Associated Primes

Section 12.2 Associated Primes

“People just kind of associate me with kicking some ass.”
―Yancy Butler

Example 12.19. Map from Quotient to Module Injective iff I Annihilates.

Let \(R\) be a ring, \(I\) be an ideal in \(R\text{,}\) and \(M\) be an \(R\)-module. To give an \(R\)-module homomorphism \(R \longrightarrow M\) is the same as choosing an element \(m\) of \(M\) (the image of \(1\) via our map) or equivalently, to choose a cyclic submodule of \(M\) (the submodule generated by \(m\)).

To give an \(R\)-module homomorphism \(R/I \longrightarrow M\) is the same as giving an \(R\)-module homomorphism \(R \longrightarrow M\) whose image is killed by \(I\text{.}\) Thus giving an \(R\)-module homomorphism \(R/I \longrightarrow M\) is to choose an element \(m \in M\) that is killed by \(I\text{,}\) meaning \(I \subseteq \ann(m)\text{.}\) The kernel of the map \(R \to M\) given by \(1 \mapsto m\) is precisely \(\ann(m)\text{,}\) so a well-defined map \(R/I \to M\) given by \(1 \mapsto m\) is injective if and only if \(I = \ann(m)\text{.}\)

Definition 12.20. Associated Prime.

Let \(R\) be a ring and \(M\) an \(R\)-module. We say that \(P \in \Spec(R)\) is an associated prime of \(M\) if \(P = \ann_R(m)\) for some \(m \in M\text{.}\) Equivalently, \(P\) is associated to \(M\) if there is an injective homomorphism \(R/P \longrightarrow M\text{.}\) We write \(\Ass_R(M)\) for the set of associated primes of \(M\text{.}\)

If \(I\) is an ideal, by the associated primes of \(I\) we (almost always) mean the associated primes of the \(R\)-module \(R/I\text{.}\)

Example 12.21. Associated Primes of \((x^2, xy)\) in \(k[[x,y]]\).

Let \(k\) be a field and let \(R = k[[x,y]]\text{.}\) Consider ideal \(I = (x^2, xy)\) and the \(R\)-module \(M = R/I.\) The element \(x + I\) in \(M\) is killed by both \(x\) and \(y\text{,}\) and thus \((x,y) \subseteq \ann (x+I)\text{.}\) On the other hand, \(x+I \neq 0\text{,}\) so \(\ann(x+I) \neq R\text{.}\) Since \((x,y)\) is the unique maximal ideal in \(R,\) we conclude that \(\ann(x+I) = (x,y)\text{.}\) In particular, \((x,y) \in \Ass(M)\text{.}\) Since \(M = R/I\text{,}\) this also says that \((x,y)\) is an associated prime of \(I\text{.}\)

Lemma 12.22. Associated Primes and Localization.

Let \(R\) be a noetherian ring and \(M\) be an \(R\)-module. A prime \(P\) is associated to \(M\) if and if and only if \(P_P \in \Ass(M_P)\text{.}\)

Proof.

localization is exact, so any inclusion \(R/P \subseteq M\) localizes to an inclusion \(R_P/P_P \subseteq M_P\text{.}\)

Conversely, suppose that \(P_P = \ann(\frac{m}{w})\) for some \(\frac{m}{w} \in M_P\text{.}\) Let \(P = (f_1, \ldots, f_n)\text{.}\) For each \(i\text{,}\) since \(\frac{f_i}{1} \frac{m}{w} = \frac{0}{1}\text{,}\) there exists \(u_i \notin P\) such that \(u_i f_i m = 0\text{.}\) Then \(u = u_1 \cdots u_n\) is not in \(P\text{,}\) since \(P\) is prime, and \(u f_i m = 0\) for all \(i\text{.}\) Since the \(f_i\) generate \(P\text{,}\) we have \(P (um) = 0\text{.}\) On the other hand, if \(r \in \ann(um)\text{,}\) then \(\frac{ru}{1} \in \ann(\frac{m}{w}) = P_P\text{.}\) We conclude that \(ru \in P_P \cap R = P\text{.}\) Since \(u \notin P\) and \(P\) is prime, we conclude that \(r \in P\text{.}\)

Lemma 12.23. Associated Primes and Quotients.

