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.


1 Answer 1


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.

Here's a helpful lemma:

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.


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