why are subextensions of Galois extensions also Galois?

An algebraic extension of fields $L|K$ is defined to be a Galois extension iff the set of elements of $L$ invariant under the action of $Aut_K L$ is $K$.

Apparently in the sequence of field extensions $L|M|K$ if $L|K$ is Galois then $L|M$ is. This is asserted to be an obvious deduction in section 3.2.1 of Robalo Delgados thesis on Galois Categories referenced in nLab-Grothendiecks Galois Theory, I don't see the obviousness...can someone clarify.

What interests me in this characterisation is that it shows one direction of the Galois correspondance, (the other being shown by Artins Lemma) without going through characterising Galois extensions as algebraic, seperable & normal extensions. In fact Delgado shows that this characterisation follows from his definition.

Edit: I've also asked this question on math.overflow as I thought most people misunderstood the question. I'd welcome attempts to formulate it in clearer terms.

Solution 1:

Go to a text on Galois Theory and look in the index for the Fundamental Theorem of Galois Theory. It says that for a Galois extension $L|K$ there is a one-to-one correspondence between {subfields between L and K} and {subgroups of $Aut_K L$}, wherein an intermediate field $M$ corresponds to the subgroup $G$ of $Aut_K L$ fixing $M$. The inverse of this correspondence takes a subgroup $G$ and sends it to the subfield of $K$ fixed by all elements of $G$. Since a Galois extension is defined by $K$ being the fixed field of $Aut_K L$, I can understand why someone would call an obvious corollary of a Fundamental Theorem "obvious", although I try to avoid using that word and would prefer "by the FTGT".

Solution 2:

Let $L \subset K$ be a Galois extension of fields. Let $ L \subset M \subset K$ be an intermediary field. Then we know a priori that $L \subset M$ is separable and $M \subset K$ is Galois. The fact both extensions is separable is straightforward. To see that $M \subset K$ is normal let $p(t) \in M[t]$ be an irreducible polynomial taking a root $\alpha \in K$. Let $q(t)$ be the minimum polynomial of $\alpha$ over $L$ since $L \subset K$ is a normal extension we know that $q(t)$ splits over $K[t]$. We also know that $p(t)$ divides $q(t)$ in $M[t]$. In particular this gives that $p(t)$ splits in $K[t]$ since it still must divide $q(t)$ in $K[t]$. Now when $L \subset M$ is a Galois extension is a more subtle question. The fundamental theorem of Galois theory is required to answer this question. It tells that $M$ is normal precisely when the subgroup of $Aut_L(K)$ fixing $M$ is normal in $Aut_L(K)$.

Solution 3:

You do not have to rephrase the question. You only have to look at the question the right way and think about the proof of the FTGT. Let $L\supseteq K$ be fields. Then $L$ is a normal separable extension of $K$ iff every irreducible polynomial of degree $d$ in $K[X]$ which has a root in $L$ has $d$ distinct roots in $L$ iff the Galois correspondence holds.

EDIT starts here.

The original answer here gives the missing equivalence. The first iff $\implies$ follows since the polynomial factors as a product of $d$ linear factors in the algebraic closure of $L$, and none of them are multiple by separability and all are in $L$ if one of them is by normality. The reverse implication $\Longleftarrow$ follows from the Lagrange Theorem (roughly 50 years before Galois) that the coefficients of an irreducible polynomial $p$ over any field $K$ are symmetric functions of the roots, and those symmetric functions will lie in the field $K\subseteq L$ iff the product of all of the linear factors of $p$ in some algebraic closure lie in $K$ are used to multiply out to $p$ so if one root of $p$ lies in $L$, and there are $d$ distinct roots, all of the roots must lie in $L$ (normal) and there are no multiple roots (separable). Now if $K\subseteq M\subseteq L$ and $q$ is the minimal polynomial in $M[X]$ of an element $\alpha\in L$, $q$ divides the minimal polynomial of $\alpha$ in $K[X]$ so if the polynomial over $K$ has distinct roots so does $q$ over $M$, and the reverse implication of this last iff is just a restriction to $K$.

My recollection from many years ago is that when I learned the proof of the FTGT as stated in terms of normal + separable, the proof immediately went to the equivalent distinct roots property. Is there any other part of the original question I am missing?