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I need your collected brainpower to help me out. This is going to be long, so grab your favorite beverage and snack. I am working through Görtz and Wedhorn's "Algebraic Geometry I" and I am currently at chapter 6, more specifically the tangent space definition. So they are giving the "usual" definition of the tangent space of a scheme $X$ at a point $x$ by

$$ T_xX:=(\mathfrak{m}_x/\mathfrak{m}_x^2)^*. $$

Now let $f:X\to Y$ be a morphism of schemes, than we have a map $f_x^\sharp:\mathcal{O}_{Y,f(x)}\to \mathcal{O}_{X,x}$ which induces a $k(x)$-linear map from $\mathfrak{m}_{f(x)}/\mathfrak{m}_{f(x)}^2\otimes_{k(f(x))}k(x) \to \mathfrak{m}_x/\mathfrak{m}_x^2$. Dualizing this map gives us a map on tangent spaces which we denote:

$$ df:T_xX \to T_{f(x)}Y\otimes_{k(f(x))}k(x). $$

For those of you who have the book in front of you: I'm now particularly interested in Example 6.4 and there especially the second part which is as following: Let $\mathbb{A}^n_k$ be the affine space over a field $k$. Here the authors don't make any restriction on the field. This is going to be important to my question.

Now let $f:\mathbb{A}^n\to \mathbb{A}^r$ be a map given by polynomials $f_1,\ldots,f_r \in k[T_1,\ldots,T_n]$ and let $x\in \mathbb{A}^n_k(k)$ be a rational point. Then the map $df_x$ is given by the Jacobian:

$$ \left( \dfrac{\partial f_i}{\partial T_j} \right)_{i,j}. $$

Here is my attempt to confirm this: first of all I think the authors secretly assume that $k$ is algebraicly closed because they assume that $x=(x_1,\ldots,x_n)$ and the associated maximal ideal is $(T_1-x_1,\ldots,T_n-x_n)$. Now let's assume this and look how the maps work. The induced map on the coordinate rings is

\begin{eqnarray} f^\sharp:k[Y_1,\ldots,Y_r] &\to & k[T_1,\ldots,T_n]\\ g &\mapsto & g(f_1,\ldots,f_r), \end{eqnarray} (I hope the notation is clear). Now going to the stalks is just localizing at the respective maximal ideals which are $(T_1-x_1,\ldots,T_n-x_n)=:\mathfrak{m}_x$ and $(Y_1-b_1,\ldots,Y_r-b_r)=:\mathfrak{m}_{f(x)}$ where $b=f(x)$. I will abuse notation a little bit here, so $\mathfrak{m}_x,\mathfrak{m}_{f(x)}$ will also denote the maximal ideals in the stalk. I think this is justified since they really are the maximal ideals under the injection given by localization.

Now the induced map on the stalk level doesn't look different from the map on the coordinate rings and since we are particularly interested in how the maps work on the maximal ideals we check this for generators of $\mathfrak{m}_{f(x)}$. We can figure that the map there works as following:

\begin{eqnarray} f^\sharp_x:\mathfrak{m}_{f(x)}&\to &\mathfrak{m}_{x}\\ Y_i-f_i(x) &\mapsto & f_i-f_i(x) \end{eqnarray}

Now comes the interesting part and my actual question. To describe the actual map I will use the Taylor-polynomial of $f_i$ which is:

\begin{eqnarray} T_nf_i(x,T)&=&f_i(x)+\langle (gradf_{i})_x,(T-x)\rangle+\mathfrak{m}_x^2 \\ &=&f_i(x)+\sum_{i=1}^n \dfrac{\partial f_i}{\partial T_i}(x)(T_i-x_i)+\mathfrak{m}_x^2 . \end{eqnarray} Now comes the part that bothers me: Using that $f_i(T)=T_nf_i(x,T)$ I get that $f_i-f_i(x)=\sum_{i=1}^n \dfrac{\partial f_i}{\partial T_i}(x)(T_i-x_i)+\mathfrak{m}_x^2 $ and so by passing to the quotient $\mathfrak{m}_x/\mathfrak{m}_x^2$ I can see how the map works and figure out that the map $df$ is indeed the Jacobian matrix.

So here is my question: My argument relies on the identity

$$ f_i(T)=T_nf_i(x,T) $$

for some large enough $n$. I know that this is true for $k=\mathbb{C}$ (and therefore for any subfield) but why should this be true for any other field? In particular you have to be very carefully in even defining this in positiv characteristic. Does anybody know about this?

Or alternatively: Is there another argument to see that $df$ is given by the Jacobian which doesn't involve the Taylor expansion?

