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Let $f:X\to Y$ be a morphism of schems, and $x\mapsto y:=f(x)$. Then we have a canonical map $\Phi:T_{X,x}\to T_{Y,y}\otimes_{k(y)}k(x)$ where $T_{X,x}:=\left( \mathfrak{m}_x/\mathfrak{m}_x^2\right)^{\vee}$(dual space) and $k(x)$ is the residue field at $x$. But I cannot imagine what this 'canonical' map is.

I guess for a given map $\sigma:\mathfrak{m}_x/\mathfrak{m}_x^2\to k(x)$, we can define a map $\rho:\mathfrak{m}_y/\mathfrak{m}_y^2\otimes_{k(y)}k(x)\to \mathfrak{m}_x/\mathfrak{m}_x^2\otimes_{k(y)}k(x)\to k(x)\otimes_{k(y)}k(x)$ where the first arrow is canonical and the second arrow is $\sigma \otimes \textrm{Id}_{k(x)}$. Then maybe $\Phi(\sigma)=\rho$.

Is this right?

User0829
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    In case you have this canonical morphism from Liu's book "Algebraic Geometry and Arithmetic Curves", then you might be interested in the erratum. He seems to acknowledge that a canonical choice of the tangent map is not possible in general. – Nils Matthes Jun 21 '13 at 20:21

2 Answers2

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Well, using Hartshorne's notation, we have a map $f_x^\#:\mathcal{O}_{Y,y}\to\mathcal{O}_{X,x}$ that restricts and descends to a map $f_x^\#:{\frak{m}}_y/{\frak{m}}_y^2\to {\frak{m}}_x/{\frak{m}}_x^2$ (since $(f_x^\#)^{-1}({\frak{m}}_x)={\frak{m}}_y$). By pulling back, we get a map $$T_{X,x}=\mbox{Hom}_{k(x)}({\frak{m}}_x/{\frak{m}}_x^2,k(x))\to\mbox{Hom}_{k(y)}({\frak{m}}_y/{\frak{m}}_y^2,k(x))\simeq\mbox{Hom}_{k(y)}({\frak{m}}_y/{\frak{m}}_y^2,k(y))\otimes k(x)$$ where $k(y)$ acts on $k(x)$ by $f_x^\#$. This last $k(x)$-vector space is then $T_{Y,y}\otimes k(x)$.

Edit As Georges points out below, this last isomorphism works whenever $k(x)$ is a finite extension of $k(y)$.

rfauffar
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  • Dear Robert, maybe I'm missing something, but the last isomorphism looks wrong. It can happen that the right hand side has dimension one, while the left hand side has dimension bigger than one (where both sides are interpreted as $k(x)$ vector spaces). – Nils Matthes Jun 21 '13 at 17:07
  • Dear Nils, I believe this is just a general fact about vector spaces (although I may be missing something); if $K/L$ is a field extension and $V$ is an $L$-vector space, then $\mbox{Hom}_L(V,K)\simeq\mbox{Hom}_L(V,L)\otimes_LK$. What counterexample do you have? – rfauffar Jun 21 '13 at 17:20
  • Dear Robert, sorry I was wrong about the counterexample, the two seem to be indeed isomorphic. One last question, though, sorry if it's easy again. Why is $f^{#}_x$ a morphism of $k(y)$-vector spaces (you seem to use that explicitly while defining the tangent map)? – Nils Matthes Jun 21 '13 at 17:54
  • Well, we can define $\lambda\in k(y)$ as acting on $\mu\in {\frak{m}}_x/{\frak{m}}_x^2$ by $\lambda\cdot\mu:=f_x^#(\lambda)\mu$. So if $\lambda\in k(y)$ and $z\in{\frak{m}}_y/{\frak{m}}_y^2$, we get that $f_x^#(\lambda z)=f_x^#(\lambda)f_x^#(z)=\lambda\cdot f_x^#(z)$. – rfauffar Jun 21 '13 at 18:01
  • No problem Nils! – rfauffar Jun 21 '13 at 18:11
  • Dear Robert, sorry for asking again, but the question of the OP is close to my heart, and I too strive for a complete understanding of the matter. Is the last isomorphism in your answer really canonical, or does it depend on the choice of a basis? – Nils Matthes Jun 21 '13 at 20:21
  • It's canonical. If you have a field extension $K/L$ and an $L$-vector space $V$, then we want an isomorphism $\mbox{Hom}_L(V,L)\otimes_LK\to\mbox{Hom}_L(V,K)$. This is given by taking an element $\varphi\otimes k$ and sending it to $k\varphi$. By choosing a basis, it is easy to see that this is an isomorphism; that is, every element of the right hand side vector space is the sum of $L$-linear functionals multiplied by elements of $K$. – rfauffar Jun 22 '13 at 03:13
  • I actually wrote the inverse isomorphism, but since it doesn't depend on the basis, all is well :) – rfauffar Jun 22 '13 at 03:16
  • let us continue this discussion in chat – Nils Matthes Jun 22 '13 at 08:45
  • Dear @Robert, your morphism $\mbox{Hom}_L(V,L)\otimes_L K\to \mbox{Hom}_L(V,K)$ is indeed canonical and injective but definitely not surjective if both $K$ and $V$ are infinite-dimensional over $L$. – Georges Elencwajg Oct 01 '13 at 21:18
  • Dear @Georges, you're right. It appears then that we need $k(x)$ to be a finite extension of $k(y)$ for the above machinery to work. – rfauffar Oct 03 '13 at 14:53
  • Dear @Robert, I'm happy we agree. I have upvoted your answer which, apart from my nitpicking, answers the question. – Georges Elencwajg Oct 03 '13 at 17:33
  • Thank you very much for your nitpicking! – rfauffar Oct 04 '13 at 01:33
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Some details for $\text{Hom}_L(V,L)\otimes_L K\simeq\text{Hom}_L(V,K)$:

  • The mapping is $\varphi\otimes\alpha\mapsto\alpha\varphi$
  • Injectivity: if $\varphi_1,\ldots,\varphi_k$ free in $\text{Hom}_L(V,L)$ and $\alpha_1,\ldots,\alpha_n$ free in $K$ (over $L$) then $\alpha_j\varphi_i$ free: taking $\lambda_{i,j}\in L$ with $\sum\lambda_{(i,j)}\alpha_j\varphi_i=0$ the for each $v\in V$ we have $\sum\lambda_{(i,j)}\alpha_j\varphi_i(b)=0$ (relation in $k$) hence by freeness of the $\alpha_j$ this gives for each $j$ $\sum_i \lambda_{(i,j)}\varphi_i(b)=0$. We then have for each $j$ \sum_i\lamda_{(i,j)}\varphi_i=0 so by freeness of the $\varphi_i$ we conclude that for each $(i,j)$ we have $\lambda_{(i,j)}=0$
  • Surjectivity (using $K/L$ finite): by dimensions: $\text{dim}_L(\text{Hom}_L(V,K))=\text{dim}_L(V)\times\text{dim}_L(K)$ and it's the same for the tensorial product.
Gabriel Soranzo
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