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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 $X$?

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  • $\begingroup$ Isn't the coordinate ring at the point $p$ the local ring? $\endgroup$
    – user40276
    Commented Feb 2, 2014 at 16:35
  • $\begingroup$ The local ring at a point is the localization of the coordinate ring by the maximal ideal of the point. @user40276 $\endgroup$ Commented Feb 2, 2014 at 17:10
  • $\begingroup$ 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). $\endgroup$
    – user40276
    Commented Feb 2, 2014 at 17:13
  • $\begingroup$ 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. $\endgroup$ Commented Feb 2, 2014 at 17:20
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    $\begingroup$ @Student That's not the coordinate ring. That's the field of rational functions. $\endgroup$
    – Zhen Lin
    Commented Feb 2, 2014 at 22:27

2 Answers 2

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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.

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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.

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