Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $X \subset \mathbb{C}^n$ be an affine variety (not irreducible). Let $Y$ be a subvariety of $X$ (again not irreducible). How can we relate the Zariski tangent space at $P \in Y$ and at $P \in X$?

(Corrected after Mariano's comments) Based on my understanding, we do have a homomorphism $T_P Y \rightarrow T_P X$ of vector spaces, but can we say something more? For example, what can we say about the dimensions of the two vector spaces $T_PY$ and $T_PX$?

share|cite|improve this question
Do you mean "not irreducible" or "not necessarily irreducible"? – Nils Matthes Jul 20 '13 at 19:36
@NilsMatthes: Yes, i mean "not necessarily irreducible" :) – Manos Jul 20 '13 at 19:37
What homomorphism $T_pX\to T_pY$ do you have in mind? Notice that there are tons of such homomorphisms, simply because we could take the zero map, say, but presumably you have in mind a natural one. – Mariano Suárez-Alvarez Jul 20 '13 at 19:39
@MarianoSuárez-Alvarez: Yes you are right, i mean the canonical one. – Manos Jul 20 '13 at 19:41
Dear @Manos, But the isomorphism is, in general, non-canonical, as it involves the choice of a basis. So if you're looking for a canonical map, choosing an isomorphism between a finite dimensional vector space and its dual is probably not going to help you. – Keenan Kidwell Jul 20 '13 at 21:10
up vote 2 down vote accepted

The natural $\Bbb{C}$-linear map $$ T_P(Y) \rightarrow T_P(X) $$ is indeed injective. This follows from the fact that it is dual to the $\Bbb{C}$-linear map $$ \mathfrak{m}_{X,P}/\mathfrak{m}_{X,P}^2 \rightarrow \mathfrak{m}_{Y,P}/\mathfrak{m}_{Y,P}^2 $$ which is surjective, since $Y$ is a subvariety of $X$. Hence one always has

$$ \dim T_P(Y) \leq \dim T_P(X) $$

and this result cannot be improved; you can have strict inequality (e.g. $X=\Bbb{A}^1$, $Y=P$ for some point $P \in \Bbb{A}^1$) and you can have equality (e.g. $X=\Bbb{A}^2$, $Y=V(y^2-x^3)$ and $P=(0,0)$).

share|cite|improve this answer
+1 Excellent, this is enlightening. – Manos Jul 20 '13 at 21:18
One more question please, to make sure i understand correctly the definitions: if all the gradients of the generators of the vanishing ideal of an affine variety $X$ of $\mathbb{C}^n$ are zero at a certain point $P$, then this means that the Zariski tangent space at this point is the entire $\mathbb{C}^n$. Right? – Manos Jul 20 '13 at 23:25
Yes, that's right, Manos. – Georges Elencwajg Jul 21 '13 at 6:00

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.