# Étale morphism induces isomorphism of tangent spaces

Let $$(A, m) \to (B, n)$$ be a flat map of local Noetherian rings with $$mB = n$$, $$B$$ of finite type over $$A$$, and $$k(B) = B / n$$ a finite separable field extension of $$k(A) = A / m$$. Then, I want to show that the map $$m / m^2 \to n / n^2$$ induces an isomorphism of (base-changed) tangent spaces: $$\text{Hom}_{k(B)}(n/n^2, k(B)) \cong \text{Hom}_{k(A)}(m/m^2, k(B))$$ But I'm running into some problems trying to manipulate the objects in question... So, I of course have a short exact sequence $$0 \to m^2 \to m \to m / m^2 \to 0$$ of $$A$$-modules, to which I apply the exact functor $$- \otimes_A B$$ to obtain $$n / n^2 \cong (m / m^2) \otimes_A B \cong (m / m^2) \otimes_{k(A)} k(B)$$. But now in applying the tensor-hom adjunction, I get $$\text{Hom}_{k(B)}(n/n^2, k(B)) \cong \text{Hom}_{k(A)}(m/m^2, \text{Hom}_{k(A)}(k(B),k(B)))$$ which is a bigger module than I want. What's going wrong?

EDIT: I finally see my issue. I was misapplying the tensor-hom adjunction; extension of scalars is left-adjoint simply to the forgetful functor of the scalar extension. The last line then comes out exactly as I would like it.

## 1 Answer

The problem is that you've used the incorrect version of the tensor-hom adjunction. Here's the correct version for $$Y$$ an $$R$$-mod, $$Z$$ an $$S$$-mod, and $$X$$ an $$R-S$$ bimodule: $$\operatorname{Hom}_S(Y\otimes_R X,Z) \cong \operatorname{Hom}_R(Y ,\operatorname{Hom}_S(X,Z))$$

You forgot to change the inner hom on your right side to being $$k(B)$$ maps instead of $$k(A)$$ maps. Once you do that, $$\operatorname{Hom}_{k(B)}(k(B),k(B))=k(B)$$ and everything works.

• Sorry for posting this after your edit - the edit did not load before I submitted my answer. If you find this extraneous, let me know and I can remove it. (Time to go submit a bug report about the edit not loading!) Oct 10 '19 at 5:51
• I'll mark your answer as correct anyway, just to put the question to rest :) Oct 10 '19 at 14:50