Processing math: 100%

The Stacks project

Lemma 15.100.4. Let A be a Noetherian ring. Let I \subset A be an ideal. Let M, N be finite A-modules. Set M_ n = M/I^ nM and N_ n = N/I^ nN. Then

  1. the systems (\mathop{\mathrm{Hom}}\nolimits _ A(M_ n, N_ n)) and (\text{Isom}_ A(M_ n, N_ n)) are Mittag-Leffler,

  2. there exists a c \geq 0 such that the kernels and cokernels of

    \mathop{\mathrm{Hom}}\nolimits _ A(M, N)/I^ n\mathop{\mathrm{Hom}}\nolimits _ A(M, N) \to \mathop{\mathrm{Hom}}\nolimits _ A(M_ n, N_ n)

    are killed by I^ c for all n,

  3. we have \mathop{\mathrm{lim}}\nolimits \mathop{\mathrm{Hom}}\nolimits _ A(M_ n, N_ n) =\mathop{\mathrm{Hom}}\nolimits _ A(M, N)^\wedge = \mathop{\mathrm{Hom}}\nolimits _{A^\wedge }(M^\wedge , N^\wedge )

  4. \mathop{\mathrm{lim}}\nolimits \text{Isom}_ A(M_ n, N_ n) = \text{Isom}_{A^\wedge }(M^\wedge , N^\wedge ).

Here {}^\wedge denotes usual I-adic completion.

Proof. Note that \mathop{\mathrm{Hom}}\nolimits _ A(M_ n, N_ n) = \mathop{\mathrm{Hom}}\nolimits _ A(M, N_ n). Choose a presentation

A^{\oplus t} \to A^{\oplus s} \to M \to 0

Applying the right exact functor \mathop{\mathrm{Hom}}\nolimits _ A(-, N) we obtain a complex

0 \xrightarrow {\alpha } N^{\oplus s} \xrightarrow {\beta } N^{\oplus t}

whose cohomology in the middle is \mathop{\mathrm{Hom}}\nolimits _ A(M, N) and such that for n \geq 0 the cohomology of

0 \xrightarrow {\alpha _ n} N_ n^{\oplus s} \xrightarrow {\beta _ n} N_ n^{\oplus t}

is \mathop{\mathrm{Hom}}\nolimits _ A(M_ n, N_ n). Let c \geq 0 be as in Lemma 15.100.1 for this A, I, \alpha , and \beta . By part (3) of the lemma we deduce the Mittag-Leffler property for (\mathop{\mathrm{Hom}}\nolimits _ A(M_ n, N_ n)). The kernel and cokernel of the maps \mathop{\mathrm{Hom}}\nolimits _ A(M, N)/I^ n\mathop{\mathrm{Hom}}\nolimits _ A(M, N) \to \mathop{\mathrm{Hom}}\nolimits _ A(M_ n, N_ n) are killed by I^ c by [art part (6) of the lemma. We find that \mathop{\mathrm{lim}}\nolimits \mathop{\mathrm{Hom}}\nolimits _ A(M_ n, N_ n) = \mathop{\mathrm{Hom}}\nolimits _ A(M, N)^\wedge by part (2) of the lemma. The equality

\mathop{\mathrm{Hom}}\nolimits _{A^\wedge }(M^\wedge , N^\wedge ) = \mathop{\mathrm{lim}}\nolimits \mathop{\mathrm{Hom}}\nolimits _ A(M_ n, N_ n)

follows formally from the fact that M^\wedge = \mathop{\mathrm{lim}}\nolimits M_ n and M_ n = M^\wedge /I^ nM^\wedge and the corresponding facts for N, see Algebra, Lemma 10.97.4.

The result for isomorphisms follows from the case of homomorphisms applied to both (\mathop{\mathrm{Hom}}\nolimits (M_ n, N_ n)) and (\mathop{\mathrm{Hom}}\nolimits (N_ n, M_ n)) and the following fact: for n > m > 0, if we have maps \alpha : M_ n \to N_ n and \beta : N_ n \to M_ n which induce an isomorphisms M_ m \to N_ m and N_ m \to M_ m, then \alpha and \beta are isomorphisms. Namely, then \alpha \circ \beta is surjective by Nakayama's lemma (Algebra, Lemma 10.20.1) hence \alpha \circ \beta is an isomorphism by Algebra, Lemma 10.16.4. \square


Comments (0)


Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.