If $A$ is generated as a $B$-algebra by $x_i$ and relations $r_j$, then $\Omega_{A/B}$ is generated over $A$ by $dx_i$ subject to $dr_j$.

This is Key Fact 21.2.3 in the November 2017 edition of Vakil's FOAG. I am trying to verify this fact but I am having trouble with one direction.

Here $\Omega_{A/B}$ is defined to be the free $A$-module generated by $da$ for $a \in A$, subject to the relations

1. $da + da' = d(a + a')$
2. $d(aa') = a \;da' + a'\;da$
3. $db = 0$ for $b \in B$.

Precisely, we want to show $\Omega_{A/B}$ is the $A$-module of $A$-linear combinations of $dx_i$, subject to (1)-(3) above and the relations $dr_j$ for all $j$.

Since $A$ is generated as a $B$-algebra by $\{x_i\}_{i \in I}$ subject to relations $\{r_{j}\}_{j \in J}$, we can view $r_j \in B[x_i]_{i \in I}$ as polynomials in $B$ so that we regard $A$ as the quotient $$A = B[x_i]_{i \in I}/(r_j)_{j \in J}.$$

So first off, $\Omega_{A/B}$ is generated by $dx_i$ as an $A$-module by the chain rule (or repeatedly applying the leibniz rule). Now we need to revise the relations. First of all, it is clear that these generators do indeed follow the relations (1)-(3) as well as $dr_j$, since $r_j = 0$ in $A$ to begin with.

Now, I am not sure how to proceed. I want to show $\Omega_{A/B}$ is the module generated by the generators and relations above. How can I show that the above relations are enough?

Thank you very much!

EDIT: I revised the question, correcting a mistake in my restatement of it.

For another attempt, we can denote $F$ the free $A$-module generated by $da$ for $a \in A$, and take $R \subseteq F$ to be the submodule generated by those relations (1)-(3). Then, we define $\Omega_{A/B}$ to be the cokernel $$0 \to R \to F \to \Omega_{A/B} \to 0.$$

Let us denote $M$ to be the module generated by $dx_i$ for $i \in I$, subject to (1)-(3) above as well as $dr_j$ for $j \in J$. We attempt to fit $M$ in the above exact sequence. For each $a \in A$, we can write $da$ as an $A$-linear combination of $dx_j$. This yields a function $\{da\:|\;a \in A\} \to M$, yielding an $A$-linear map $F \to M$. It is easy to see that this map is surjective. Now we just need to show that the kernel of this map is $R$.

Solution 1:

Question: "This is Key Fact 21.2.3 in the November 2017 edition of Vakil's FOAG. I am trying to verify this fact but I am having trouble with one direction. Here $\Omega_{A/B}$ is defined to be the free $A$-module generated by $da$ for $a \in A$, subject to the relations

1. $da + da' = d(a + a')$
2. $d(aa') = a \;da' + a'\;da$
3. $db = 0$ for $b \in B$.

Precisely, we want to show $\Omega_{A/B}$ is the $A$-module of $A$-linear combinations of $dx_i$, subject to (1)-(3) above and the relations $dr_j$ for all $j$."

Answer: When you define $\Omega:=\Omega^1_{A/B}$ as in 1-3 it is immediate that $\Omega$ is generated by $\overline{dx}$ for $x\in A$.

Note: Here you should "overline" the $dx$ to make clear to the reader you are speaking about equivalence classes. Take any element $\omega \in\Omega$. Since

$$\Omega:=\oplus_{x\in A}Adx/U,$$

where $U$ is the submodule generated by the elements in $1-3$ in your question, it follows (by definition) that there is a finite set of elements $a_i,x_i \in A$ with $\omega=\overline{\sum_i a_idx_i}= \sum_i a_i\overline{dx_i}$. This is beacuse the direct sum $\oplus_{x\in A}Adx$ is the set of finite sums $\sum_{i=1}^k a_idx_i$ for $a_i,x_i \in A$. Hence $\Omega$ is generated by the elements $\overline{dx}$ as left $A$-module for varying $x\in A$.

Solution 2:

Let $A = B[x_i]_{i \in I}/(r_j)_{j \in J}$. We would like to compute $\Omega_{A/B}$ using the conormal sequence. Let $I = (r_j)_{j \in J}$ be the ideal in $B[x_i]_{i \in I}$. There we recall the conormal exact sequence of $A$-modules$$I/I^2 \stackrel{\delta}\to A \otimes_B \Omega_{B[x_i]/B} \to \Omega_{A/B} \to 0.$$ Now we observe that $\Omega_{B[x_i]/B} = \bigoplus_{i \in I} Bdx_i$ canoncially so that the middle term of the above exact sequence becomes $A \otimes_B \bigoplus_{i \in I} Bdx_i = \bigoplus_{i \in I} Adx_i$. Hence, we just need to show that the image of $\delta$ is generated by $dr_i$. This is easy, though, since $\delta([f]) = 1 \otimes df$ by definition. This the guarantees that since $I/I^2$ is generated by $[r_i]$, we have that that the image of $\delta$ is generated by $\delta([r_i]) = 1 \otimes dr_i \sim dr_i$ where the $\sim$ is the identification $A \otimes_B \bigoplus_{i \in I} Bdx_i = \bigoplus_{i \in I} Adx_i$.