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Let $X = \text{Spec}(k[X_{1}, \ldots, X_{n}] / I)$ be an affine scheme that may be non reduced. Therefore, we can consider its associated reduced affine scheme $X_{red} = \text{Spec}(k[X_{1}, \ldots, X_{n}] / \sqrt{I})$. Clearly, there is a closed embedding $X_{red} \subseteq X$ which is an homeomorphism.

Let now $TC_{x}X$ be the tangent cone to the scheme $X$ at a closed point $x \in X(k)$. This is, $$ TC_{x}X = \text{Spec}(\text{gr}(\mathcal{O}_{X,x})),$$ where $\mathcal{O}_{X,x}$ is the stalk of the structure sheaf of the scheme $X$ at the point $x \in X(k)$, and $$\text{gr}(\mathcal{O}_{X, x}) = \bigoplus_{p \geq 0} \mathfrak{m}^{p} / \mathfrak{m}^{p+1},$$ being $\mathfrak{m}$ the maximal ideal of the local ring $\mathcal{O}_{X,x}$.

Recall that if $T_{x}X$ is the tangent Zariski space of $X$ at the point $x \in X(k)$, we can regard $T_{x}X$ as the set of $k$-rational points of the scheme $$ T_{x}X^{\text{sch}} = \text{Spec}(\text{Sym} (\mathfrak{m} / \mathfrak{m}^{2})),$$ with $\text{Sym} (\mathfrak{m} / \mathfrak{m}^{2})$ the symmetric algebra on $\mathfrak{m} / \mathfrak{m}^{2} \simeq T_{x}X^{*}$.

Now, using the reduced affine scheme $X_{red}$ I have the following two questions:

  1. I know that there is a canonically closed inmersion $TC_{x}X \subseteq T_{x}X^{\text{sch}}$. Now, using that there is a surjective morphism of local rings $$ \mathcal{O}_{X, x} \to \mathcal{O}_{X_{red},x}$$ because of the fact that $\mathcal{O}_{X_{red}, x} = \mathcal{O}_{X,x} / \sqrt{ 0}$, being $\sqrt{0}$ the nilradical of $\mathcal{O}_{X,x}$. Is it also true that we have a canonical inmersion $TC_{x}X \subseteq T_{x}X_{red}^{\text{sch}}$? Or at least, is it true that $TC_{x}X (k) \subseteq T_{x}X_{red}^{\text{sch}}(k)$ ? I know that $T_{x}X_{red}^{\text{sch}} \subseteq T_{x}X^{\text{sch}}$, but I don't know how to continue.
  2. Suppose we have a locally closed subset $Y \subset X$ and that we equip $Y$ with the reduced locally closed subscheme structure. In this case, if there is $x \in X(k) \cap Y(k)$ such that $TC_{x}X(k) = T_{x}Y^{\text{sch}}(k)$, can we say affirm $T_{x}Y^{\text{sch}}(k) = T_{x}X_{red}^{\text{sch}}(k)$? I think this is true because $TC_{x}X(k) = T_{x}Y^{\text{sch}}(k)$ means that $TC_{x}X(k)$ is ''linear'', but I am not totally sure about how to proceed.

Any help is kindly welcomed. Most of the definitions and facts used are based on Mumford's Red book and on the book The geometry of schemes of Eisenbud and Harris.

gal16
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  • Please include motivation for your study of the tangent cone for a non-reduced scheme and its relation with $TC_x(X)$ with references. – hm2020 Jan 11 '23 at 12:00
  • @hm2020 the context is the study of deformations of Lie algebras. Particularly, my first question is the Remark that appears on page 21 (which is page 258 of the journal) of this paper by G. Rauch: http://www.numdam.org/item/AIF_1972__22_1_239_0/ . – gal16 Jan 11 '23 at 13:03
  • If the base field $k$ is algebraically closed, you may for any maximal ideal $I \subseteq A$ define the cotangent space $I/I^2$ as the fiber $\Omega^1_{A/k}\otimes_A A/I \cong I/I^2$ of the module of Kahler differentials $\Omega^1_{A/k}$. There is a similar construction of the "tangent cone". – hm2020 Jan 11 '23 at 13:52
  • If $J \subseteq A\otimes_k A$ is the "ideal of the diagonal", you may define $Gr(J):=\oplus_n J^n/J^{n+1}$ and $\pi: C(X):=Spec(Gr(J)) \rightarrow X:=Spec(A)$ is by detinition the "global tangent cone". It has the property that the fiber $\pi^{-1}(x) \cong TC_xX$ is the tangent cone you speak of above. This construction has several "functorial properties". If $nil(A):=I_1 \subseteq A$ is the nilradical, you get a sequence $k \rightarrow A \rightarrow A/nil(A):=B$ and there are exact sequences relating the Kahler differentials of $A$ to that of $B$. There are similar sequences for $C(X)$. – hm2020 Jan 11 '23 at 13:57
  • In the case of the Kahler differentials there is a surjection $\Omega^1_A/I_1(\Omega^1_A) \rightarrow \Omega^1_B \rightarrow 0$, hence the Kahler differentials of the reduced ring $A/nil(A)$ is a quotient of the Kahler differentials of $A$. – hm2020 Jan 11 '23 at 14:11
  • Moreover: If $X/k$ is regular of finite type over $k$, it follows for any closed point $I:=\mathfrak{m}\in X$ there is an isomorphism $Sym^l(I/I^2) \cong I^l/I^{l+1}$, hence there is an isomorphism $\oplus_l Sym^l(I/I^2) \cong \oplus_l I^l/I^{l+1} :=Gr(I)$. Hence $Spec(Gr(I)) \cong Spec(Sym^*(I/I^2))$. – hm2020 Jan 11 '23 at 14:24
  • @hm2020 thanks! the point I don't understand is how is the exact sequence for the cone and how this implies that $TC_{x}X \subseteq T_{x}X_{red}$. And, what does it happen when $k$ is not algebraically closed? I would be very pleased if you could explain it a bit more – gal16 Jan 11 '23 at 14:57
  • there is a thread here which has some information. https://math.stackexchange.com/questions/4125495/about-the-definition-of-tangent-space-and-tangent-cone/4126576#4126576 – hm2020 Jan 11 '23 at 14:59
  • @hm2020 and the argument in your last comment (when $X/k$ is regular) applies to prove the second question I asked in the post, I think. – gal16 Jan 11 '23 at 14:59

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