In my algebraic geometry course we've seen the notion of tangent cone and tangent space at the origin, $T_0X$ and $C_0 X$ respectively, as follows $$T_0 X = V(f_1: f \in I(X)) \quad \text{and} \quad C_0Y = V(f^{in}: f \in J)$$ for $Y = V(J)$ with $J$ an ideal of $k[x_1, \ldots, x_n]$, $f_1$ the linear term of $f$ and $f^{in}$ the lowest degree term. We can see that it is important in the definition of tangent space that $I(X)$ is the radical ideal with $X$ as zero locus. Indeed, for $(x)$ and $(x^2)$ in $k[x]$, then $V(x) = V(x^2) = \{0\}$ but $x^2$ has no linear term. I was wondering if the choice of the ideal $J$ in the definition of tangent cone is also important. It seems not but I have some troubles to write down a formal proof, any help ?


If you're working with just classical varieties (not schemes), it's true that the tangent cone is insensitive to the ideal of definition of your variety. To be more precise, let $I$ be a radical ideal and let $J$ be any other ideal with $\sqrt{J}=I$. Let $I^{in}$ be the ideal generated by the initial terms of elements of $I$, and similarly for $J^{in}$ and $J$. We'll show that $\sqrt{I^{in}}=\sqrt{J^{in}}$.

As $J\subset \sqrt{J}=I$, we have that $J^{in}\subset I^{in}$ and thus $\sqrt{J^{in}}\subset \sqrt{I^{in}}$, while it remains to prove the reverse inclusion. Since initial ideals are homogeneous and radicals of homogeneous ideals are homogeneous, we need only concern ourselves with homogeneous elements. Suppose $f\in \sqrt{I^{in}}$ is a homogeneous element, so we can write $f^n$ as the leading term of some element $\alpha\in I$. But $\alpha^p\in J$ for some $p\geq0$, and initial terms and powers commute for elements of $k[x_1,\cdots,x_n]$ so we have the result. (I've left some details a little sketchy - please leave a comment if you get stuck expanding some of them.)


Comment: "We can see that it is important in the definition of tangent space that I(X) is the radical ideal with X as zero locus."

Response: Let $A$ be a commutative ring and let $A_{red}:=A/nil(A)$ where $nil(A)$ is the nilradical. Let $\mathfrak{m} \subseteq A$ be a maximal ideal with $\overline{\mathfrak{m}}:=\mathfrak{m}A_{red}$ with corresponding point $x\in S:=Spec(A)$.

You may define the tangent cone $C_x(S)$ at $x$ as

$$C1.\text{ } C_x(S):=Spec(Gr(\mathfrak{m}))$$


$$C2.\text{ }Gr(\mathfrak{m}):= \oplus \mathfrak{m}^n/\mathfrak{m}^{n+1}:=A/\mathfrak{m} \oplus \mathfrak{m}/\mathfrak{m}^2\oplus \cdots.$$

Question: "I was wondering if the choice of the ideal J in the definition of tangent cone is also important. It seems not but I have some troubles to write down a formal proof, any help?"

Answer: Since the cotangent space $\mathfrak{m}/\mathfrak{m}^2$ depends on the reduced structure a similar result holds for the tangent cone. You may define the tangent cone in terms of the local ring: Let $S:=Spec(A)$ and let $\mathfrak{m} \in S$ be a maximal ideal. It follows

$$\mathcal{O}_{S, \mathfrak{m}} \cong A_{\mathfrak{m}}$$ and there is an inclusion

$$ \mathfrak{m}_{\mathfrak{m}} \subseteq \mathcal{O}_{S,\mathfrak{m}}$$ and we may define

$$C3.\text{ }Gr(\mathfrak{m}_{\mathfrak{m}}):= \oplus_n \mathfrak{m}_{\mathfrak{m}}^n/\mathfrak{m}_{\mathfrak{m}}^{n+1}.$$

there is a canonical isomorphism

$$\mathfrak{m}_{\mathfrak{m}}^n/\mathfrak{m}_{\mathfrak{m}}^{n+1} \cong \mathfrak{m}^n/\mathfrak{m}^{n+1}$$

and an isomorphism of rings

$$Gr(\mathfrak{m}) \cong Gr(\mathfrak{m}_{\mathfrak{m}}),$$

hence the definition in $C2$ may be done using the local ring at $\mathfrak{m}$. The definition in $C3$ is intrinsic since it is defined using the local ring and the local ring does not depend on an embedding of $S$ into some affine space: If $A$ is a finitely generated $k$-algebra for some field $k$ you may choose a set of generators $a_1,..,a_n \in A$ and a presentation

$$A\cong k[x_1,..,x_n]/I$$

giving an embedding

$$i:S \rightarrow \mathbb{A}^n_k$$

as a closed sub-scheme of affine $n$-space. The local ring of $S$ at $x$ and the cone $C_x(S)$ does not depend on the embedding $i$ - it depends on the maximal ideal $\mathfrak{m}_{\mathfrak{m}}$.

As you have observed: A similar statement is true for the tangent space.

If your variety/scheme $S \subseteq \mathbb{A}^n_k$ contains the origin $(0)$ with corresponding ideal $\mathfrak{m}:=(x_1,..,x_n)$, you must prove that your definition of $C_0(S)$ agrees with the definition in $C1$.

If $A$ is a $k$-algebra with $k$ a field and if $I \subseteq A \otimes_k A$ is the ideal of the diagonal. Assume $I^n/I^{n+1},A\otimes_k A/I^{n+1}$ is a projective $A$-module for all $n\geq 1$. It follows the module $I^n/I^{n+1}$ has the property that

$$ I^n/I^{n+1}\otimes_A A/\mathfrak{m}\cong \mathfrak{m}^n/\mathfrak{m}^{n+1}$$

for any maximal ideal $\mathfrak{m}$ with $A/\mathfrak{m} \cong k$.

If you define $Gr(I):= \oplus_n I^n/I^{n+1}$ you get a map

$$\pi: Spec(Gr(I)) \rightarrow Spec(A)$$ with the property that for any such maximal ideal $\mathfrak{m}\in Spec(A)$ it follows

$$\pi^{-1}(\mathfrak{m}) \cong Gr(\mathfrak{m}).$$

Hence $Spec(Gr(I))$ has the tangent cone $C_x(S)$ as fibers.

This is similar to the definition of the cotangent sheaf $\Omega:=\tilde{\Omega^1_{A/k}}$: There is an isomorphism

$$ \Omega_x \otimes_{\mathcal{O}_{S,x}} \kappa(x) \cong \mathfrak{m}/\mathfrak{m}^2$$

where $\mathfrak{m}$ is the ideal of the $k$-rational point $x$.

Note: The maximal ideals $\mathfrak{m}$ and $\overline{\mathfrak{m}}$ have the same residue field. On page 303 in Mumford's "The red book.." you will find this discussed and a relation between the tangent cone and "blowing up" a variety/scheme at a point.

  • $\begingroup$ I am really sorry but I am not very confortable with the notion of Schemes etc., is there no more "intuitive" way to understand that ? $\endgroup$ – Falcon 2 days ago

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