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Tag 00N1

Chapter 10: Commutative Algebra > Section 10.101: What makes a complex exact?

Proposition 10.101.10. In Situation 10.101.1, suppose $R$ is a local Noetherian ring. The following are equivalent

  1. $0 \to R^{n_e} \to R^{n_{e-1}} \to \ldots \to R^{n_0}$ is exact at $R^{n_e}, \ldots, R^{n_1}$, and
  2. for all $i$, $1 \leq i \leq e$ the following two conditions are satisfied:
    1. $\text{rank}(\varphi_i) = r_i$ where $r_i = n_i - n_{i + 1} + \ldots + (-1)^{e-i-1} n_{e-1} + (-1)^{e-i} n_e$,
    2. $I(\varphi_i) = R$, or $I(\varphi_i)$ contains a regular sequence of length $i$.

Proof. If for some $i$ some matrix coefficient of $\varphi_i$ is not in $\mathfrak m$, then we apply Lemma 10.101.2. It is easy to see that the proposition for a complex and for the same complex with a trivial complex added to it are equivalent. Thus we may assume that all matrix entries of each $\varphi_i$ are elements of the maximal ideal. We may also assume that $e \geq 1$.

Assume the complex is exact at $R^{n_e}, \ldots, R^{n_1}$. Let $\mathfrak q \in \text{Ass}(R)$. Note that the ring $R_{\mathfrak q}$ has depth $0$ and that the complex remains exact after localization at $\mathfrak q$. We apply Lemmas 10.101.3 and 10.101.6 to the localized complex over $R_{\mathfrak q}$. We conclude that $\varphi_{i, \mathfrak q}$ has rank $r_i$ for all $i$. Since $R \to \bigoplus_{\mathfrak q \in \text{Ass}(R)} R_\mathfrak q$ is injective (Lemma 10.62.19), we conclude that $\varphi_i$ has rank $r_i$ over $R$ by the definition of rank as given in Definition 10.101.5. Therefore we see that $I(\varphi_i)_\mathfrak q = I(\varphi_{i, \mathfrak q})$ as the ranks do not change. Since all of the ideals $I(\varphi_i)_{\mathfrak q}$, $e \geq i \geq 1$ are equal to $R_{\mathfrak q}$ (by the lemmas referenced above) we conclude none of the ideals $I(\varphi_i)$ is contained in $\mathfrak q$. This implies that $I(\varphi_e)I(\varphi_{e-1})\ldots I(\varphi_1)$ is not contained in any of the associated primes of $R$. By Lemma 10.14.2 we may choose $x \in I(\varphi_e)I(\varphi_{e - 1})\ldots I(\varphi_1)$, $x \not \in \mathfrak q$ for all $\mathfrak q \in \text{Ass}(R)$. Observe that $x$ is a nonzerodivisor (Lemma 10.62.9). According to Lemma 10.101.8 the complex $0 \to (R/xR)^{n_e} \to \ldots \to (R/xR)^{n_1}$ is exact at $(R/xR)^{n_e}, \ldots, (R/xR)^{n_2}$. By induction on $e$ all the ideals $I(\varphi_i)/xR$ have a regular sequence of length $i - 1$. This proves that $I(\varphi_i)$ contains a regular sequence of length $i$.

Assume (2)(a) and (2)(b) hold. We claim that for any prime $\mathfrak p \subset R$ conditions (2)(a) and (2)(b) hold for the complex $0 \to R_\mathfrak p^{n_e} \to R_\mathfrak p^{n_{e - 1}} \to \ldots \to R_\mathfrak p^{n_0}$ with maps $\varphi_{i, \mathfrak p}$ over $R_\mathfrak p$. Namely, since $I(\varphi_i)$ contains a nonzero divisor, the image of $I(\varphi_i)$ in $R_\mathfrak p$ is nonzero. This implies that the rank of $\varphi_{i, \mathfrak p}$ is the same as the rank of $\varphi_i$: the rank as defined above of a matrix $\varphi$ over a ring $R$ can only drop when passing to an $R$-algebra $R'$ and this happens if and only $I(\varphi)$ maps to zero in $R'$. Thus (2)(a) holds. Having said this we know that $I(\varphi_{i, \mathfrak p}) = I(\varphi_i)_\mathfrak p$ and we see that (2)(b) is preserved under localization as well. By induction on the dimension of $R$ we may assume the complex is exact when localized at any nonmaximal prime $\mathfrak p$ of $R$. Thus $\mathop{\rm Ker}(\varphi_i)/\mathop{\rm Im}(\varphi_{i + 1})$ has support contained in $\{\mathfrak m\}$ and hence if nonzero has depth $0$. As $I(\varphi_i) \subset \mathfrak m$ for all $i$ because of what was said in the first paragraph of the proof, we see that (2)(b) implies $\text{depth}(R) \geq e$. By Lemma 10.101.9 we see that the complex is exact at $R^{n_e}, \ldots, R^{n_1}$ concluding the proof. $\square$

