# Proving exactness of the conormal sequence

Problem: Let $\phi \colon A \to B$ be a surjective homomorphism of $R$-algebras with kernel $I$. I want to show that the conormal sequence $$I/I{}^2 \longrightarrow B \otimes_A \Omega_{A/R} \longrightarrow \Omega_{B/R} \longrightarrow 0$$ is an exact sequence of $B$-modules, where $\Omega_{A/R}$ is the module of Kähler-Differentials (see below for a construction) of $A$ with differential $d$ and the maps are given by $[f] \mapsto 1 \otimes d(f)$ and $b \otimes d(a) \mapsto bd(\phi(a))$ respectively.

My Attempt: It's pretty clear that the middle map is surjective, so all that is left to show is $$(*) \qquad \ker\left(B \otimes_A \Omega_{A/R} \longrightarrow \Omega_{B/R} \right) = \text{im}\left(I/I{}^2 \longrightarrow B \otimes_A \Omega_{A/R} \right).$$ Construct $\Omega_{A/R}$ as the quotient of the free $A$-module with generators $\widetilde{d}(a)$ for $a \in A$ divided by the submodule $\text{Rel}_{A/R}$ generated by the elements $$\widetilde{d}(a + a') - \widetilde{d}(a) - \widetilde{d}(a'), \qquad \widetilde{d}(r \cdot 1_A), \qquad \widetilde{d}(aa') - a\widetilde{d}(a') - a'\widetilde{d}(a)$$ for all $a,a'\in A$, $r \in R$, let $d$ be the composition $A \to \oplus_{a \in A} \widetilde{d}(a)A \to \Omega_{A/R}$.

Using the right-exactness of the tensor product and the same construction for $\Omega_{B/R}$ we get the following commutative diagram with exact rows

$$B \otimes_A \text{Rel}_{A/R} \longrightarrow B \otimes_A \bigoplus\nolimits_{a \in A} \widetilde{d}(a)A \longrightarrow B \otimes_A \Omega_{A/R} \longrightarrow 0$$ $$\hspace{2.5cm} \downarrow \hspace{3cm} \downarrow$$ $$\hspace{.25cm}\text{Rel}_{B/R}\hspace{.25cm} \longrightarrow \hspace{.25cm}\bigoplus\nolimits_{b \in B} \widetilde{d'}(b)B\hspace{.25cm} \longrightarrow \hspace{.25cm}\Omega_{B/R}\hspace{.25cm} \longrightarrow 0.$$

A diagram chase shows that for $a \in A$ and $b \in B$ we get $$b \otimes d(a) \in \ker\left(B \otimes_A \Omega_{A/R} \to \Omega_{B/R}\right) \iff b \widetilde{d}(\phi(a)) \in \text{Rel}_{B/R}.$$

So we are pretty close to $(*)$. Am I done yet? Is it enough to consider these elementary tensor elements (instead of linear combinations of elements of $B$). Also I don't really know how $B \otimes_A \text{Rel}_{A/R}$ looks like.

• What do you mean by $G_{B/R}$? Also, have you seen the standard textbook accounts, such as Prop. 16.3 in Eisenbud, or Thm. 25.2 in Matsumura's Commutative Ring Theory? I also think it's hard to write down what the kernel of $B \otimes_A \Omega_{A/R} \to \Omega_{B/R}$ is using your method, since it's hard to describe the image of $I/I^2$ in $B \otimes_A \Omega_{A/R}$, compared to just trying to describe the cokernel of the map $I/I^2 \to B \otimes_A \Omega_{A/R}$, like Eisenbud does. – Takumi Murayama May 9 '15 at 8:27
• @TakumiMurayama Thanks, $G_{B/R}$ was supposed to be $\text{Rel}_{B/R}$, I fixed it in the post. I read the part in Eisenbud and will have a look at Matsumura's book, thanks for the reference. However I don't really understand how describing the cokernel of this map helps yet, because then I would have to show that the mapping involving the cokernel is the same as the one given in my question, and I'd be back to start, wouldn't I? – legacytron May 9 '15 at 9:48
• Also It's not clear to me why the cokernel is given as Eisenbud describes it. – legacytron May 9 '15 at 9:56

## 1 Answer

Actually, I think what you wrote down produces a proof. I think what follows is essentially the content of the proof in Eisenbud.

In your equation $(*)$, it's clear that $\supseteq$ holds, since any element of the form $1 \otimes d(f)$ for $f \in I$ maps to $d(\phi(f)) = d(0) = 0$; in the other direction, it suffices to show $I$ maps surjectively onto $\ker(B \otimes_A \Omega_{A/R} \to \Omega_{B/R})$.

Consider the diagram $$\require{AMScd}\begin{CD} @. B \otimes_A \mathrm{Rel}_{A/R} @>>> B \otimes_A \bigoplus_{a \in A} \tilde{d}(a)A @>>> B \otimes_A \Omega_{A/R} @>>> 0\\ @. @VV{\alpha}V @VV{\beta}V @VV{\gamma}V\\ 0 @>>> \mathrm{Rel}_{B/R} @>>> \bigoplus_{b \in B} \tilde{d}'(b)B @>>> \Omega_{B/R} @>>> 0 \end{CD}$$ where I've completed the bottom row to a short exact sequence by definition of $\Omega_{B/R}$. The only non-obvious map is $\beta$, which maps $b \otimes \tilde{d}(a) \mapsto b\tilde{d'}(\phi(a))$. Before we apply the snake lemma, we make the following modification: the short exact sequence $0 \to B \otimes \tilde{d}(0)A \to B \otimes \tilde{d}(0)A \to 0$ can be spliced off of the first row, and analogously $0 \to \tilde{d}(0)B \to \tilde{d}(0)B \to 0$ can from the second row, giving the new diagram $$\begin{CD} @. B \otimes_A \overline{\mathrm{Rel}}_{A/R} @>>> B \otimes_A \bigoplus_{a \in A \setminus \{0\}} \tilde{d}(a)A @>>> B \otimes_A \Omega_{A/R} @>>> 0\\ @. @VV{\alpha}V @VV{\beta}V @VV{\gamma}V\\ 0 @>>> \overline{\mathrm{Rel}}_{B/R} @>>> \bigoplus_{b \in B \setminus \{0\}} \tilde{d}'(b)B @>>> \Omega_{B/R} @>>> 0 \end{CD}$$ where now $\beta$ maps $b \otimes \tilde{d}(a) \mapsto b\tilde{d'}(\phi(a))$ if $a \notin I$, and if $a \in I$, then the image is $0$.

We know $\alpha$ is surjective by definition of $\Omega_{B/R}$ and by surjectivity of $\phi$, and $\beta$ has kernel isomorphic to $I$, where the map is given by $f \mapsto 1 \otimes \tilde{d}(f)$. Thus, $I \to \ker\gamma \to 0$ is exact by the snake lemma, and we get the opposite inclusion in $(*)$.

• Thanks! I very much like your solution as it's also a good opportunity for me to get used to the snake lemma. – legacytron May 10 '15 at 18:54