Map between Zariski tangent spaces(?) 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?
 A: 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 $k(y)$.
A: 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.