If \(P\) is prime, \(\Ass_R(R/P)=\{ P \}\text{.}\)

Proof.

For any nonzero \(r + P \in R/P\text{,}\) we have \(\ann_R(r+P) = \{ s \in R \ | \ rs \in P\} = P\) by definition of prime ideal.

Example 12.24. Associated Primes of \((x^{2})\) in \(k[[x]]\).

Let \(k\) be a field and \(R = k[[x]]\text{.}\) Consider the ideal \(I = (x^2)\text{,}\) and the \(R\)-module \(M = R/I\text{.}\) If \(P\) is a prime ideal in \(R\) and \(P\) is associated to \(M\text{,}\) then \(P = \ann(r+I)\) for some \(r \in R\text{.}\)

Solution.

Since \(I = \ann(M)\text{,}\) \(P\) must contain \(I\text{;}\) since \(P\) is prime and \(x^2 \in P\text{,}\) we conclude that \(P \supseteq (x)\text{.}\) On the other hand, \((x)\) is a maximal ideal, so \(P = (x)\text{.}\) Thus \(\Ass(M) \subseteq \{ (x) \}\text{.}\) Moreover, by an argument similar to the one we used in , we can show that \((x) = \ann(x+I)\text{.}\) Therefore, \(\Ass(M) = \{ (x) \}\text{,}\) and \((x)\) is the unique associated prime of \((x^2)\text{.}\) However, \((x^2)\) is not a prime ideal.

Lemma 12.25. Graded Primes.

If \(R\) is a \(\Z\)-graded ring, then any ideal \(I\) with the property

\begin{equation*} \text{ for any homogeneous elements }r,s\in R, \quad rs\in I \Rightarrow r\in I \ \text{or} \ s\in I \end{equation*}

is prime.

Proof.

We need to show that this property implies that for any \(a,b\in R\) not necessarily homogeneous, \(ab\in I\) implies \(a\in I\) or \(b\in I\text{.}\) We do this by induction on the number of nonzero homogeneous components of \(a\) plus the number of nonzero homogeneous components of \(b\text{.}\) This is not interesting if \(a = 0\) or \(b=0\text{,}\) so the base case is when this is two. In that case, both \(a\) and \(b\) are homogeneous, so the hypothesis already gives us this case. For the induction step, write \(a=a' + a_m\) and \(b=b'+b_n\text{,}\) where \(a_m,b_n\) are the nonzero homogeneous components of \(a\) and \(b\) of largest degree, respectively. We have \(ab=(a'b'+a_m b' + b_n a') + a_m b_n\text{,}\) where \(a_m b_n\) is either the largest homogeneous component of \(ab\) or zero. Either way, \(a_m b_n\in I\text{,}\) so \(a_m\in I\) or \(b_n\in I\text{;}\) without loss of generality, we can assume \(a_m\in I\text{.}\) Then \(a b= a' b + a_m b\text{,}\) and \(ab, a_m b\in I\text{,}\) so \(a' b\in I\text{,}\) and the total number of homogeneous pieces of \(a'b\) is smaller, so by induction, either \(a'\in I\) so that \(a\in I\text{,}\) or else \(b\in I\text{.}\)

Definition 12.26. Module Zerodivisor.

Let \(M\) be an \(R\)-module. An element \(r \in R\) is a zerodivisor on \(M\) if \(rm = 0\) for some nonzero \(m \in M\text{.}\) We denote the set of zerodivisors of \(M\) by \(\mathcal{Z}(M)\).

Lemma 12.27. Every Ann Contained in Ass Prime.

If \(R\) is noetherian, and \(M\) is an arbitrary \(R\)-module, then for any nonzero \(m \in M\text{,}\) \(\ann_R(m)\) is contained in an associated prime of \(M\text{.}\) If \(R\) and \(M\) are graded and \(m\) is a homogeneous element, then \(\ann_R(m)\) is contained in a homogeneous prime.

Proof.

