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Postmodern Algebra

Sam Macdonald

Section 16.1 Depth and Grade

Definition 16.1.

Let \(R\) be a ring, \(M\) an \(R\)-module, and \({\bf x}=x_1,\dots,x_n\in R\text{.}\) We say \({\bf x}\) is a regular sequence on \(M\), or simply an \(M\)-sequence, if \((x_1,\dots, x_n)M\neq M\) and \(x_i\) is regular (i.e., a non-zero-divisor) on \(M/(x_1,\dots,x_{i-1})M\) for each \(i=1,\dots,n\text{.}\) An \(M\)-sequence \({\bf x}\) is called maximal if \({\bf x}\) cannot be extended to a longer \(M\)-sequence.

Example 16.2.

Example 49. Let \(R=k[x_1,\dots,x_n]\) be a polynomial ring over the field \(k\text{.}\) Then \(x_1,\dots,x_n\) is a maximal \(R\)-sequence.

Proof.

For part (a), note that if \(x_1,\dots,x_n\) is an \(M\)-sequence then
\begin{equation*} 0\subset (x_1)M\subset (x_1,x_2)M\subset \cdots \subset (x_1,\dots,x_n)M \end{equation*}
is a strictly ascending chain of submodule of \(M\text{.}\) For if \((x_1,\dots,x_i)M=(x_1,\dots,x_{i+1})M\) for some \(i\text{,}\) then \(x_{i+1}M\subseteq (x_1,\dots,x_i)M\text{.}\) As \(M\neq (x_1,\dots,x_i)M\text{,}\) then \(x_{i+1}\) is a zero-divisor on \(M/(x_1,\dots,x_i)M\text{,}\) a contradiction. Consequently, as \(M\) is Noetherian, this chain must terminate. Hence, there are no \(M\)-sequences of infinite length. The same argument (using ACC) proves (b) as well. 

Example 16.4.

Let \(R=k[x,y,x]\text{.}\) Then \(x,y,z\) is a maximal \(R\)-sequence.

Example 16.5.

Let \(R=k[x,y,z]_{(x,y,z)}\) and \(M=R/(xy,yz)\text{.}\) Then \(y-z\) is a maximal \(M\) -sequence. To see this, note that \((xy,xz)=(y)\cap (x,z)\) is a irredundant primary decomposition of \(I\text{.}\) Thus, \((y)\) and \((x,z)\) are the associated primes of \(M\text{.}\) Since \(y-z\) is not in either associated prime, it is a non-zero-divisor on \(M\text{.}\) Also, \(M\neq (y-z)M\text{,}\) so \(y-z\) is \(M\)-regular. Note that \(M/(y-z)M\cong k[x,y,z]_{(x,y,z)}/(xy, xz, y-z)\cong k[x,y]_{(x,y)}/(xy, y^2)\text{.}\) Since \((x,y)x=0\) in \(k[x,y]_{(x,y)}/(xy, y^2)\text{,}\) \((x,y)\) consists of zero-divisors on \(M/(y-z)M\text{.}\) Thus, \(y-z\) is a maximal \(M\)-sequence.
Here we summarize some essential facts about primary decompositions for modules, and then use them to prove Krull’s Intersection Theorem.

Definition 16.6.

Let \(M\) be an \(R\)-module. A submodule \(Q\) of \(M\) is called primary if \(Q\subsetneq M\) and for every \(r\in R\) multiplication by \(r\) on \(M/Q\) is either injective or nilpotent; i.e., \(r\) is either a non-zero-divisor on \(M/Q\) or \(r^nM\subseteq Q\) for some \(n\text{.}\)

Remark 16.7.

Let \(Q\) be a primary submodule of \(M\text{.}\) Then \(p:=\sqrt{\operatorname{Ann}_R M/Q}\) is a prime ideal of \(R\text{.}\) We say that \(Q\) is a \(p\)-primary submodule of \(M\text{.}\)

Proof.

See Atiyah-Macdonald. ◻

Proof.

