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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.

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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.

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  • $\begingroup$ 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!) $\endgroup$
    – KReiser
    Oct 10 '19 at 5:51
  • $\begingroup$ I'll mark your answer as correct anyway, just to put the question to rest :) $\endgroup$
    – Feryll
    Oct 10 '19 at 14:50

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