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  • $\begingroup$ I think that my answer in this question math.stackexchange.com/questions/1652671/… is useful. I shall wait for a your response. $\endgroup$ Jun 29, 2016 at 22:11
  • $\begingroup$ @Armandoj18eos Thank's a lot for commenting! I'm not quite sure how I can use your answer though. If you look at the question I'm looking at closed points of $\mathbb{A}^n_k$. I'm aware that this method brakes down much earlier when considering non closed points. However you also seem to use a formal taylor expansion in your argument. My question is just: how is this formal taylor expansion justified. Especially in positiv characteristic you have to be carefull what you even mean by this. $\endgroup$
    – Maik Pickl
    Jun 30, 2016 at 8:11
  • $\begingroup$ Polynomials start life having Taylor expansions at $0$ and you can "center" at any other point by viewing $x-a$ as the variable. And I don't think there's any problem with picking off the linear terms using partials, since those derivatives can't be zero anyway. The higher terms are slightly worrisome but you don't need them here. $\endgroup$
    – Hoot
    Jun 30, 2016 at 8:38
  • $\begingroup$ @hoot I need the higher terms in order to substitute $f_i(T)=T_nf_i(x,T)$. Or don't I? If not I would be happy if you could explain how to justify using only the linear terms in the expansion. $\endgroup$
    – Maik Pickl
    Jun 30, 2016 at 8:42
  • $\begingroup$ It just seems to me that the upshot of your Taylor stuff is that you can write $f_i(T)$ as a linear combination of monomials in the $T_j - x_j$. You need the "constant term" to be $f_i(x)$ and the coefficient of the linear term $T_j - x_j$ to be $(\partial f_i/\partial x_j)(x)$ but this can be checked. $\endgroup$
    – Hoot
    Jun 30, 2016 at 8:56

1 Answer 1

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I don't think you need to worry about whether $f_i(a)=T_nf_i(x,a)$- all you need to worry about is that this is true modulo $\mathfrak{m}_x^2$. This is easily verifiable: in positive characteristic, the difference between $f$ and $T_nf$ is a polynomial in $x^p$, so since all $p\geq 2$, we have that $f-T_nf\in \mathfrak{m}_x^2$.

Edited to add the following, from our chat:

Consider $\overline{f}$, the image of $f$ in $\mathcal{O}_{X,x}/\mathfrak{m}_x^2$. This can be written $b+\sum c_i(y_i-x_i)+\mathfrak{m}_x^2$ for $b,c_i\in\mathcal{O}_{X,x}^*$. When we take the derivative of this with respect to $y_i$, $b$ vanishes (as it is a unit), we get $c_i$ and any term in $\mathfrak{m}_x^2$ ends up in $\mathfrak{m}_x$. Passing to $\mathfrak{m}_x/\mathfrak{m}_x^2$ is equivalent to evaluating at $x$, which gives us the desired equality.

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  • $\begingroup$ First of all thanks a lot for answering! I agree with what you are saying. But as far as I see it, that $f-T_nf \in \mathbb{m}_x^2$ is a claim of yours. You didn't proof anything.What you say is easy enough if $a=0$, but what if $a\neq 0$? Have you thought about on how to define $T_nf$ in positiv characteristic? After all you divide by $p$ for large enough $n$. I don't want to sound rude, sorry. In the comments above it was pointed out that you can use the Lefschetz principle. But I would still be interested in something elementary. If this is completely obvious to you I would be happy if you $\endgroup$
    – Maik Pickl
    Jul 1, 2016 at 8:22
  • $\begingroup$ could explain it some more. $\endgroup$
    – Maik Pickl
    Jul 1, 2016 at 8:22
  • $\begingroup$ I meant $x=0$ and $x\neq0$. Too late to edit, sorry. $\endgroup$
    – Maik Pickl
    Jul 1, 2016 at 8:35
  • $\begingroup$ I thought that $f-T_nf\in\mathfrak{m}_x^2$ was a claim you made- this comes in your post right before "Now comes the part that bothers me...". It's clear anyways- consider the image $\overline{f}$ of $f$ in the quotient $\mathcal{O}_{X,x}/\mathfrak{m}_x^2$. We can write this as $f(x)+c_i(y_i-x_i)+\mathfrak{m}_x^2$ and we have all we need once we observe that the $c_i$ are exactly the derivative in the $y_i$ direction. $\endgroup$
    – KReiser
    Jul 1, 2016 at 13:34
  • $\begingroup$ You are right. I claimed that this is the case. But I was aware that I only understand this in certain situations. Can you explain to me how you observe that the $c_i$ are the partial derivatives? Also why is $f(x)-\sum c_i x_i$ the constant term of $\overline{f}$? $\endgroup$
    – Maik Pickl
    Jul 1, 2016 at 13:39

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