Tangent spaces of affine algebraic varieties at singular points

Let $X$ be an affine algebraic over the algebraically closed field $k$ and let $\mathcal{O}(X)$ be the ring of its regular functions. Let us assume that $X$ is irreducible and let $x\in X$.

There are different ways to define the tangent space $\operatorname{T}_x X$ of $X$ at $x$, one way being to let $\operatorname{T}_x X := \operatorname{Der}_k(\mathcal{O}(X),k_x)$, where $k_x$ is the field $k$, considered as an $\mathcal{O}(X)$-module via the map $f\mapsto f(x)$.

Now, one says that $X$ is smooth at $x$ iff $\dim_k(\operatorname{T}_x X) = \dim X$, where the right hand side denotes the Krull-dimension of $X$ as a topological space.

Questions:

I was wondering what one could say about $\dim_k (\operatorname{T}_x X)$ in general, i.e. at non-smooth points. So let $y\in X$ be a singular point.

1) Can $\dim_k(\operatorname{T}_y X)$ be smaller than $\dim X$, or is it always larger?

2) And if $X\subset k^n$ is a subvariety of the affine $n$-space, is $\dim_k (\operatorname{T}_x X) \le n$ true for each $y\in X$?

So you see that I'm just learning to play with tangent spaces. Thank you for your help!

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I'd leave this as a comment but I don't have the reputation. The answer to (2) is yes. In fact, given any affine variety $Y$, a subvariety $X \subset Y$ will have $$\dim T_y X \leq \dim T_y Y.$$

The reason is because the functor called "Send an augmented algebra $A \to k$ to $\text{Der}_k(A,k)$" is contravariant, and it sends surjections to injections. That is, $(A \to k) \mapsto \text{Der}_k(A,k)$ is a contravariant, left exact functor. In particular, if you have a surjection of rings $\mathcal{O}(Y) \to \mathcal{O}(X)$, this induces an injection of vector spaces $\text{Der}(\mathcal{O}(X),k) \to \text{Der}(\mathcal{O}(Y),k)$. This proves the dimension inequality.

Here's a proof that the $\text{Der}(\bullet,k)$ assignment takes surjections to injections. First, given an onto map $A \to B$ of algebras that respects their augmentations, choose any derivation $d$ on $B$. Then this defines a derivation $d_A$ on $A$ by the formula $$d_A(f) := d([f])$$ where $[f]$ is the image of $f$ in $B$. This is a derivation on $A$ because the map $A \to B$ respected the augmentations $A, B \to k$.

Finally, this map $\text{Der}(B,k) \to \text{Der}(A,k)$ is injective because if the induced derivation $d_A$ is zero on all $f \in A$, then $d$ is zero on all $[f]$ by definition.

Edit: I thought about (1). It's true, but not obviously true. The only proof I can think of is a little complicated.

Heuristically, here's what you want to do. Given an ascending chain of prime ideals $0 \subset \mathfrak{p}_1 \subset \ldots \subset \mathfrak{p}_d \subset \mathcal{O}(X)$, examine the corresponding sequence of surjective ring maps $$\mathcal{O}(X) \to \mathcal{O}(X)/\mathfrak{p}_1 \to \ldots \to \mathcal{O}(X)/\mathfrak{p}_d.$$ From above we know that $\text{Der}(\mathcal{O}(X)/\mathfrak{p}_i,k)$ has dimension $\leq$ than $\text{Der}(\mathcal{O}(X)/\mathfrak{p}_{i+1},k)$ and we need to prove that this $\leq$ is actually a strict inequality $<$. This would do the trick. Geometrically it's intuitive -- each successive prime ideal gives a new direction one can move in, so one takes the derivation corresponding to the directional derivative in that direction, -- but algebraically, I've found it hard to prove.

An actual proof is as follows. It makes use of a non-trivial theorem from Atiyah-MacDonald, which is that the Krull dimension of a local Noetherian ring (for instance, the ring of $X$ localized at $x \in X$) is equal to the minimum number $s$ of generators one needs to generate an ideal that's primary with respect to the unique maximal ideal $m$. It turns out that if $X$ is irreducible, the Krull dimension of $X$ is the same as the Krull dimension of $\mathcal{O}(X)$ localized at any point $p$. Since the space of derivations is clearly equal to the dual of $m/m^2$, and by Nakayama, a basis for $m/m^2$ generates $m$, we have that $s \leq \dim m/m^2$. Hence $$\dim_{\text{Krull}}(X) = s \leq \dim m/m^2 = \dim T_p X$$

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Awesome answer! Thank you very much! – Sh4pe Dec 30 '12 at 11:54

Answer to 1) Yes $\dim_k(T_yX)$ is always larger than or equal to $\dim_y(X)$ with equality being achieved for $y$ smooth points (2) Yes it is true for each $y \in X$

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Could you please elaborate why both times "yes" is the answer? – Sh4pe Dec 29 '12 at 18:38