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}=\mbo...
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).
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?
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)?
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 $f_x^\#(\lambda z)=f_x^\#(\lambda)f_x^\#(z)=\lambda\cdot f_x^\#(z)$.
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?
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 $K$.
I actually wrote the inverse isomorphism, but since it doesn't depend on the basis, all is well :)
well, is this really an isomorphism? For a concrete example, take $K=\Bbb{Q}$ $V=L=\Bbb{R}$. Then the identity map $\id: \Bbb{R} \rightarrow \Bbb{R}$ cannot be induced from a $\Bbb{Q}$-linear map $f: \Bbb{R} \rightarrow \Bbb{Q}$, no?
Discussion between Nils Matthes and R…
Discussion between Nils Matthes and R…