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I am learning Linear Algebraic Groups without enough knowledge on Algebraic Groups. I see the definition for the differential of a morphism on page 42 of James Humphreys' Linear Algebraic Groups (GTM 21):

Let $\phi: X \rightarrow Y$ be a morphism of (irreducible) varieties. If $x \in X$, $y = \phi(x)$, then $\phi^*$ maps $(\mathcal{o}_y, \mathcal{m}_y)$ into $(\mathcal{o}_x, \mathcal{m}_x)$. By composition with $\phi^*$, a linear function x on $\mathcal{m}_x/ \mathcal{m}_x^2$ therefore induces a linear function on $d \phi _x($x$)$ on $\mathcal{m}_y/ \mathcal{m}_y^2$. The resulting map $d \phi_x: \mathcal{T}(X)_x \rightarrow \mathcal{T}(Y)_y$ is evidently $K$-linear. We call it the differential of $\phi$ at $x$.

Here, $K$ is an algebraically closed field. $\phi^*: K[Y] \rightarrow K[X]$, $f \mapsto f \circ \phi$. $\mathcal{o}_x$ (resp. $\mathcal{o}_y$) represents the local ring of regular functions at $x$ (resp. $y$) with maximal ideal $\mathcal{m}_x$ (resp. $\mathcal{m}_y$).

I think, if $\phi^*$ maps $(\mathcal{o}_y, \mathcal{m}_y)$ into $(\mathcal{o}_x, \mathcal{m}_x)$, and x is a linear function on $\mathcal{m}_x/ \mathcal{m}_x^2$, then the composition x$\circ \phi^*$ will map $\mathcal{m}_y/ \mathcal{m}_y^2$ to $\mathcal{m}_x/ \mathcal{m}_x^2$. Why would it be an element of $\mathcal{T}(Y)_y$, i.e., a map from $\mathcal{m}_y/ \mathcal{m}_y^2$ to itself?

Are there any mistakes in my understanding? Or are there any equivalent definitions that are easier to follow?

Sincere thanks.

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This is proposition 1.7.14 in my (incomplete) notes here. –  Zhen Lin Jul 22 '11 at 15:51
    
I am reading it. It is very useful. Thank you very much. –  ShinyaSakai Jul 27 '11 at 1:22

2 Answers 2

up vote 2 down vote accepted

The definition I prefer is the following: a tangent vector $v$ to a point $p : k[X] \to k$ on $X$ is a morphism $k[X] \to k[\epsilon]/\epsilon^2$ such that quotienting by $\epsilon$ gives the point $p$ back. (It sends a function $f \in k[X]$ to $f(p) + \epsilon f_v(p)$ where $f_v$ denotes the directional derivative in the direction $v$.) The collection of all tangent vectors at $p$ is the tangent space, and composing with a morphism $k[Y] \to k[X]$ turns a tangent vector on $X$ into a tangent vector on $Y$ in a completely transparent way. (This definition, as it turns out, also works for smooth manifolds.)

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Thank you very much. –  ShinyaSakai Jul 27 '11 at 1:32

If $\lambda$ is an element in $T_x=(m_x/m_x^2)^\ast$ the composition $\lambda\circ\phi^*$ is a linear functional on $m_{\phi(x)}/m_{\phi(x)}^2$, i.e. an element of $T_y$ where $y=\phi(x)$.

The differential $d\phi_x$ maps $\lambda$ to $\lambda\circ\phi^*$.

This is nothing but the fact that passing to the dual switches the arrows, so to speak. If you have a linear map $f:V\rightarrow W$, the dual map goes $f^*:W^*\rightarrow V^*$.

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Thank you very much for answering my question. As I said in the question, if $\phi^*$ maps $(\mathcal{o}_y, \mathcal{m}_y)$ into $(\mathcal{o}_x, \mathcal{m}_x)$, and x (or $\lambda$) is a linear function on $\mathcal{m}_x/ \mathcal{m}_x^2$, then the composition x$\circ \phi^*$ (or $\lambda \circ \phi^*$) will map $\mathcal{m}_y/ \mathcal{m}_y^2$ to $\mathcal{m}_x/ \mathcal{m}_x^2$. Why would it be an element of $\mathcal{T}(Y)_y$, i.e., a map from $\mathcal{m}_y/ \mathcal{m}_y^2$ to itself? I think I don't follow you very well. –  ShinyaSakai Jul 27 '11 at 1:30
    
No, if $\lambda:(m_x/m_x^2)\rightarrow{\Bbb R}$ the composition $\lambda\circ\phi^*$ maps $m_y/m_y^2$ to $\Bbb R$. –  Andrea Mori Jul 27 '11 at 8:26
    
I got it. Thank you very much. –  ShinyaSakai Jul 27 '11 at 12:47

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