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I am trying to do the first exercises of the Mazza-Voevodsky-Weibel book Lecture Notes on Motivic Cohomology, which are some elementary results about correspondences between schemes.

Recall that if $X$ and $Y$ are smooth schemes over $k$, a correspondence $Z=\sum_{i}n_i Z_i \in \mathrm{Cor}_k(X, Y)$ is a formal sum of irreducible closed subsets of $X \times Y$, with each projection morphism $\mathrm{pr}_1 : Z_i \rightarrow X$ finite and surjective.

Here I am trying to prove the following (Exercise 1.11, page 21) :

Let $k$ be a perfect field, $L/k$ a finite Galois extension of Galois group $G$, and $S=\mathrm{spec}(k)$, $T=\mathrm{spec}(L)$. Then there exists a canonical isomorphism $\mathrm{Cor}k(T, T) \approx > \mathbb{Z}[G]$ and under this identification, the morphism $T > \rightarrow S \rightarrow T$ is the element $\sum_{g \in G}{g} \in > \mathbb{Z}[G]$. Moreover, for all smooth scheme $Y$ over $k$, $\mathrm{Cor}_k(S, Y) \approx > \mathrm{Cor}_k(T, Y)^{G}$.

Here is my attempt :

An element $g \in G$, i.e. $L \xrightarrow{g} L$, induces a scheme morphism $T \xrightarrow{\tilde{g}} T$, whose graph $\Gamma_{\tilde{g}} \subset T \times_k T$ is a correspondence in $\mathrm{Cor}(T, T)$.
But now if I take $Z=\sum_{i}n_i Z_i \in \mathrm{Cor}_k(T, T)$, I'd like to assign to each $Z_i$ a group element $g_i \in G$, and keep the same integer coefficients. Since $Z_i$ is a closed subscheme, I write it as $Z_i=V(I_i) \subset \mathrm{spec}(L \otimes _k L)$, and $V(I_i) \approx \mathrm{spec}(L \otimes _k L / I_i)$, so I get a morphism $k \rightarrow L \otimes _k L / I_i$. Then I don't know how to get an isomorphism $L \rightarrow L$.
My understanding is that the elementary correspondences $Z_i$ are supposed to represent graphs of morphisms $T \rightarrow T$, but they are not all of this form. Also am I not supposed to use the fundamental theorem of Galois Theory somewhere?

For the second part, to each $Z \subset S \times _k Y \approx Y$ we can assign by base change $Z_L \in T \times _k Y$, and the map $Z \rightarrow Z_L$ is injective because $L$ is flat as a $k$-module ($L$ is a vector space on $k$). I don't know how to prove surjectivity.

I am still at a very basic level but I'm trying my best to understand all this. Thanks in advance for answers!

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Could the person who voted this down please give a reason why? –  S. Carnahan Mar 16 '13 at 5:41

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