    The code snippet corresponding to this tag is a part of the file algebra.tex and is located in lines 23555–23571 (see updates for more information).

    \begin{proposition}
    \label{proposition-what-exact}
    In Situation \ref{situation-complex}, suppose $R$ is
    a local Noetherian ring. The following are equivalent
    \begin{enumerate}
    \item $0 \to R^{n_e} \to R^{n_{e-1}} \to \ldots \to R^{n_0}$
    is exact at $R^{n_e}, \ldots, R^{n_1}$, and
    \item for all $i$, $1 \leq i \leq e$
    the following two conditions are satisfied:
    \begin{enumerate}
    \item $\text{rank}(\varphi_i) = r_i$ where
    $r_i = n_i - n_{i + 1} + \ldots + (-1)^{e-i-1} n_{e-1} + (-1)^{e-i} n_e$,
    \item $I(\varphi_i) = R$, or $I(\varphi_i)$ contains a
    regular sequence of length $i$.
    \end{enumerate}
    \end{enumerate}
    \end{proposition}
    
    \begin{proof}
    If for some $i$ some matrix coefficient of $\varphi_i$
    is not in $\mathfrak m$, then we apply Lemma \ref{lemma-add-trivial-complex}.
    It is easy to see that the proposition for a complex and
    for the same complex with a trivial complex added to it
    are equivalent. Thus we may assume that all matrix entries
    of each $\varphi_i$ are elements of the maximal ideal.
    We may also assume that $e \geq 1$.
    
    \medskip\noindent
    Assume the complex is exact at $R^{n_e}, \ldots, R^{n_1}$.
    Let $\mathfrak q \in \text{Ass}(R)$.
    Note that the ring $R_{\mathfrak q}$ has depth $0$
    and that the complex remains exact after localization at $\mathfrak q$.
    We apply Lemmas \ref{lemma-exact-depth-zero-local} and
    \ref{lemma-trivial-case-exact} to the localized complex
    over $R_{\mathfrak q}$. We conclude that
    $\varphi_{i, \mathfrak q}$ has rank $r_i$ for all $i$.
    Since $R \to \bigoplus_{\mathfrak q \in \text{Ass}(R)} R_\mathfrak q$
    is injective (Lemma \ref{lemma-zero-at-ass-zero}), we conclude that
    $\varphi_i$ has rank $r_i$ over $R$ by the definition of rank as given
    in Definition \ref{definition-rank}. Therefore we see that
    $I(\varphi_i)_\mathfrak q = I(\varphi_{i, \mathfrak q})$
    as the ranks do not change. Since all of the ideals
    $I(\varphi_i)_{\mathfrak q}$, $e \geq i \geq 1$
    are equal to $R_{\mathfrak q}$ (by the lemmas referenced above)
    we conclude none of the ideals $I(\varphi_i)$ is contained in $\mathfrak q$.
    This implies that $I(\varphi_e)I(\varphi_{e-1})\ldots I(\varphi_1)$
    is not contained in any of the associated primes
    of $R$. By Lemma \ref{lemma-silly} we may choose
    $x \in I(\varphi_e)I(\varphi_{e - 1})\ldots I(\varphi_1)$,
    $x \not \in \mathfrak q$ for all $\mathfrak q \in \text{Ass}(R)$.
    Observe that $x$ is a nonzerodivisor (Lemma \ref{lemma-ass-zero-divisors}).
    According to Lemma \ref{lemma-div-x-exact-one-less}
    the complex $0 \to (R/xR)^{n_e} \to \ldots \to (R/xR)^{n_1}$ is exact
    at $(R/xR)^{n_e}, \ldots, (R/xR)^{n_2}$. By induction
    on $e$ all the ideals $I(\varphi_i)/xR$ have a regular
    sequence of length $i - 1$. This proves that $I(\varphi_i)$
    contains a regular sequence of length $i$.
    
