# How to think of the Zariski tangent space

The Zariski tangent space at a point $\mathfrak m$ is defined as the dual of $\mathfrak m/\mathfrak m ^2$. While I do appreciate this definition, I find it hard to work with, because we are not given an isomorphism from $\mathfrak m/\mathfrak m^2$ to $(\mathfrak m/\mathfrak m ^2)^\vee$ (which I'd wish for at least in the finite dimensional case so that I could put my hands on something concrete).

So my question is: how does one go from this abstract definition to actually writing down what is the $T_{X,p}$ as a scheme? To take a simple case, we might consider $$X=k[x,y,z]/(x+y+z^2,x+y+z^3); \qquad p=(x-0,y-0,z-0)$$

Then, the cotangent space is easy to calculate. It is the plane cut out by $x+y$, i.e. it is the scheme $k[x,y,z]/(x+y)$. But what is the tangent space as a scheme?

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Let me just comment that I think the questions "How to think of the Zariski tangent space" and "What is the tangent space as a scheme?" are more or less orthogonal. –  user64687 Feb 14 '14 at 9:41

I'm a little confused as to why you'd want to consider the tangent space as a scheme. The scheme structure, if any, would come from the fact that it's a vector space, not because there is some natural scheme structure on it. For example, in topology, one doesn't often consider the cotangent space to be a manifold.

That said, one can ask for a scheme structure on the tangent bundle of a variety (or, more generally, the relative cotangent space of a map $X\to Y$ of schemes). This parallels exactly what one does in the case of topology--one considers the cotangent bundle of a manifold as a manifold.

To define the tangent bundle is a bit involved. For a variety $X/k$ the cotangent bundle is $\mathcal{Spec}(\text{Symm }\Omega_{X/k})$ where $\Omega_{X/k}$ is the cotangent sheaf. This looks a little confusing, but it's because I'm making some identifications. Namely, the tangent sheaf is the dual $\Omega_{X/k}^\vee$, and then the vector bundle associated to that is $\mathcal{Spec}(\text{Symm }(\Omega_{X/k}^\vee)^\vee)$ which is the same thing (in the case $X$ is a variety) as what I wrote above.

I think what you may be asking though is not what the scheme structure of the tangent space is, but what is the vector space structure. For an affine finite type $k$-scheme, and a $k$-rational point (i.e. one of the form $(x-a,y-b,z-c)$) there is a very natural way to describe the space.

Namely, let $X=k[x_1,\ldots,x_n]/(f_1,\ldots,f_r)$ be our affine finite type $k$-scheme and $p=(a_1,\ldots,a_n)=(x_1-a_1,\ldots,x_n-a_n)$ be our point. We obtain a linear map $J_p:k^n\to k^r$ defined by the Jacobian map:

$$J_p=\begin{pmatrix}\frac{\partial f_1}{\partial x_1}(p) & \cdots & \frac{\partial f_1}{\partial x_n}(p)\\ \vdots & \ddots & \vdots\\ \frac{\partial f_r}{\partial x_1}(p) & \cdots & \frac{\partial f_r}{\partial x_n}(p)\end{pmatrix}$$

Then, one can show that $T_{X,p}$ is isomorphic to $\ker J_p$.

This is a good exercise, one I leave to you. I will outline the idea though. First, prove the proposition for $r=0$ (i.e. $X=\mathbb{A}^n$). Then, identify any $X$ (written as above) as the zero set of a map $f:\mathbb{A}^n\to\mathbb{A}^r$. This will allow you to write an "exact sequence" $X\to\mathbb{A}^n\to\mathbb{A}^r$. This will actually be an exact sequence when you move to ideal land. You can then show that $T_{X,p}$ will be the kernel of the induced map $T_{\mathbb{A}^n,p}\to T_{\mathbb{A}_n,f(p)}$ which, when you identify these spaces with $k^n$ and $k^r$ (as you should have in the first step) will just be the map $J_p$.

One can actually identify the tangent space of an affine finite type $k$-scheme $X$ as the kernel of the Jacobian (defined appropriately) for any $p\in X$ where $p$ is a closed point with $k(p)/k$ separable. It fails in the non-separable case: think about $\text{Spec}(\mathbb{F}_p(T^{\frac{1}{p}}))/\mathbb{F}_p$.

