Computing the presentation of a morphism of a $k$-algebras

Question

To compute the Kahler differentials of a ring map $A\to B$, one writes $B = A[\{x_i\}]/(\{f_j\})$ and then we find $\Omega_{B/A} = \bigoplus Bdx_i/\sum Bdf_j$.

This raises the question (for me), how can I find the presentation $B = A[x_i]/(f_j)$ in general? Here's a simple example.

Consider the normalization of the cuspidal cubic $\mathbb A^1 = \operatorname{Spec}k[t]\to X = \operatorname{Spec}k[x,y]/(y^2-x^3)$ given by $x\mapsto t^3,y\mapsto t^2$.

To compute $\Omega_{\mathbb A^1/X}$, I want to write $k[t]$ as $(k[x,y]/(y^2-x^3))[u_i]/(g_i)$ and compute the differentials as above. Apparently, the answer is $k[t]dt/(tdt)$.

How can I do this systematically? I'm happy to see an answer specific to morphisms of affine $k$-varieties, but more generality is great too.

Answer 1

OP asked for a specific example so this is to explain the answer given in the post. This turned out to be longer than a comment so I'm posting it here instead.

First off, let me imagine $k[x,y]/(y^2-x^3)\cong k[t^2,t^3]$. Then $k[t^2,t^3]\to k[t]$ gives the map from the affine line to the cuspidal cubic. To write $k[t]$ as a $k[t^2,t^3]$ algebra, we write $B:=k[t]=k[t^2,t^3][u]/(u^2-t^2,u^3-t^3)$. Then $d(u^2-t^2)=2udu$ and $d(u^3-t^3)=3u^2du$. The reason $d(t^2)=d(t^3)=0$ is because $t^2,t^3\in k[t^2,t^3]=:A$ which are killed off.

The relations $2udu=0$ and $3u^2du=0$ imply that we only need a single relation: $udu=0$. So we see that viewing $u$ as the variable $t$, we have $\Omega_{B/A}=Bdu/(udu)=k[t]dt/(tdt)$ as in OP's post.

Second off, is OP's question about finding the presentations in general, there are computer programs designed for this such as Macaulay2's presentation(module) function. Generally, the examples one computes are when $\Omega_{B/A}$ comes from $A$-algebra morphisms $A\to B$ associated to $\operatorname{Spec}(B)\to \operatorname{Spec}(A)$ being closed immersions, open immersions, etale, and so on. Another method to circumvent the computation of presentations is to use the following.

First Fundamental Exact Sequence (Matsumura's Commutative Ring Theory Theorem 25.1): For $k\to A\to B$ ring homorphisms, there is an exact sequence $$\Omega_{A/k}\otimes_AB\to^\alpha\Omega_{B/k}\to \Omega_{B/A}\to 0.$$ The map $\alpha$ being $d_{A/k}a\otimes b\to b(d_{B/k}a)$ and the send map sending $d_{B/k}b\to d_{B/A}b$. Here, the subscripts indicates which module of differentials I am in. Furthermore, if there are some smoothness hypotheses, then this is actually left exact.

Applied to the example, we would get $k[t^2,t^3]dt^2\oplus k[t^2,t^3]dt^3\otimes_A k[t]\to k[t]dt\to \Omega_{B/A}\to 0$. The definition of the first map says the image is generated by the relations $d(t^2)=2tdt$ and $d(t^3)=3t^2dt$. So the image, a $B$-submodule, is generated by $tdt$.

Personally, I find Matsumura's presentation in Chapter 25 Derivations and Differentials to be really good so if you want more practice and details, this is where I would start.