# About the definition of tangent space and tangent cone.

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

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

where

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

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.

• 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 ? – Falcon 2 days ago