The Stacks project

Lemma 42.68.6. Let $(R, \mathfrak m, \kappa )$ be a local ring. For every short exact sequence

\[ 0 \to K \to L \to M \to 0 \]

of finite length $R$-modules there exists a canonical isomorphism

\[ \gamma _{K \to L \to M} : \det \nolimits _\kappa (K) \otimes _\kappa \det \nolimits _\kappa (M) \longrightarrow \det \nolimits _\kappa (L) \]

defined by the rule on nonzero symbols

\[ [e_1, \ldots , e_ k] \otimes [\overline{f}_1, \ldots , \overline{f}_ m] \longrightarrow [e_1, \ldots , e_ k, f_1, \ldots , f_ m] \]

with the following properties:

  1. For every isomorphism of short exact sequences, i.e., for every commutative diagram

    \[ \xymatrix{ 0 \ar[r] & K \ar[r] \ar[d]^ u & L \ar[r] \ar[d]^ v & M \ar[r] \ar[d]^ w & 0 \\ 0 \ar[r] & K' \ar[r] & L' \ar[r] & M' \ar[r] & 0 } \]

    with short exact rows and isomorphisms $u, v, w$ we have

    \[ \gamma _{K' \to L' \to M'} \circ (\det \nolimits _\kappa (u) \otimes \det \nolimits _\kappa (w)) = \det \nolimits _\kappa (v) \circ \gamma _{K \to L \to M}, \]
  2. for every commutative square of finite length $R$-modules with exact rows and columns

    \[ \xymatrix{ & 0 \ar[d] & 0 \ar[d] & 0 \ar[d] & \\ 0 \ar[r] & A \ar[r] \ar[d] & B \ar[r] \ar[d] & C \ar[r] \ar[d] & 0 \\ 0 \ar[r] & D \ar[r] \ar[d] & E \ar[r] \ar[d] & F \ar[r] \ar[d] & 0 \\ 0 \ar[r] & G \ar[r] \ar[d] & H \ar[r] \ar[d] & I \ar[r] \ar[d] & 0 \\ & 0 & 0 & 0 & } \]

    the following diagram is commutative

    \[ \xymatrix{ \det \nolimits _\kappa (A) \otimes \det \nolimits _\kappa (C) \otimes \det \nolimits _\kappa (G) \otimes \det \nolimits _\kappa (I) \ar[dd]_{\epsilon } \ar[rrr]_-{\gamma _{A \to B \to C} \otimes \gamma _{G \to H \to I}} & & & \det \nolimits _\kappa (B) \otimes \det \nolimits _\kappa (H) \ar[d]^{\gamma _{B \to E \to H}} \\ & & & \det \nolimits _\kappa (E) \\ \det \nolimits _\kappa (A) \otimes \det \nolimits _\kappa (G) \otimes \det \nolimits _\kappa (C) \otimes \det \nolimits _\kappa (I) \ar[rrr]^-{\gamma _{A \to D \to G} \otimes \gamma _{C \to F \to I}} & & & \det \nolimits _\kappa (D) \otimes \det \nolimits _\kappa (F) \ar[u]_{\gamma _{D \to E \to F}} } \]

    where $\epsilon $ is the switch of the factors in the tensor product times $(-1)^{cg}$ with $c = \text{length}_ R(C)$ and $g = \text{length}_ R(G)$, and

  3. the map $\gamma _{K \to L \to M}$ agrees with the usual isomorphism if $0 \to K \to L \to M \to 0$ is actually a short exact sequence of $\kappa $-vector spaces.

