quick question on Galois theory

Question

I am reading a book on Galois theory and, as per usual (why is that?), all sorts of unproven properties start to magically appear in the proofs of the couple of theorems that really matter. The author tries to prove that, when $L:K$ is a finite separable extension of degree $n$, then there are precisely $n$ distinct $K-$ monomorphisms of $L$ into a normal closure $N$ of $L:K$. In the proof, the following is claimed: if $z_1 \in L\setminus K$, then $N$ is also a normal closure of the extension $L:K(z_1)$. So, why is that? Seems far from trivial to me (the exact statement is needed in order to apply an induction hypothesis). Many thanks in advance for any clues provided.

Answer 1

it is not clear because it is false!for example consider L=K(z_1) and $L/K$ is not normal.$Q\subset Q(\sqrt[4]{2})$

Answer 2

One way (the primary way?) to approach this kind of question is to carefully write out the definitions and then attempt to verify the statement.

So:

A normal closure of $L/K$ is a normal extension $N$ of $K$ containing $L$, and which is minimal with respect to this property, i.e. any proper intermediate extension between $N$ and $L$ is not normal over $K$.

Now $L$ is also an extension of $K(z_1)$, as is $N$, and since $N$ is normal over $K$, it is also normal over $K(z_1)$. So to check if $N$ is a normal closure of $L$ over $K(z_1)$, we have to consider the following problem: if $N \supsetneq M \supseteq L$; is it possible that $M$ is normal over $K(z_1)$?

Now we see that the answer could be yes, because it's "easier" for $M$ to be normal over $K(z_1)$ then it is for $M$ to be normal over $K$. E.g. what if $M = L = K(z_1)$?

This is what is described in ali's answer, and could certainly happen. As he notes, if $L/K$ is not normal, then $N$ will be a proper extension of $L$. But $L/L$ is a normal extension, so $N$ is not the normal closure of $L$ over $L$. And since $L/K$ was assumed separable, we can write $L = K(z_1)$; ali gives the explicit example of $L = \mathbb Q(2^{1/4})$ over $K =\mathbb Q$.

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Note though that if $N$ contains $L$ and is normal over $K$, then $N$ is normal over $K(z_1)$. This suggests a way to modify what is to be proved so that the induction procedure of replacing $K$ by $K(z_1)$ can be correctly applied.

Namely, rather than just prove the theorem for $N$ a normal closure of $L$ over $K$, one should prove the theorem for any $N$ containing $L$ which is normal over $K$. This has the advantage of giving a statement that is more flexible: now it is okay to replace $K$ by $K(z_1)$ in the inductive argument.

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It is in fact an easy exercise to check that the theorem holds in the form "$N$ is a normal closure of $L$" if and only if it holds in the form "$N$ is any normal extension of $K$ containing $L$". (Use the fact that (a) normal closures are normal extensions, and (b) any normal extension containing $L$ does contain a normal closure of $L$.)

So my suggested reformulation isn't very drastic from the point of view of the statement to be proved. But, it does give a version of the statement which is amenable to proof by induction.

This is a common phenomenon when setting up non-trivial inductive arguments: often some care has to be taken to phrase the statement in a form that is amenable to the anticipated framework of the induction. In this particular case, the author doesn't seem to have taken that care.