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In my Galois Theory notes we have the following theorem:

Let $L/K$ be a finite extension. TFAE:

(ii) For any extension $M/L$ and any $K$-homomorphism $\sigma:L \rightarrow M$, we have $\sigma(L)=L$.

(iii) $L/K$ is normal.

The proof of (ii)$\Rightarrow $(iii) reads

Suppose (ii) holds. Given any $\alpha \in L$, let $M/L$ be such that $f_\alpha^K$ splits in $M$, i.e. $f_\alpha^K=\prod (X-\alpha_i)$ for some $\alpha_i \in M$. We need to show $\alpha_i \in L$.

Take a $K$-homomorphism $\hat{\sigma}: K(\alpha) \rightarrow M$, $\alpha \mapsto \alpha_i$. Extend to a $K$-homomorphism $\sigma:L \rightarrow M$.

Why can we make this extension of $\hat{\sigma}$?

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You can use the following, which does not invoke the existence of algebraic closures. I apologize that my choice of letters is going to clash with yours...

Theorem. Let $F$ and $L$ be fields, and let $\sigma\colon F\to L$ be a field isomorphism. Let $S$ be a set of nonconstant polynomials in $F[x]$, and let $\sigma S$ be the corresponding set of polynomials in $L[x]$. If $K$ is a splitting field of $S$ over $F$, and $M$ is a splitting field of $\sigma S$ over $L$, then $\sigma$ extends to an isomorphism $\tau\colon K\to M$.

For your purposes, it suffices to establish the theorem in the case where $S$ is a single polynomial $f(x)$.

We proceed by induction on $[K:F]$. If $[K:F]=1$, then $K=F$, so $f$ splits over $F$; hence $\sigma f$ splits over $L$, so $M=L$, and thus $\sigma\colon K\to M$ is an isomorphism already.

Assume the result is true whenever the degree of the splitting field of $f$ over the ground field is strictly less than $n$, and that $[K:F]=n\gt 1$. So $f$ has some irreducible factor $g$ of degree greater than $1$. Let $\alpha\in K$ be a root of $g(x)$. Then $1\lt \deg(g)\leq \deg(f)$, and $\sigma g$ is an irreducible factor of $\sigma f$. Let $\beta$ be a root of $\sigma g$ in $M$.

Consider $F(\alpha) \cong F[x]/(g(x)) \cong L[x]/(\sigma g(x)) \cong L(\beta)$; the isomorphism induced by $\sigma$ restricts to $\sigma$ on $F$, and maps $\alpha$ to $x+(g(x))$ to $x+(\sigma g(x))$ to $\beta$. Thus, we get an isomorphism $\rho\colon F(\alpha)\to L(\beta)$ with $\rho|_F = \sigma$ and $\rho(\alpha)=\rho(\beta)$.

Now, $K$ is a splitting field of $f(x)$ over $F(\alpha)$ (since $\alpha$ is one of the roots of $f(x))$, and $M$ is a splitting field of $\sigma f(x)$ over $L(\beta)$. Since $$[K:F(\alpha)] = [K:F]/[F(\alpha):F] = \frac{n}{\deg(g)}\lt n,$$ so by the induction hypothesis, $\rho$ extends to an isomorphism $\tau\colon K\to M$ with $\tau|_{F(\alpha)} = \rho$. Since $\rho|_F = \sigma$, it follows that $\tau|_{F}=\sigma$, which proves the inductive step.

(For the case of arbitrary $S$, one uses Zorn's Lemma over all triples $(E,N,\rho)$ with $F\subseteq E\subseteq K$, $L\subseteq N\subseteq M$, $\rho\colon E\to N$ an isomorphism that extends $\sigma$; and then proves that a maximal element in fact has $E=K$, $N=M$.)

Now, in your case, we apply the theorem as follows: we let $M$ be the splitting field of $f_{\alpha}^K$ over $L$, which means it is a splitting field of $f_{\alpha}^K$ over $K(\alpha)$. We let $K(\alpha)$ play the role of both $F$ and $L$ in the theorem, and let $\hat{\sigma}$ play the role of $\sigma$ in the theorem.. We let your $M$ play the roles of both $K$ and $M$ in the theorem. Then $\hat{\sigma}$ extends to an isomorphism $M\to M$, which in turn restricts to a homomorphism $\sigma\colon L\to M$. This gives you the $\sigma$ you want.

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You use the following fact: if $L/K$ is algebraic then any embedding of $K$ into an algebraically closed field, $M$, can be extended to an embedding of $L$ in $M$. In your case, $M$ can be chosen to be algebraically closed (for example, take $M$ to be the algebraic closure of $L$). Since $L/K(\alpha)$ is algebraic you can extend $\hat{\sigma}$ to $\sigma$.

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In my notes we do this proof before mentioning algebraic closures, so I don't think this is the reason the lecturer uses. But thanks very much anyway, I'll look into that. –  MA390 Apr 29 '12 at 12:00
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