Proof. The significance of taking nonzero symbols in the explicit description of the map $\gamma _{K \to L \to M}$ is simply that if $(e_1, \ldots , e_ l)$ is an admissible sequence in $K$, and $(\overline{f}_1, \ldots , \overline{f}_ m)$ is an admissible sequence in $M$, then it is not guaranteed that $(e_1, \ldots , e_ l, f_1, \ldots , f_ m)$ is an admissible sequence in $L$ (where of course $f_ i \in L$ signifies a lift of $\overline{f}_ i$). However, if the symbol $[e_1, \ldots , e_ l]$ is nonzero in $\det _\kappa (K)$, then necessarily $K = \langle e_1, \ldots , e_ k\rangle $ (see proof of Lemma 42.68.3), and in this case it is true that $(e_1, \ldots , e_ k, f_1, \ldots , f_ m)$ is an admissible sequence. Moreover, by the admissible relations of type (b) for $\det _\kappa (L)$ we see that the value of $[e_1, \ldots , e_ k, f_1, \ldots , f_ m]$ in $\det _\kappa (L)$ is independent of the choice of the lifts $f_ i$ in this case also. Given this remark, it is clear that an admissible relation for $e_1, \ldots , e_ k$ in $K$ translates into an admissible relation among $e_1, \ldots , e_ k, f_1, \ldots , f_ m$ in $L$, and similarly for an admissible relation among the $\overline{f}_1, \ldots , \overline{f}_ m$. Thus $\gamma $ defines a linear map of vector spaces as claimed in the lemma.

By Lemma 42.68.5 we know $\det _\kappa (L)$ is generated by any single symbol $[x_1, \ldots , x_{k + m}]$ such that $(x_1, \ldots , x_{k + m})$ is an admissible sequence with $L = \langle x_1, \ldots , x_{k + m}\rangle $. Hence it is clear that the map $\gamma _{K \to L \to M}$ is surjective and hence an isomorphism.

Property (1) holds because

\begin{eqnarray*} & & \det \nolimits _\kappa (v)([e_1, \ldots , e_ k, f_1, \ldots , f_ m]) \\ & = & [v(e_1), \ldots , v(e_ k), v(f_1), \ldots , v(f_ m)] \\ & = & \gamma _{K' \to L' \to M'}([u(e_1), \ldots , u(e_ k)] \otimes [w(f_1), \ldots , w(f_ m)]). \end{eqnarray*}

Property (2) means that given a symbol $[\alpha _1, \ldots , \alpha _ a]$ generating $\det _\kappa (A)$, a symbol $[\gamma _1, \ldots , \gamma _ c]$ generating $\det _\kappa (C)$, a symbol $[\zeta _1, \ldots , \zeta _ g]$ generating $\det _\kappa (G)$, and a symbol $[\iota _1, \ldots , \iota _ i]$ generating $\det _\kappa (I)$ we have

\begin{eqnarray*} & & [\alpha _1, \ldots , \alpha _ a, \tilde\gamma _1, \ldots , \tilde\gamma _ c, \tilde\zeta _1, \ldots , \tilde\zeta _ g, \tilde\iota _1, \ldots , \tilde\iota _ i] \\ & = & (-1)^{cg} [\alpha _1, \ldots , \alpha _ a, \tilde\zeta _1, \ldots , \tilde\zeta _ g, \tilde\gamma _1, \ldots , \tilde\gamma _ c, \tilde\iota _1, \ldots , \tilde\iota _ i] \end{eqnarray*}

(for suitable lifts $\tilde{x}$ in $E$) in $\det _\kappa (E)$. This holds because we may use the admissible relations of type (c) $cg$ times in the following order: move the $\tilde\zeta _1$ past the elements $\tilde\gamma _ c, \ldots , \tilde\gamma _1$ (allowed since $\mathfrak m\tilde\zeta _1 \subset A$), then move $\tilde\zeta _2$ past the elements $\tilde\gamma _ c, \ldots , \tilde\gamma _1$ (allowed since $\mathfrak m\tilde\zeta _2 \subset A + R\tilde\zeta _1$), and so on.

Part (3) of the lemma is obvious. This finishes the proof. $\square$

Comments (0)

Post a comment

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.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 02PA. Beware of the difference between the letter 'O' and the digit '0'.