The Stacks project

Proof. The assumption means, after a change of basis of $R^{n_ i}$ and $R^{n_{i-1}}$ that the first basis vector of $R^{n_ i}$ is mapped via $\varphi _ i$ to the first basis vector of $R^{n_{i-1}}$. Let $e_ j$ denote the $j$th basis vector of $R^{n_ i}$ and $f_ k$ the $k$th basis vector of $R^{n_{i-1}}$. Write $\varphi _ i(e_ j) = \sum a_{jk} f_ k$. So $a_{1k} = 0$ unless $k = 1$ and $a_{11} = 1$. Change basis on $R^{n_ i}$ again by setting $e'_ j = e_ j - a_{j1} e_1$ for $j > 1$. After this change of coordinates we have $a_{j1} = 0$ for $j > 1$. Note the image of $R^{n_{i + 1}} \to R^{n_ i}$ is contained in the subspace spanned by $e_ j$, $j > 1$. Note also that $R^{n_{i-1}} \to R^{n_{i-2}}$ has to annihilate $f_1$ since it is in the image. These conditions and the shape of the matrix $(a_{jk})$ for $\varphi _ i$ imply the lemma. $\square$

Proof. Pick $x \in R$, $x \not= 0$, with $\mathfrak m x = 0$. Let $i$ be the biggest index such that $n_ i > 0$. If $i = 0$, then the statement is true. If $i > 0$ denote $f_1$ the first basis vector of $R^{n_ i}$. Since $xf_1$ is not mapped to zero by exactness of the complex we deduce that some matrix coefficient of the map $R^{n_ i} \to R^{n_{i - 1}}$ is not in $\mathfrak m$. Lemma 10.102.2 then allows us to decrease $n_ e + \ldots + n_1$. Induction finishes the proof. $\square$

Proof. This is a special case of Lemma 10.102.3 because an Artinian local ring has depth $0$. $\square$

Proof. We may assume the complex is the direct sum of trivial complexes. Then for each $i$ we can split the standard basis elements of $R^{n_ i}$ into those that map to a basis element of $R^{n_{i-1}}$ and those that are mapped to zero (and these are mapped onto by basis elements of $R^{n_{i + 1}}$ if $i > 0$). Using descending induction starting with $i = e$ it is easy to prove that there are $r_{i + 1}$-basis elements of $R^{n_ i}$ which are mapped to zero and $r_ i$ which are mapped to basis elements of $R^{n_{i-1}}$. From this the result follows. $\square$

Proof. Denote $F_\bullet$ the complex with terms $F_ i = R^{n_ i}$ and differential given by $\varphi _ i$. Then we have a short exact sequence of complexes

Applying the snake lemma we get a long exact sequence

The lemma follows. $\square$

Proof. Let $H = \mathop{\mathrm{Ker}}(M_ i \to M_{i-1})/\mathop{\mathrm{Im}}(M_{i + 1} \to M_ i)$ be the cohomology group in question. We may break the complex into short exact sequences $0 \to M_ e \to M_{e-1} \to K_{e-2} \to 0$, $0 \to K_ j \to M_ j \to K_{j-1} \to 0$, for $i + 2 \leq j \leq e-2$, $0 \to K_{i + 1} \to M_{i + 1} \to B_ i \to 0$, $0 \to K_ i \to M_ i \to M_{i-1}$, and $0 \to B_ i \to K_ i \to H \to 0$. We proceed up through these complexes to prove the statements about depths, repeatedly using Lemma 10.72.6. First of all, since $\text{depth}(M_ e) \geq e$, and $\text{depth}(M_{e-1}) \geq e-1$ we deduce that $\text{depth}(K_{e-2}) \geq e - 1$. At this point the sequences $0 \to K_ j \to M_ j \to K_{j-1} \to 0$ for $i + 2 \leq j \leq e-2$ imply similarly that $\text{depth}(K_{j-1}) \geq j$ for $i + 2 \leq j \leq e-2$. The sequence $0 \to K_{i + 1} \to M_{i + 1} \to B_ i \to 0$ then shows that $\text{depth}(B_ i) \geq i + 1$. The sequence $0 \to K_ i \to M_ i \to M_{i-1}$ shows that $\text{depth}(K_ i) \geq 1$ since $M_ i$ has depth $\geq i \geq 1$ by assumption. The sequence $0 \to B_ i \to K_ i \to H \to 0$ then implies the result. $\square$

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. $\square$