Proof. Let \(\displaystyle{N=\bigcap_{n=1}^{\infty} I^nM}\text{.}\) We claim that \(IN=N\text{.}\) If \(IN=M\text{,}\) there is nothing to prove. Suppose \(IN\subsetneq M\) and let \(IN=Q_1\cap Q_2\cap \cdots \cap Q_r\) be a primary decomposition of \(IN\subset M\text{.}\) Then for each \(i\text{,}\) \(IN\subseteq Q_i\text{.}\) If \(N\not\subset Q_i\) then \(I\) consists of zero-divisors on \(M/Q_i\text{.}\) As \(Q_i\) is a primary submodule of \(M\text{,}\) we must have \(I^nM\subseteq Q_i\) for some \(n\text{.}\) But \(N\subseteq I^nM\text{,}\) so \(N\subseteq Q_i\text{,}\) a contradiction. Thus, \(N\subseteq Q_1\cap\cdots \cap Q_r=IN\text{.}\) Consequently, \(IN=N\text{.}\) By a homework exercise, this implies there exists \(s\in I\) such that \((1-s)N=0\text{.}\) ◻

Definition 16.10.

Let \(R\) be a ring. The Jacobson radical of \(R\), denoted \(J(R)\text{,}\) is the intersection of all maximal ideals of \(R\text{.}\) It is easily seen that if \(r\in J(R)\) then \(1-r\) is a unit.

Proof.

Apply Krull’s Intersection Theorem and use that \(1-s\) is a unit for every \(s\in I\text{.}\)

Proof.

It suffices to show that if \(x,y\in J(R)\) is an \(M\)-sequence then so is \(y,x\text{.}\) First note that \((y,x)M=(x,y)M\neq M\text{.}\) We next show that \(y\) is regular on \(M\text{:}\) suppose \(yu=0\) for some \(u\in M\text{.}\) Then \(yu\in xM\) so \(u\in xM\text{.}\) Write \(u=xu_1\) where \(u_1\in M\text{.}\) Then \(0=yu=xyu_1\text{,}\) so \(yu_1=0\text{.}\) Repeating the argument above, we get \(u_1\in xM\) and hence \(u\in (x)^2M\text{.}\) Continuing in this way, we obtain that \(u\in \bigcap_{n\geqslant 1} (x)^nM\text{.}\) As \((x)\subseteq J(R)\) we have \(\bigcap_{n\geqslant 1} (x)^nM=0\) and thus \(u=0\text{.}\) This shows \(y\) is regular on \(M\text{.}\) Now assume \(xv\in yM\) for some \(v\in M\text{.}\) Then \(xv=yw\) for some \(w\in M\text{.}\) Since \(y\) is regular on \(M/xM\text{,}\) we obtain that \(w=xz\) for some \(z\in M\text{.}\) Consequently, \(xv=xyz\text{.}\) As \(x\) is regular on \(M\text{,}\) we then have \(v=yz\in yM\text{.}\) Hence, \(x\) is regular on \(M/yM\text{.}\)

Definition 16.13.

Let \(R\) be a Noetherian ring, \(I\) an ideal of \(R\text{,}\) and \(M\) a finitely generated \(R\)-module such that \(IM\neq M\text{.}\) Then the grade of \(I\) on \(M\), denote \(\operatorname{grade}_I M\) or \(\operatorname{grade}(I,M)\text{,}\) is defined to be the supremum of the lengths of all \(M\)-sequences contained in \(I\text{.}\)

Remark 16.14.

Note that it is not clear from the definition that \(\operatorname{grade}_I M<\infty\text{.}\) Although every maximal \(M\)-sequence is finite, the supremum of the lengths of such sequences might be infinite.

Convention 16.15.

Let \(R\) be a ring and \(L\) and \(M\) \(R\)-modules. Define
\begin{equation*} g(L,M):=\inf \{n\mid \operatorname{Ext}^n_R(L,M)\neq 0\}. \end{equation*}
Note that \(g(L,M)\geqslant 0\) for all \(R\)-modules \(L\) and \(M\text{.}\) (Recall \(\inf \emptyset=\infty\text{.}\))

Proof.

