The Stacks project

Lemma 10.74.1. Let $R$ be a Noetherian ring. Let $I \subset R$ be an ideal contained in the Jacobson radical of $R$. Let $N \to M$ be a homomorphism of finite $R$-modules. Suppose that there exists arbitrarily large $n$ such that $N/I^ nN \to M/I^ nM$ is a split injection. Then $N \to M$ is a split injection.

Proof. Assume $\varphi : N \to M$ satisfies the assumptions of the lemma. Note that this implies that $\mathop{\mathrm{Ker}}(\varphi ) \subset I^ nN$ for arbitrarily large $n$. Hence by Lemma 10.51.5 we see that $\varphi $ is injection. Let $Q = M/N$ so that we have a short exact sequence

\[ 0 \to N \to M \to Q \to 0. \]


\[ F_2 \xrightarrow {d_2} F_1 \xrightarrow {d_1} F_0 \to Q \to 0 \]

be a finite free resolution of $Q$. We can choose a map $\alpha : F_0 \to M$ lifting the map $F_0 \to Q$. This induces a map $\beta : F_1 \to N$ such that $\beta \circ d_2 = 0$. The extension above is split if and only if there exists a map $\gamma : F_0 \to N$ such that $\beta = \gamma \circ d_1$. In other words, the class of $\beta $ in $\mathop{\mathrm{Ext}}\nolimits ^1_ R(Q, N)$ is the obstruction to splitting the short exact sequence above.

Suppose $n$ is a large integer such that $N/I^ nN \to M/I^ nM$ is a split injection. This implies

\[ 0 \to N/I^ nN \to M/I^ nM \to Q/I^ nQ \to 0. \]

is still short exact. Also, the sequence

\[ F_1/I^ nF_1 \xrightarrow {d_1} F_0/I^ nF_0 \to Q/I^ nQ \to 0 \]

is still exact. Arguing as above we see that the map $\overline{\beta } : F_1/I^ nF_1 \to N/I^ nN$ induced by $\beta $ is equal to $\overline{\gamma _ n} \circ d_1$ for some map $\overline{\gamma _ n} : F_0/I^ nF_0 \to N/I^ nN$. Since $F_0$ is free we can lift $\overline{\gamma _ n}$ to a map $\gamma _ n : F_0 \to N$ and then we see that $\beta - \gamma _ n \circ d_1$ is a map from $F_1$ into $I^ nN$. In other words we conclude that

\[ \beta \in \mathop{\mathrm{Im}}\Big(\mathop{\mathrm{Hom}}\nolimits _ R(F_0, N) \to \mathop{\mathrm{Hom}}\nolimits _ R(F_1, N)\Big) + I^ n\mathop{\mathrm{Hom}}\nolimits _ R(F_1, N). \]

for this $n$.

Since we have this property for arbitrarily large $n$ by assumption we conclude that the image of $\beta $ in the cokernel of $\mathop{\mathrm{Hom}}\nolimits _ R(F_0, N) \to \mathop{\mathrm{Hom}}\nolimits _ R(F_1, N)$ is zero by Lemma 10.51.5. Hence $\beta $ is in the image of the map $\mathop{\mathrm{Hom}}\nolimits _ R(F_0, N) \to \mathop{\mathrm{Hom}}\nolimits _ R(F_1, N)$ as desired. $\square$

Comments (3)

Comment #1938 by Brian Conrad on

The same proof works verbatim with a noetherian ring (not necessarily local) and any ideal in the Jacobson radical. This applies in particular for a noetherian complete with respect to an ideal . It seems nice to avoid locality when setting up the general adic-style theorems whenever possible.

P.S. This is Tag 02HO, a rather heinous mixture of "0" and "O" that made it hard to pass the "I am not a bot" test until looking closely at the URL for this page to figure out what was going on. Can you make a computer search on all tags containing both 0 and O and making the tag typography for those show the distinction in the letters more clearly?

Comment #1940 by on

@Brian: That is actually a very good idea, I will implement this in the next round of bugfixing, which should happen soon. Thanks!

Comment #1995 by on

@#1938 Thanks very much. I have made the corresponding changes here.

There are also:

  • 3 comment(s) on Section 10.74: An application of Ext groups

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