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Hi Youcis! Thank you for your answer, but I didn't ask about the scheme structure on the cotangent space, -- I said the scheme structure on the Tangent space! I mentioned that I knew how to find the cotangent space (as I did for a particular case, and as you wrote down in general), but since I do not have any isomorphism from the cotangent space to the tangent space I can't even get to the tangent space! The reason I assume there is some scheme structure on it is because Vakil suggests some exercises concerning the scheme-theoretic intersection of tangent spaces. –  Rodrigo Feb 14 '14 at 16:45
@Rodrigo I wrote cotangent when I meant tangent--it was a typo. What I wrote above was for the tangent space. Look carefully at my work vs. Vakil's. I really am describing the dual of the cokernel of this Jacobian. Just to reiterate, while this may look like the computation of the cotangent space that Vakil does with the Jacobian, it's actually the dual of this computation, and so a computation of the tangent space. You also are mistaken about the scheme theoretic intersection. He asks you to show something like $T_{Z\cap Y,p}=T_{Z,p}\cap T_{Y,p}$ where $Z\cap Y$ is the –  Alex Youcis Feb 14 '14 at 20:15
scheme theoretic intersection taken in some ambient scheme $X$, and $T_{Z,p}\cap T_{Y,p}$ is the intersection as vector spaces, where you are thinking of these as embedded in $T_{X,p}$ by part a) of that exercise. He is not thinking about the scheme structure of the tangent space. –  Alex Youcis Feb 14 '14 at 20:16
Oh! I did notice that you took the kernel whereas he took the cokernel, but I didn't realize that this difference amounted to writing equations for the tangent vs cotangent space. I first checked that you are right by the following argument: the tangent space should be spanned by vectors that are orthogonal to each one of the differentials $df_i$. This is exactly the kernel of the matrix you wrote. But now that I follow through your more rigorous outline I am convinced that these are the equations of the tangent space and not of the (isomorphic) cotangent space as well. Thank you! –  Rodrigo Feb 14 '14 at 22:39
@Rodrigo You're welcome :) –  Alex Youcis Feb 14 '14 at 23:04

I'd like to briefly explain how the Zariski tangent space can be worked with and where the abstract definition comes from. A good introduction to this is Shafarevich's book on affine and projective varieties. He first shows that if an affine variety $X$ is defined by an ideal $I$, then the tangent space (say at $0$, assuming $0$ is in $X$) is defined by the zero set of the polynomials $\{dG:G\in I\}$, where $dG$ is the differential of $G$ at $0$; that is, the zero set of all homogeneous elements of degree 1 that appear in the polynomials in the ideal $I$. For example, the tangent space to $x-y+x^2+y^4=0$ at $0$ is defined by the equation $x-y=0$. Shafarevich shows that the variety defined by these equations is naturally what one would expect the tangent space to be.

He then goes on to show that abstractly, as a vector space, this space is isomorphic to $(\frak{m}_0/\frak{m}_0^2)^\vee$ in a natural way. Now, why pick $(\frak{m}_0/\frak{m}_0^2)^\vee$ instead of $\frak{m}_0/\frak{m}_0^2$? For functorial reasons. In topology, a map $f:X\to Y$ between two manifolds induces a map $T_pX\to T_{f(p)}Y$ for a point $p\in X$, where $T_pX$ denotes the tangent space of $X$ at $p$. If we take $\frak{m}_p/\frak{m}_p^2$ as the definition of tangent space and we have a map $f:X\to Y$ between two varieties, then we only get the pullback map $f^*:{\frak{m}}_{f(p)}/{\frak{m}}_{f(p)}^2\to \frak{m}_0/\frak{m}_0^2$ and not the other way around, as we would expect. That's why taking the dual space works perfectly.

In practice, if you want explicit equations for the tangent space of a variety, then the first method is very concrete. If you want more theoretical calculations, then the abstract method is the way to go. In general for a scheme $X$ over a field $k$, you can also think of the tangent space at $p$ as $\mbox{Mor}((\mbox{Spec}(k[\epsilon]/\epsilon^2),(\epsilon)),(X,p))$.

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