The set of ideals \(S := \{ \ann_R(m) \ | \ m\in M, m \neq0\}\) is nonempty, and any element in \(S\) is contained in a maximal element. Note in fact that any element in \(S\) must be contained in a maximal element of \(S\text{.}\) Let \(I=\ann(m)\) be any maximal element, and let \(rs\in I\text{,}\) \(s\notin I\text{.}\) We always have \(\ann(sm)\supseteq \ann(m)\text{,}\) and equality holds by the maximality of \(\ann(m)\) in \(S\text{.}\) Then \(r(sm)=(rs)m=0\text{,}\) so \(r\in \ann(sm)=\ann(m)=I\text{.}\) We conclude that \(I\) is prime, and therefore it is an associated prime of \(M\text{.}\)

The same argument above works if we take \(\{ \ann_R(m) \ | \ m\in M, m \neq 0 \ \text{homogeneous} \}\text{,}\) using [provisional cross-reference: cite] [[Mathematics/Commutative Algebra/Results/Lemma - Graded Primes|Lemma]].

Theorem 12.28. Associated Primes Nonzero.

If \(R\) is noetherian, and \(M\) is an arbitrary \(R\)-module, then

\begin{equation*} \Ass(M)= \varnothing \Longleftrightarrow M=0. \end{equation*}

If \(R\) and \(M\) are \(\Z\)-graded and \(M\neq 0\text{,}\) then \(M\) has an associated prime that is homogeneous.

Proof.

Even if \(R\) is not noetherian, \(M = 0\) implies \(\Ass(M) = \es\) by definition. So we focus on the case when \(M \neq 0\text{.}\) If \(M \neq 0\text{,}\) then \(M\) contains a nonzero element \(m\text{,}\) and \(\ann(m)\) is contained in an associated prime of \(M\text{.}\) In particular, \(\Ass(M) \neq 0\text{.}\) In the graded setting, [provisional cross-reference: cite] [[Mathematics/Commutative Algebra/Results/Lemma - Every Ann Contained in Ass Prime|Lemma]] gives us a homogeneous associated prime.

Theorem 12.29. Zerodivisors Associated Primes.

If \(R\) is noetherian, and \(M\) is an arbitrary \(R\)-module, then

\begin{equation*} \bigcup\limits_{P \in \Ass(M)} P = \mathcal{Z}(M). \end{equation*}

Proof.

If \(r \in \mathcal{Z}(M)\text{,}\) then by definition we have \(r \in \ann(m)\) for some nonzero \(m \in M\text{.}\) Since \(\ann(m)\) is contained in some associated prime of \(M\text{,}\) then \(r\) is also contained in some associated prime of \(M\text{.}\) On the other hand, if \(P\) is an associated prime of \(M\text{,}\) then by definition all elements in \(P\) are zerodivisors on \(M\text{.}\)

For the graded case, replace the set of zerodivisors with the annihilators of homogeneous elements. Such annihilator is homogeneous, since if \(m\) is homogeneous, and \(fm=0\text{,}\) writing \(f=f_{a_1}+\dots+f_{a_b}\) as a sum of homogeneous elements of different degrees \(a_i\text{,}\) then \(0=fm=f_{a_1}m+\dots+f_{a_b}m\) is a sum of homogeneous elements of different degrees, so \(f_{a_i}m=0\) for each \(i\text{.}\)

Lemma 12.30. Associated Primes and SES.

\begin{equation*} \begin{CD} 0@>>>L@>>>M@>>>N@>>>0 \end{CD} \end{equation*}

is an exact sequence of \(R\)-modules, then \(\Ass(L) \subseteq \Ass(M) \subseteq \Ass(L) \cup \Ass(N)\text{.}\)

Proof.

