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I need some help to understanding why Galois extensions of commutative rings are faithful. The definitions I'm using for Galois extensions is the one below.

Let $R \rightarrow T$ be a ring homomorphism and $G$ a group of $R$-algebra homomorphims of $S$. Consider the inclusion of $R$ into the subring of $T$ fixed by $G$ $$i : R \rightarrow T^G$$ and the map $$h : T \otimes_R T \rightarrow \prod_G T$$ defined by $x \otimes y \mapsto (xg(y))_{g \in G}$. We say that $R \rightarrow T$ is a $G$-Galois extension if both of these maps are isomorphisms.

My main reference is Cyclic Galois Extensions of Commutative Rings by Greither and right now I am trying to figure out how the proof of lemma 1.9 in chapter 0 works. So basically the proposition is this:

Any Galois extension $R \rightarrow T$ is faithfully flat.

Greither's very brief proof goes like this: We've shown earlier that $T$ is finitely generated projective as an $R$-module, so it must be flat. It is left to prove that it is faithful as an $R$-module; we do this by showing that $T/\mathfrak{m}T \neq 0$ for all maximal ideals $\mathfrak{m}$ in $R$. Then he uses Nakayama's lemma to conclude that it suffices to prove that $T_\mathfrak{m} \neq 0$. Great, I understand the proof up to this point. But then he says: "But $R \subset T$ and localization preserves monomorphisms, so we are done."

Can anyone help me to understand what he means by this? As far as I understand he doesn't explicitly use that $R \rightarrow T$ is Galois. Of course, he uses it to show that Galois extensions are finitely generated projective, so it is still there implicitly. But I don't see any reason why this proof wouldn't work just assuming that $T$ is finitely generated projective. Is that correct? Or am I missing something?

Also: is it possible to show faithfulness directly by showing that $M\otimes_R T=0 $ implies that $M=0$ for any $R$-module $M$? I tried this approach myself (just playing around with the isomorphisms $i: R \rightarrow T^G$ and $h : T \otimes_R T \rightarrow \prod_G T$), but I never really got any good results. Does anyone know if such a proof exist?

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The map $R \to T$ is a monomorphism (of $R$-modules), and localization preserves monomorphisms, so the localized map $R_m \to T_m$ is also a monomorphism. But $R_m$ is nonzero (since it has a quotient $R/m$ which is nonzero), hence so is $T_m$.

This proof does not work only assuming that $T$ is finitely generated projective as an $R$-module because that doesn't imply that $R \to T$ is a monomorphism. For an explicit counterexample, take $R = k_1 \times k_2, T = k_1$ where the $k_i$ are fields.

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  • $\begingroup$ Great, thanks I understand now! A follow up question: in the proof, is it important to know that $T$ is finitely generated, or would the proof work only knowing that it is projective? $\endgroup$ – Nelly L Dec 14 '15 at 11:11
  • $\begingroup$ @Nelly: it's needed for the appeal to Nakayama's lemma. $\endgroup$ – Qiaochu Yuan Dec 14 '15 at 17:31
  • $\begingroup$ Yes, of course it is. Thanks! $\endgroup$ – Nelly L Dec 14 '15 at 18:21

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