Let \(f:L\to M\) be a homomorphism. Then \(xf(L)=f(xL)=f(0)=0\text{.}\) As \(x\) is regular on \(M\text{,}\) we see that \(f(L)=0\text{.}\) Hence, \(\operatorname{Hom}_R(L,M)=0\) and so \(g>0\text{.}\) Now applying \(\operatorname{Hom}_R(L,-)\) to the s.e.s. \(0\to M\xrightarrow{x} M\to M/xM\to 0\) we get the long exact sequence
\begin{equation*} \cdots\to \operatorname{Ext}^i_R(L,M)\xrightarrow{x} \operatorname{Ext}^i_R(L,M)\to \operatorname{Ext}^i_R(L,M/xM) \to \operatorname{Ext}^{i+1}_R(L,M)\xrightarrow{x} \cdots \end{equation*}
Since \(xL=0\) we have \(x\operatorname{Ext}^i_R(L,M)=0\) for all \(i\) (cf. Grifo’s 915 notes, Exercise 73(c).) Thus, for each \(i\) we have an exact sequence
\begin{equation*} 0\to \operatorname{Ext}^i_R(L,M)\to \operatorname{Ext}^i_R(L,M/xM) \to \operatorname{Ext}^{i+1}_R(L,M)\to 0. \end{equation*}
Both conclusions follow easily from this sequence.

Proof.

Notice that the right-hand-side is \(g:=g(R/I,M)\text{.}\) Suppose \(r=0\text{.}\) Since \(IM\neq M\text{,}\) it must be that \(I\) consists of zero-divisors on \(M\text{.}\) Thus, \(I\) is contained in the union of the associated primes of \(M\) (Grifo 905, Theorem 6.27). By the prime avoidance lemma (Grifo 905, Theorem 3.29), \(I\subseteq p\) for some associated prime \(p\) of \(M\text{.}\) Then the composition \(R/I\to R/p\hookrightarrow M\) is nonzero. Hence, \(\operatorname{Hom}_R(R/I,M)=\operatorname{Ext}^0_R(R/I,M)\neq 0\) and \(g=0=r\text{.}\) Proceeding by induction on \(r\text{,}\) we may assume \(r>0\) and that the result holds for all finitely generated \(R\)-modules \(N\) with \(IN\neq N\) and having a maximal \(N\)-sequence of length at most \(r-1\text{.}\) Let \(N=M/x_1M\text{.}\) Then \(IN\neq N\) and \(x_2,\dots, x_r\) is a maximal \(N\)-sequence contained in \(I\text{.}\) Hence, \(r-1=g(R/I,N)=g-1\) by Proposition Proposition 16.16. Thus, \(r=g\text{,}\) which completes the proof.

Proof.

For the first equality, observe that any sequence \({\bf y}\) in \(I\) is an \(M/xM\)-sequence if and only if its image \({\bf \overline{y}}\) in \(I/(x)\) is an \(M/xM\)-sequence. For the second equality, we have by Theorem Theorem 16.17 and Proposition Proposition 16.16,
\begin{equation*} \operatorname{grade}_I M/xM=g(R/I,M/xM)=g(R/I,M)-1=\operatorname{grade}_I M-1. \end{equation*}

Definition 16.19.

Let \((R,m)\) be a local ring and \(M\) a finitely generated \(R\)-module. Then the depth of \(M\), denoted \(\operatorname{depth}M\text{,}\) is defined to be \(\operatorname{grade}_m M\text{;}\) i.e., the length of the longest \(M\)-sequence from \(R\text{.}\)

Remark 16.20.

Let \(R\) be a local ring and \(M\) a finitely generated \(R\)-module. By Theorem Theorem 16.17, we have
\begin{equation*} \operatorname{depth}M=\inf \{n \mid \operatorname{Ext}^n_R(R/m,M)\neq 0\}. \end{equation*}
Note: By convention, the depth of the zero module is infinity, since \(\inf \emptyset = \infty\text{.}\)

Proof.

