# Questions about tangent and cotangent bundle on schemes

In differential geometry, for a smooth manifold $M$ we have the definition of the tangent bundle and the cotangent bundle and then $k$-forms are defined to be (smooth) sections of the $k$-exterior power of the latter.

In algebraic geometry, say $X$ is the affine space $\mathbb{A}_k^n=\operatorname{Spec}(k[x_1,\ldots,x_n])$, we can talk about the Zariski tangent space at a point $x$ which is by definition $T_xX=(\mathfrak{m}_x/\mathfrak{m}_x^2)^*$ ($\mathfrak{m}_x$ is the maximal ideal $\mathcal{O}_{X,x}$). I have read that this definition is not the best one unless the point $x$ is $k$-rational point. For other points there exists the concept of relative tangent space (see for example https://www.math.ucdavis.edu/~blnli/buildings/bag.pd, page 151).

1) My first question is: Is there exists the notion of tangent bundle on $X$ as a sheaf?

Comming back to differential geometry, I have read on the wikipedia a nice definition of the cotangent bundle: let $M$ be a smooth manifold and let $\mathcal{I}$ be the sheaf of of germs of smooth functions on M×M which vanish on the diagonal $\Delta(M)\subseteq M\times M$, and define the contangent bundle $T^*M$ as the sheaf $\Delta^*(\mathcal{I}/\mathcal{I}^2)$ (where $\Delta:M\rightarrow M\times M$ is the diagonal map). This makes me suspect that a possible definition of the contangent space on a separated scheme $X$ over a field $k$ will be the same.

2) So my second question is: is this definition of cotangent bundle correct in algebraic geometry?.

3) (More urgent) I make these questions because while reading a paper I found this part: "Let $h:Y\rightarrow X=\operatorname{Spec}(k[x_1,\ldots,x_n])$ be a morphism of integral schemes and consider the divisor of $h^*(dx_1\wedge\ldots\wedge dx_n)$ ... ". I would appreciate a lot if you help me to interpret this part. What is exactly $dx_1\wedge\ldots \wedge dx_n$?, why can we obtain a divisor from $h^*(dx_1\wedge\ldots\wedge dx_n)$?.

Thanks in advance!

Diego

## 1 Answer

(1) Yes, definitely; and you'll be able to check that the fiber of $\Omega_{X/k}$ t a $k$-rational point is the old cotangent space.

(2) It's more or less the same, and you don't need the scheme (or morphism, relatively speaking) to be separated. Look at Hartshorne II.8, Vakil 21.2, or the Stacks Project for details. It's a nice global construction and the fact that it gives the expected bundle in differential geometry is helpful for motivation, but I mostly see it as a clean way of gluing together the modules of relative differentials coming from commutative algebra.

(3) The $k[x_1, \dots, x_n]$-module $\Omega_{k[x_1, \dots, x_n]/k}$ is free on the symbols $dx_i$, and $dx_1 \wedge \cdots \wedge dx_n$ is thus a section of the top exterior power, which corresponds to a invertible sheaf on $\mathbb{A}^n_k$ (this has to be globally trivial, but that doesn't matter for what I'm going to say). Pull back the section to get a section of the invertible sheaf $h^*\Omega_{\mathbb{A}^n_k/k}$ and take the divisor of that.

In general from a morphism of $k$-schemes $f\colon X \to Y$ we do get a pull-back $f^*\Omega_{Y/k} \to \Omega_{X/k}$ that you would expect from differential geometry, but unless the source in your example is smooth of dimension $n$ then I don't see how to get a divisor out of this.