    \medskip\noindent
    Assume (2)(a) and (2)(b) hold. We claim that for any prime
    $\mathfrak p \subset R$ conditions (2)(a) and (2)(b)
    hold for the complex
    $0 \to R_\mathfrak p^{n_e} \to R_\mathfrak p^{n_{e - 1}} \to \ldots \to
    R_\mathfrak p^{n_0}$ with maps $\varphi_{i, \mathfrak p}$
    over $R_\mathfrak p$. Namely, since $I(\varphi_i)$ contains a
    nonzero divisor, the image of $I(\varphi_i)$ in $R_\mathfrak p$
    is nonzero. This implies that the rank of $\varphi_{i, \mathfrak p}$
    is the same as the rank of $\varphi_i$: the rank as defined above
    of a matrix $\varphi$ over a ring $R$ can only drop when passing
    to an $R$-algebra $R'$ and this happens if and only $I(\varphi)$
    maps to zero in $R'$. Thus (2)(a) holds. Having said this
    we know that $I(\varphi_{i, \mathfrak p}) = I(\varphi_i)_\mathfrak p$
    and we see that (2)(b) is preserved under localization as well.
    By induction on the dimension of $R$ we may assume the complex
    is exact when localized at any nonmaximal prime $\mathfrak p$ of $R$.
    Thus $\Ker(\varphi_i)/\Im(\varphi_{i + 1})$ has support contained in
    $\{\mathfrak m\}$ and hence if nonzero has depth $0$.
    As $I(\varphi_i) \subset \mathfrak m$ for all $i$ because
    of what was said in the first paragraph of the proof, we
    see that (2)(b) implies $\text{depth}(R) \geq e$.
    By Lemma \ref{lemma-acyclic} we see
    that the complex is exact at $R^{n_e}, \ldots, R^{n_1}$
    concluding the proof.
    \end{proof}

    Comments (7)

    Comment #2226 by David Savitt on September 6, 2016 a 8:05 pm UTC

    I think there's a gap in this argument, at the application of [00MW] and [00MY] in the second paragraph. The issue is that localization can (in theory) decrease the rank. So while certainly $I((\varphi_i)_{\mathfrak q}) = R_{\mathfrak q}$, this need not be the same as $I(\varphi_i)_{\mathfrak q}$. I think you want to pass to $Q(R)$ instead. This doesn't decrease ranks since you're only inverting nonzerodivisors. It has depth 0, so it follows from [00MW] and [00MY] that each $I(\varphi_i)Q(R)=Q(R)$. Thus each $I(\varphi_i)$ contains a nonzerodivisor and then you're home free.

    Comment #2228 by David Savitt on September 6, 2016 a 8:15 pm UTC

    Sorry, you have to be more careful than this because $Q(R)$ needn't be local, but I think the idea is basically correct.

    Comment #2229 by Johan (site) on September 6, 2016 a 9:28 pm UTC

    Yes, somebody else made the same comment on email on April 1. But I think it is OK, bc the ranks are determined by (1). In other words, in the presence of (1) the disaster you mention cannot happen. OK?

    If you agree I will change the text to explain this better. If there is still a problem, then I will completely rewrite the proof.

    Comment #2231 by David Savitt on September 6, 2016 a 10:52 pm UTC

    I don't immediately see how that argument is supposed to go, since the statement about ranks is deduced from [00MW], i.e. only after a suitable localization. So it seems to me that it goes the other way: first you prove the existence of a nonzerodivisor in each $I(\varphi_i)$, this implies that the ranks are stable under localization, and then you get (1). But maybe I'm missing a trick?

    Comment #2232 by Johan (site) on September 6, 2016 a 11:17 pm UTC

    OK, sure, but if we have (1) and (2) for $R_\mathfrak q$ for all $\mathfrak q \in \text{Ass}(R)$, then we immediately have the statement about ranks over $R$ bc the map $R \to \bigoplus_{\mathfrak q \in \text{Ass}(R)} R_\mathfrak q$ is injective. This works because we have our silly definition of rank in terms of nonvanishing of minors and because the ranks we want are predetermined by the $n_i$ as stated in Lemma 00MW. Then the only thing left is to show (2) which we do by induction as in the current proof.

    Anyway, this should be the easy direction of the Proposition and hence we should do a much better job of explaining the proof. Thanks very much for pointing this out!

    Comment #2233 by David Savitt on September 7, 2016 a 12:38 am UTC

    OK, great! I agree with this. By the way, I think you can simply remove "local" from the statement. The argument you've explained for the 'only-if' direction doesn't use it at all, and for the 'if' direction, you just observe that (1)+(2) together are preserved by localization (given (2), since each $I(\varphi_i)$ has a nonzerodivisor). [Maybe also requires deleting "local" from [00LL] where it isn't necessary either.]

    Comment #2268 by Johan (site) on November 3, 2016 a 1:52 pm UTC

    OK, I finally made the required edits to the proof. I am not sure that you can generalize to the nonlocal case so I am leaving the statement of the proposition as is for now. The change is here.

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