$K$-homomorphisms from an étale $K$-algebra to a field

I am somewhat confused by the proof of Theorem 1.5.4 (p.22) (Grothendieck's version of the main theorem of Galois Theory) in Szamuely's Galois Groups and Fundamental Groups. The theorem establishes an (anti)equivalence of categories between the category of finite étale $K$-algebras and the category of finite sets equipped with a continuous left $\operatorname{Gal}(K)$-action, where $\operatorname{Gal}(K) = \operatorname{Gal}(K_s : K)$ is the absolute Galois group of $K$ and $K_s$ is a separable closure of $K$.

A $K$-algebra $A$ is a finite étale $K$-algebra if $A\cong\prod_{i=1}^n L_i$ where each $L_i$ is a seperable extension of $K$. The main point in the proof is that there is a bijection between $\operatorname{Hom}_K (A,K_s)$ and $\coprod_{i=1}^n \operatorname{Hom}_K (L_i, K_s)$, where $\operatorname{Hom}_K$ denotes the set of $K$-homomorphisms.

To do this, Szamuely claims that every $K$-homomorphism $\phi: A\rightarrow K_s$ induces an injection $L_i\rightarrow K_s$ for precisely one factor $L_i$. I find his argument a bit confusing, but this is how I see things; there are homomorphisms $A\xleftarrow{\alpha} K \xrightarrow{\beta} K_s$ since these are both $K$-algebras, and since $A\cong \prod_{i=1}^n L_i$ where each $L_i : K$ is a separable extension we obtain (by composition with the projections) homomorphisms $\alpha_i : K\rightarrow L_i$ for each $i$. Now $\alpha (K)$ is a subfield of $A$, and we have $\phi\circ\alpha = \beta$ since $\phi$ is a $K$-homomorphism, so $\phi(\alpha (K)) = \beta (K)$. Therefore $\phi$ induces an injection on some subfield of $A$ containing $\alpha (K)$. The problem is, it seems like Szamuely regards each $L_i$ as a subfield of $A$ and can then argue that precisely one $L_i$ contains $\alpha (K)$ to cook up an injection $L_i \rightarrow K_s$, when in fact they aren't even subrings, since they don't contain the identity of $A$ (i.e. he seems to be treating $A$ like a coproduct rather than a product).

So does anyone have a rigorous and more detailed explanation of this argument than that which features in the book? Many thanks.

2 Answers

This is geometrically obvious:

Translating from algebra to affine $K$-schemes you have to determine the $\operatorname {Spec}(K)$-morphisms $$\operatorname {Spec}(K_s)\to \operatorname {Spec}(\prod_{i=1}^n L_i)=\coprod _{i=1}^n\operatorname {Spec}(L_i)$$ Since $\operatorname {Spec}(K_s)$ is a one-point scheme and the right hand scheme is a discrete $n$-point scheme, you get $$\operatorname {Hom}_{\operatorname {Spec}(K)}\bigg ( \operatorname {Spec}(K_s), \coprod _{i=1}^n\operatorname {Spec}(L_i) \bigg )=\coprod _{i=1}^n\operatorname {Hom}_{\operatorname {Spec}(K)}\bigg (\operatorname {Spec}(K_s), \operatorname {Spec}(L_i) \bigg )$$ (beware that this last equality is clearly not obtained by the categorical definition of $\coprod$ but by the scheme-theoretic situation!)
Translating back to algebra yields what you want, namely
$$\operatorname {Hom}_K ( \prod _{i=1}^n L_i,K_s )=\coprod _{i=1}^n\operatorname {Hom}_K(L_i,K_s )$$

• Great answer, thank you! This really is the first time something has been made clearer to me by using schemes, so I'm very happy you explained it like this. Commented Dec 9, 2015 at 21:54
• Dear Alex, you are welcome. And I'm very happy about your comment: welcome to the merry Schemes Fan Club ! Commented Dec 9, 2015 at 22:01

As a complement to Georges's answer - perhaps the following is useful?

In algebraic terms, the underlying point we need is:

If $$A = A_1 \times \cdots \times A_n$$ is a finite product of commutative unitary rings, then every ideal $I\subset A$ is of the form $I = I_1\times \cdots I_n$ - where the $I_i$ need not be necessarily proper. [Proof:
$$1 = (1,0,\cdots,0 ) + \cdots + (0,\cdots, 0,1),$$ so we can express any $I$ as a sum of ideals $p_i(I)$, where $p_i$ are the projection maps; $p_i(I)$ are ideals, because the $p_i$ are onto. On the other hand, going the other way, the direct product $I_1\times \cdots \times I_n$ is an ideal of $A$.]

Then: Suppose $F$ is a field over $K$. Every $K$-algebra morphism $L_i\to F$, gives rise to a $K$-algebra morphism $A \to F$. Conversely, a $K$-algebra morphism $A \to F$ must kill a maximal ideal $\mathcal m_i$ of $A$, and the set of the latter are in bijection with the set of fields $L_i$: $A/\mathcal m_i = L_i$.

• Another nice answer; it's very helpful to have two viewpoints on this statement. Thank you! Commented Dec 10, 2015 at 14:14