If \(R/P\) includes in \(L\text{,}\) then composition with the inclusion \(L\hookrightarrow M\) gives an inclusion \(R/P \hookrightarrow M\text{.}\) Therefore, \(\Ass(L) \subseteq \Ass(M)\text{.}\)

Now let \(P \in \Ass(M)\text{,}\) and let \(m \in M\) be such that \(P=\ann(m)\text{.}\) First, note that \(P \subseteq \ann(rm)\) for all \(r \in R\text{.}\)

Thinking of \(L\) as a submodule of \(M\text{,}\) suppose that there exists \(r \notin P\) such that \(rm \in L\text{.}\) Then

\begin{equation*} s(rm) = 0 \Longleftrightarrow (sr)m = 0 \implies sr \in P \implies s \in P. \end{equation*}

So \(P = \ann(rm)\text{,}\) and thus \(P \in \Ass(L)\text{.}\)

If \(rm \notin L\) for all \(r \notin P\text{,}\) let \(n\) be the image of \(m\) in \(N\text{.}\) Thinking of \(N\) as \(M/L\text{,}\) if \(rn = 0\text{,}\) then we must have \(rm \in L\text{,}\) and by assumption this implies \(r \in P\text{.}\) Since \(P = \ann(m) \subseteq \ann(n)\text{,}\) we conclude that \(P = \ann(n)\text{.}\) Therefore, \(P \in \Ass(N)\text{.}\)

Example 12.31. Strict Containment in Ass SES.

If \(M\) is a module with at least two associated primes, and \(P\) is an associated prime of \(M\text{,}\) then

\begin{equation*} \begin{CD} 0@>>>R/P@>>>M \end{CD} \end{equation*}

is exact, but \(\{ P \} = \Ass(R/P) \subsetneq \Ass(M)\text{.}\)

Example 12.32. Strict Containment in Ass SES (2).

Let \(R = k[x]\text{,}\) where \(k\) is a field, and consider the short exact sequence of \(R\)-modules

\begin{equation*} \begin{CD} 0@>>>(x)@>>>R@>>>R/(x)@>>>0 \end{CD} \end{equation*}

Then one can check that: - \(\Ass(R/(x)) = \Ass(k) = \{ (x) \}\text{.}\) - \(\Ass(R) = \Ass((x)) = \{ (0) \}\text{.}\) In particular, \(\Ass(R) \subsetneq \Ass(R/(x)) \cup \Ass((x))\text{.}\)

Corollary 12.33. Ass Direct Sum.

Let \(A\) and \(B\) be \(R\)-modules. Then \(\Ass(A \oplus B) = \Ass(A) \cup \Ass(B)\text{.}\)

Proof.

Apply this [provisional cross-reference: cite] [[Mathematics/Commutative Algebra/Results/Lemma - Associated Primes and SES|Lemma]] to the short exact sequence

\begin{equation*} \begin{CD} 0@>>>A@>>>A\oplus B@>>>B@>>>0 \end{CD} \end{equation*}

We obtain \(\Ass(A) \subseteq \Ass(A \oplus B) \subseteq \Ass(A) \cup \Ass(B)\text{.}\) Repeat with

\begin{equation*} \begin{CD} 0@>>>B@>>>A\oplus B@>>>A@>>>0 \end{CD} \end{equation*}

to conclude that \(\Ass(B) \subseteq \Ass(A \oplus B)\text{.}\) So we have shown both \(\Ass(A) \subseteq \Ass(A \oplus B)\) and \(\Ass(B) \subseteq \Ass(A \oplus B)\text{,}\) so \(\Ass(A) \cup \Ass(B) \subseteq \Ass(A \oplus B)\text{.}\) Since we have also already shown \(\Ass(A \oplus B) \subseteq \Ass(A) \cup \Ass(B)\text{,}\) we must have \(\Ass(A \oplus B) = \Ass(A) \cup \Ass(B)\text{.}\)

Definition 12.34. Shift.

Let \(R\) and \(M\) be \(T\)-graded, and \(t\in T\text{.}\) The shift of \(M\) by \(t\) is the graded \(R\)-module \(M(t)\) with graded pieces \(M(t)_i:=M_{t+i}\text{.}\) This is isomorphic to \(M\) as an \(R\)-module, when we forget about the graded structure.

Theorem 12.35. Filtration.

Let \(R\) be a noetherian ring, and \(M\) is a finitely generated module. There exists a filtration of \(M\)

\begin{equation*} M=M_t \supsetneq M_{t-1} \supsetneq M_{t-2} \supsetneq \cdots \supsetneq M_1 \supsetneq M_0 = 0 \end{equation*}

such that \(M_i / M_{i-1} \cong R/P_i\) for primes \(P_i\in \Spec(R)\text{.}\) This is a prime filtration of \(M\text{.}\)

If \(R\) and \(M\) are \(\Z\)-graded, there exists a prime filtration of \(M\) where the quotients \(M_i / M_{i-1} \cong (R/P_i)(t_i)\) are graded modules, the \(P_i\) are homogeneous primes, and \(t_i \in \Z\text{.}\)

Proof.

