The Stacks project

Proof. If for some $i$ some matrix coefficient of $\varphi _ i$ is not in $\mathfrak m$, then we apply Lemma 10.102.2. 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 10.102.3 and 10.102.6 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 10.63.19), we conclude that $\varphi _ i$ has rank $r_ i$ over $R$ by the definition of rank as given in Definition 10.102.5. 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 10.15.2 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 10.63.9). According to Lemma 10.102.7 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 will prove that (1) holds by induction on $\dim (R)$. If $\dim (R) = 0$, then we must have $I(\varphi _ i) = R$ for $1 \leq i \leq e$ by (2)(b). Since the coefficients of $\varphi _ i$ are contained in the maximal ideal this can happen only if $r_ i = 0$ for all $i$. By (2)(a) we conclude that $e = 0$ and (1) holds. Assume $\dim (R) > 0$. 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 if $I(\varphi )$ 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 $\mathop{\mathrm{Ker}}(\varphi _ i)/\mathop{\mathrm{Im}}(\varphi _{i + 1})$ 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 10.102.8 we see that the complex is exact at $R^{n_ e}, \ldots , R^{n_1}$ concluding the proof. $\square$