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In Qing Liu 4.2 (page 126) is write that the tangent map $T_{f,x}:T_{X,x}\to T_{Y,y}\otimes_{\mathcal{O}_{X,x}}k(x)$ is $k(x)$-linear. I don't see why.

I see that by definition $T_{X,x}\to T_{Y,y}$ obtain by duality from $\mathfrak{m}_{Y,y}/\mathfrak{m}_{Y,y}^2\to \mathfrak{m}_{X,x}/\mathfrak{m}_{X,x}^2$ is $k(y)$-linear (for the structur obtain by restriction of scalar of $\mathfrak{m}_{X,x}/\mathfrak{m}_{X,x}^2$) but I don't see why it's induce the conclusion: with the notation $\varphi:k(y)\to k(x)$ and $F:T_{X,x}\to T_{Y,y}$, I have for all $u\in T_{X,x}$ and $\alpha\in k(y)$, $T(\varphi(\alpha)u)=\alpha T(u)$, so $T_{f,x}(\varphi(\alpha)u)=T(\varphi(\alpha)u)\otimes1=\alpha T(u)\otimes1=\alpha(T(u)\otimes1)=T(u)\otimes\varphi(\alpha)$. So if we have $\beta=\varphi(\alpha)\in k(x)$, $T_{f,x}(\beta u)=T(u)\otimes\beta=\beta(T_{f,x}(u))$ and I obtain the linearity. But I don't see the reason for why $k(y)\to k(x)$ should be surjective ie why $\beta=\varphi(\alpha)$.

I have seen the errata that require $T_{Y,y}$ has finite dimension over $k(y)$ (e.g. $Y$ is locally Noetherian) or $f$ is locally of finite type but I don't see link with my problem.

I have seen the question Map between Zariski tangent spaces(?) but it seems to be a problem with the isomorphism $\mbox{Hom}_{k(y)}({\frak{m}}_y/{\frak{m}}_y^2,k(x))\simeq\mbox{Hom}_{k(y)}({\frak{m}}_y/{\frak{m}}_y^2,k(y))\otimes k(x)$ wich is not clear to me.

Community
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Gabriel Soranzo
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1 Answers1

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If $\mathfrak{m}_y / \mathfrak{m}_y^2$ is finite-dimensional over $k(y)$ (for example when $Y$ is locally noetherian), then (by linear algebra) the canonical homomorphism $$\hom_{k(y)}(\mathfrak{m}_y / \mathfrak{m}_y^2,k(y)) \otimes_{k(y)} k(x) \to \hom_{k(y)}(\mathfrak{m}_y / \mathfrak{m}_y^2,k(x))$$ is an isomorphism. This also holds when $k(x)$ is finite-dimensional over $k(y)$, for example when $f$ is locally of finite type.

Now, the $k(y)$-linear map $\mathfrak{m}_y / \mathfrak{m}_y^2 \to \mathfrak{m}_x / \mathfrak{m}_x^2 $ induces a $k(x)$-linear map $\mathfrak{m}_y / \mathfrak{m}_y^2 \otimes_{k(y)} k(x) \to \mathfrak{m}_x / \mathfrak{m}_x^2$, hence by dualizing a $k(x)$-linear map from $T_x$ to $$\hom_{k(x)}(\mathfrak{m}_y / \mathfrak{m}_y^2 \otimes_{k(y)} k(x),k(x)) \cong \hom_{k(y)}(\mathfrak{m}_y / \mathfrak{m}_y^2,k(x)) \cong \hom_{k(y)}(\mathfrak{m}_y / \mathfrak{m}_y^2,k(y)) \otimes_{k(y)} k(x),$$ that is $T_y \otimes_{k(y)} k(x)$.

Martin Brandenburg
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  • Thanks, all is clear now. But please don't say "elmentary" ;-) – Gabriel Soranzo Oct 02 '13 at 19:40
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    Well it is elementary when you generalize it so that the essential emerges, and the unimportant waste disappears. If $S$ is a $R$-algebra and $M,N$ are $R$-modules, you can look at $\hom_R(M,N) \otimes_R S \to \hom_S(M \otimes_R S, N \otimes_R S)$. The class of $M$ for which this is an isomorphism is closed unter direct sums and contains $R$, hence it contains all finitely generated free modules. This suffices here. Moreover, it is closed unter direct summands, hence contains all f.g. projective modules. If $S/R$ is flat, it is closed under cokernels, hence contains finitely presented modules. – Martin Brandenburg Oct 02 '13 at 22:12
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    You can identify the map with $\hom_R(M,N) \otimes_R S \to \hom_R(M,N \otimes_R S)$, and this map even makes sense when $S$ is just some $R$-module. Now play the same game, the class of "good" $S$ is closed under direct sums and contains $R$, hence it contains f.g. free $S$. These "elementary" isomorphisms can be found in Bourbaki's Algèbre, but I think usually they are proven using diagrams and exact sequences. In my opinion the proofs using closure properties are easier. – Martin Brandenburg Oct 02 '13 at 22:15