# Relationship between two Galois Theorems.

The classical Galois theorem states that

For $$K\subset L$$ a finite Galois extension of fields, any intermediate extension $$K\subset M\subset L$$ satisfies $$\mathrm{Fix}(\mathrm{Hom}_M(L,L))\subset M$$ and any subgroups $$H$$ of $$\mathrm{Hom}_K(L,L)$$ satisfies $$\mathrm{Hom}_{\mathrm{Fix}(H)}(L,L)\subset H$$, where $$\mathrm{Hom}_M$$ means the $$M$$-algebra morphisms.

Another version of the theorem, which in the book Galois Theories, chapter 2, is a generalization according to the authors:

First define the category $$\mathrm{Split}_K(L)$$ to be the $$K$$-algebra $$A$$ with the property that $$A$$ is finite over $$K$$ and the minimal polynomial of any $$a\in A$$ splits to linear factors in $$L$$. As usual let $$K\subset L$$ be a finite Galois extension. The book states that

The contravariant functor $$F: \mathrm{Split}_K(L)\to \mathrm{Hom}_K(L,L)\mathrm{-Sets}$$ defines by $$F(A)=\mathrm{Hom}_K(A,L)$$ induces a categorical equivalence.

But how does this equivalence contains the classical version of Galois theorem?

Let $$G= \mathrm{Hom}_K(L,L)$$. Taking $$H$$ a subgroup of $$G$$ then the quotient $$G/H$$ as $$G\mathrm{-Set}$$ is isomorphic to some $$\mathrm{Hom}_K(A,L)$$ so we have a surjective homomorphism from $$\mathrm{Hom}_K(L,L)$$ to $$\mathrm{Hom}_K(A,L)$$ which by the categorical equivalence corresponds exactly to one mono arrow $$A\to L$$.

But does it mean that I need to show every mono is injective in the category $$\mathrm{Split}_K(L)$$? I have attempted to prove this using varies results in integral extensions but can not obtain anything.

It seems to be trivial but I am lacking the background of categorical language. Any help would be appreciated.

Assume $$a\in A$$ is sent to $$0$$. Let $$\mu_a$$ denote its minimal polynomial: then $$K[x]/(\mu_a)\in Split_K(L)$$.

Indeed, let $$P\in K[x]$$, and let $$Q\in K[x]$$ denote its minimal polynomial ($$P$$ being seen as an element of $$K[x]/(\mu_a)$$). We want to show that over $$L$$, $$Q$$ splits as a product of monomials.

Let $$B$$ be a $$K$$-algebra, $$b\in B$$, and $$\mu_b$$ its minimal polynomial over $$K$$. Then $$\mu_b$$ is the minimal polynomial over $$L$$ of $$b\otimes 1\in B\otimes_K L$$.

Proof : clearly $$\mu_b$$ kills $$b\otimes 1$$. Let $$n$$ denote the degree of $$\mu_b$$. Then $$1,b,...,b^{n-1}$$ are linearly independent over $$K$$, so $$1\otimes 1, b\otimes 1,..., b^{n-1}\otimes 1$$ are linearly independent over $$L$$ (note that $$K$$ is a field here !). It follows that the minimal polynomial of $$b\otimes 1$$ over $$L$$ has degree at least $$n$$, and divides $$\mu_b$$ : it must be $$\mu_b$$.

It follows that over $$L$$, $$Q$$ must be the minimal polynomial of $$1\otimes P\in L\otimes_K K[x]/(\mu_a) \cong L[x]/(\mu_a) = L[x]/\prod_i (x-\alpha_i) \cong \prod_i L$$.

But any element of $$\prod_i L$$ has a minimal polynomial over $$L$$ which splits as a product of linear polynomials, so $$Q$$ splits as such over $$L$$.

The claim about $$K[x]/(\mu_a)$$ follows.

But now, $$\mu_a(0) = 0$$ : indeed, if we let $$f: A\to L$$ denote our morphism, $$0=f(0)= f\mu_a(a) = \mu_a(f(a))= \mu_a(0)$$, so $$\mu_a$$ is divisible by $$x$$ in $$L$$, therefore it is so in $$K$$.

Therefore, the morphism $$K[x]/(\mu_a)\to A, x\mapsto 0$$ is well-defined, just as $$K[x]/(\mu_a)\to A, x\mapsto a$$.

In particular, if $$A\to L$$ is a monomorphism, since the two composites $$K[x]/(\mu_a)\to L$$ agree (they both send $$x$$ to $$0$$), the two morphisms $$K[x]/(\mu_a)\to A$$ must already agree : $$a=0$$, so $$A\to L$$ is injective.

There might be an easier solution to prove that mono $$\implies$$ injective in $$Split_K(L)$$, but I don't immediately see it.