Kato May 10 '12 at 23:10

It suffices to prove that every irreducible polynomial in $K[X]$ splits in E(why?). Let $f(X)$ be an irreducible polynomial in $K[X]$. By the assumption, $f(X)$ has a root in $E$. Since $E$ is separable over $K$, $f(X)$ has no multiple root in an algebraic closure of $E$. Let $\alpha_1, ..., \alpha_n$ be all the roots of $f(X)$ in an algebraic closure of $E$. Put L = $K(\alpha_1, ..., \alpha_n)$. Since $L/K$ is separable, by the primitive element theorem, there exists an element $\omega$ in $L$ such that $L = K(\omega)$. Let $g(X)$ be the minimal polynomial of $\omega$ in $K[X]$. By the assumption, $g(X)$ has a root $\lambda$ in E. Since $L/K$ is a Galois extension, $L = K(\omega) = K(\lambda)$. Hence $L ⊂ E$ as desired.
• I think I'm missing something. Why is it enough to prove that every irreducible polynomial in $K[X]$ splits in $E$? – Nils Matthes May 10 '12 at 11:03
• @Nils Let $f(X)$ be a nonconstant polynomial in $E[X]$. Let $F$ be a splitting field of $f(X)$ over $E$. Let $\alpha$ be a root of $f(X)$ in $F$. It suffices to prove that $\alpha$ is contained in $E$. Since $E$ is algebraic over $K$, $\alpha$ is algebraic over $K$. Let $g(X)$ be the minimal polynomial of $\alpha$ in $K[X]$. By the assumption, $g(X)$ splits in $E$. Hence $\alpha$ is contained in $E$ as desired. – Makoto Kato May 10 '12 at 12:11