First, if either module is zero the result trivially holds. So assume \(M\) and \(N\) are nonzero and let \(d=\operatorname{dim}M\text{.}\) If \(d=0\) then \(M\) has finite length. Since \(\operatorname{Ext}^i_R(R/m,N)=0\) for all \(i<\operatorname{depth}N\) by Remark Remark 16.20, we have \(\operatorname{Ext}^i_R(M,N)=0\) for all \(i<\operatorname{depth}N\) by Lemma [cross-reference to target(s) "bass3" missing or not unique]. Hence, the result holds for the case \(d=0\text{.}\)
Suppose \(d>0\) and assume the result holds for all finitely generated modules of dimension less than \(d\text{.}\) Consider a filtration of \(M\text{:}\)
\begin{equation*} 0=M_0\subset M_1\subset \cdots \subset M_n=M \end{equation*}
where \(M_j/M_{j-1}\cong R/p_j\) for some primes \(p_1,\dots,p_n\) of \(R\text{.}\) (See Grifo 905 :Theorem 6.33.) Note that each \(p_j\) contains \(\operatorname{Ann}_R M\text{,}\) so \(\operatorname{dim}R/p_j\leqslant d\) for all \(j\text{.}\) Hence, if we show \(\operatorname{Ext}^i_R(R/p_j,N)=0\) for all \(i<\operatorname{depth}N-\operatorname{dim}R/p_j\) and all \(j\text{,}\) then we’ll have \(\operatorname{Ext}^i_R(M_j/M_{j-1},N)=0\) for all \(i<\operatorname{depth}N-d\) and all \(j\text{.}\) Using the long exact sequences on \(\operatorname{Ext}\) arising from the short exact sequences
\begin{equation*} 0\to M_{j-1}\to M_j\to M_j/M_{j-1}\to 0, \end{equation*}
we can conclude \(\operatorname{Ext}^i_R(M_j,N)=0\) for all \(i<\operatorname{depth}N-d\text{.}\) Since \(M_n=M\text{,}\) we’ll be done.
Thus, assume \(M\cong R/p\) for some prime \(p\) with \(\operatorname{dim}R/p=d>0\text{.}\) Since \(p\subsetneq m\text{,}\) let \(x\in m\setminus p\) and consider the short exact sequence
\begin{equation*} 0\to R/p\xrightarrow{x} R/p\to R/(p,x)\to 0. \end{equation*}
Since \(\operatorname{dim}R/(p,x)\leqslant d-1\text{,}\) we have \(\operatorname{Ext}^i_R(R/(p,x), N)=0\) for all \(i<\operatorname{depth}N-d+1\) by the induction hypothesis. Rewriting this, we have \(\operatorname{Ext}^{i+1}_R(R/(p,x),N)=0\) for all \(i<\operatorname{depth}N-d\text{.}\) From the long exact sequence on \(\operatorname{Ext}\text{,}\) we have
\begin{equation*} \operatorname{Ext}^i_R(R/p,N)\xrightarrow{x} \operatorname{Ext}^i_R(R/p,N)\to \operatorname{Ext}^{i+1}_R(R/(p,x),N). \end{equation*}
By Nakayama’s Lemma, we see that \(\operatorname{Ext}^i_R(R/p,N)=0\) for all \(i<\operatorname{depth}N-d\text{.}\) This completes the proof.

Proof.

Let \(p\in \operatorname{Ass}_R M\text{.}\) Then \(\operatorname{Ext}^0_R(R/p,M)=\operatorname{Hom}_R(R/p,M)\neq 0\text{.}\) By Theorem Theorem 16.21, \(\operatorname{depth}M\leqslant\operatorname{dim}R/p\text{.}\) ◻

Example 16.23.

Let \(k\) be a field and \(R=k[x_1,\dots,x_{d+1}]/(x_1,\dots,x_{d+1})(x_{d+1})\text{.}\) Let \(m=(x_1,\dots,x_{d+1})R\text{.}\) Then \(R_m\) has depth zero and dimension \(d\text{.}\) To see \(\operatorname{depth}R_m=0\text{,}\) note that \(mx_{d+1}=0\) in \(R_m\text{,}\) so \(mR_m\) consists of zero-divisors. For \(\operatorname{dim}R_m\text{,}\) note that \(\operatorname{dim}k[x_1,\dots,x_{d+1}]_m\) is a \((d+1)\)-dimensional domain, so \(\operatorname{dim}R_m/I\leqslant d\) for any nonzero ideal \(I\text{.}\) On the other hand, \(R_m/(x_{d+1})R_m\cong k[x_1,\dots,x_d]_{(x_1,\dots,x_d)}\text{,}\) which has dimension \(d\text{.}\) Thus, \(\operatorname{dim}R_m\geqslant d\text{.}\)
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