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I read alternative definitions of the Zariski tangent space.

Let $k$ be a field.

Definition 1: The Zariski tangent space to a $k$-rational point $p$ of an affine variety $X\subset\overline k^n$, is the set of $k$-derivations of the coordinate ring $k[X]$ of $X$ at the point $p$.

Definition 2: The Zariski tangent space to a point $p$ of an affine variety $X/k\subset\overline k^n$, defined over $k$, is the set of $k$-derivations of the local ring $O_p(X)$ of $p$.

If $X$ is defined over $k$, each derivation from definition one can be extended by linearity to a derivation of the local ring of $p$, but is this correspondence surjective? Furthermore, if $X$ is an arbitrary variety, how does definition two extend to $k$-rational points of $X$?

superAnnoyingUser
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  • Isn't the coordinate ring at the point $p$ the local ring? – user40276 Feb 02 '14 at 16:35
  • The local ring at a point is the localization of the coordinate ring by the maximal ideal of the point. @user40276 – superAnnoyingUser Feb 02 '14 at 17:10
  • Yes, but isn't $k[X]$ at $p$ the localization at $p$? So you mean that in one you use a prime ideal and in the other you use the maximal ideal that contains the prime ideal (the closure). – user40276 Feb 02 '14 at 17:13
  • The localization of which ring and by which ideal would it be?... I don't think the coordinate ring is a localization. It is a factor of the ring of polynomials by the ideal of the variety. – superAnnoyingUser Feb 02 '14 at 17:20
  • Yes, but how you distinguish the coordinate rings at different points by your definition? – user40276 Feb 02 '14 at 17:22
  • You take the localization by the maximal ideal generated by the point. If $p=(p_1,\dots,p_n)$, then $m_p=\langle x_1-p_1+I(X),\dots,x_n-p_n+I(X) \rangle$ and $O_p(X)$ turns out to be $k[x]_{m_p}$ If you have any more questions, please ask them in a separate post. – superAnnoyingUser Feb 02 '14 at 17:26
  • So what's your definition of the coordinate ring at $p$? Are working with the maximal spectrum? – user40276 Feb 02 '14 at 17:29
  • @Student It doesn't make sense to talk about the coordinate ring of an arbitrary variety – only affine ones. – Zhen Lin Feb 02 '14 at 22:04
  • @ZhenLin In some russian books, I've ever saw coordinate ring being used as the name for the structure sheaf, though it's weird. – user40276 Feb 02 '14 at 22:17
  • @ZhenLin It makes perfect sense if you define the coordinate ring in terms of equivalence classes of regular functions. Moreover, the coordinate ring of a variety coincides with the coordinate ring of any of its open subsets. – superAnnoyingUser Feb 02 '14 at 22:26
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    @Student That's not the coordinate ring. That's the field of rational functions. – Zhen Lin Feb 02 '14 at 22:27
  • Pardon me, Mr.\Mrs. @Lin. Whatever I said about the "coordinate ring" in my last comment is true about the "local ring.” Still one can define the coordinate ring of a projective variety as the factor of the polynomial ring by the homogenous ideal of the variety. – superAnnoyingUser Feb 03 '14 at 18:26
  • @user40276 Since your comment "Yes, but how you distinguish the coordinate rings at different points by your definition?" I interchanged the terms "coordinate ring" and "local ring." Henceforth, what I said about the "coordinate ring," I meant about the "local ring." – superAnnoyingUser Feb 03 '14 at 18:31
  • @user40276 To answer your question, you don't distinguish between the coordinate ring of a variety at different points of the variety, because you only define it for a variety as a whole. – superAnnoyingUser Feb 03 '14 at 18:33
  • Ok, but if you don't distinguish, the tangent space is the same at each point. – user40276 Feb 06 '14 at 14:32
  • That is not true, @user40276. You consider derivations at a point and that's how you distinguish them. But once again, please, take this issue to its own post. – superAnnoyingUser Feb 06 '14 at 16:30

2 Answers2

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$\def\Der{\operatorname{Der}}$Let $k$ be an algebraically closed field. We want to show that the map $$ \Der_k(\mathcal{O}_{X,p},k)\to\Der_k(k[X],k), $$ induced by precomposition by $k[X]\to\mathcal{O}_{X,p}$, is a bijection ($k$ is equipped with the $\mathcal{O}_{X,p}$-module structure given by evaluation at $p$). But $\mathcal{O}_{X,p}\cong k[X]_{\mathfrak{m}_p}$, where $\mathfrak{m}_p=\{f\in k[X]\mid f(p)=0\}$. Thus the result follows from this.

Elías Guisado Villalgordo
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The way I defined the tangent space in my question only works for affine varieties $X/k$ defined over the non-algebraically closed field $k$. For the general variety $X$ one defines the tangent space at a point $p$ as the dual to the factor $ m_p(X)/m_p(X)^2$ of the maximal ideal $m_p(X)$ of $p$ in $X$ by its square $m_p(X)^2$

The two definitions from the question are equivalent when $X/k$ is defined over $k$ and $p$ is a $k$-rational point of $X$, in the sense that the spaces are isomorphic. However, the correspondence I suggested does not extend by linearity. Rather it may extend by continuity, given suitable topologies. Note here a purely algebraic construction of an isomorphism. Any extension of derivations at a point $p$, needs to preserve the Leibniz rule such that if $f=\frac{f_1}{f_2}\in O_p(X)$ then

$$\overline D_p(f_1) = \overline D_p(f_2f) = \overline D_p(f_2)f(p) + f_2(p)\overline D_p(f)$$

This gives an explicit formula for the continuation $\overline D_p:O_p(X)\to k$ of $D_p:k[X]\to k$, namely

$$\overline D_p(\frac{f_1}{f_2})=\frac{D_p(f_1)f_2(p) - f_1(p)D_p(f_2)}{f_2(p)^2}$$

Easily one can see that this is an injective homomorphism and its inverse is the restriction of derivations to the coordinate ring.

superAnnoyingUser
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