If \(M\neq 0\text{,}\) then \(M\) has at least one associated prime, by , so there is an inclusion \(R/P_1 \subseteq M\text{.}\) Let \(M_1\) be the image of this inclusion. If \(M/M_1\neq 0\text{,}\) it has an associated prime, so there is an \(M_2 \subseteq M\) such that \(R/P_2 \cong M_2/M_1 \subseteq R/M_1\text{.}\) Continuing this process, we get a strictly ascending chain of submodules of \(M\) where the successive quotients are of the form \(R/P_i\text{.}\) If we do not have \(M_t=M\) for some \(t\text{,}\) then we get an infinite strictly ascending chain of submodules of \(M\text{,}\) which contradicts that \(M\) is a noetherian module.

In the graded case, if \(P_i\) is the annihilator of an element \(m_i\) of degree \(t_i\text{,}\) we have a degree-preserving map \((R/P_i)(t_i)\cong R m_i\) sending the class of \(1\) to \(m_i\text{.}\)

Example 12.36. Building Prime Filtration.

Let’s build a prime filtration for the module \(M = R/I\text{,}\) where \(I = (x^2,yz)\) and \(R = \mathbb{Q}[x,y,z]\text{.}\) With a little help from Macaulay2, we find that

i4 : associatedPrimes M

o4 = {ideal (y, x), ideal (z, x)}

o4 : List

So our first goal is to find \(m \in M\) such that \(\operatorname{ann}(m) = (x,z)\) or \(\operatorname{ann}(m) = (x,y)\text{.}\) Let’s start from \((x,z)\text{.}\) To find such an element, we can start by searching for all the elements killed by \((x,z)\text{:}\)

i5 : I : ideal"x,z"
                      2
o5 = ideal (y*z, x*y, x )

o5 : Ideal of R

Now \(yz\) and \(x^2\) are both \(0\) in \(M\text{,}\) so the submodule of \(M\) generated by \(xy\) is precisely the set of elements killed by \((x,z)\text{.}\) Is \(\operatorname{ann}(R \cdot xy) = (x,z)\text{?}\)

i6 : ann ((I + ideal(x*y))/I)

o6 = ideal (z, x)

o6 : Ideal of R

Yes, it is! So our prime filtration starts with

\begin{equation*} M_0 = 0 \subseteq M_1 = \frac{I + (xy)}{I}, \end{equation*}

where our computations so far show that \(\operatorname{ann}(M_1) = (x,z)\text{.}\) For step \(2\text{,}\) we start from scratch, and compute the associated primes of \(M/M_1 \cong R/(I+(y))\text{:}\)

i7 : associatedPrimes (R^1/(I + ideal"xy"))

o7 = {ideal (y, x), ideal (z, x)}

o7 : List

Unfortunately, we will again have to find another element killed by \((x,z)\text{.}\) So we repeat the process:

i8 : (I + ideal"xy") : ideal"x,z"
                2
o8 = ideal (y, x )

o8 : Ideal of R

i9 : ann((I + ideal"y")/(I + ideal"xy"))

o9 = ideal (z, x)

o9 : Ideal of R

So in \(M_1\text{,}\) \(\operatorname{ann}(y) = (x,z)\text{,}\) so we can take the submodule generated by \(y\) for our next step, so our prime filtration for now looks like

\begin{equation*} M_0 = 0 \underset{R/(x,z)}{\subseteq} M_1 = \frac{I + (xy)}{I} \underset{R/(x,z)}{\subseteq} M_2 = \frac{I + (y)}{I}. \end{equation*}

So now we repeat the process with \(M/M_2 \cong R/(I + (y))\text{:}\)

i10 : associatedPrimes (R^1/(I + ideal"y"))

o10 = {ideal (y, x)}

o10 : List

i11 : (I + ideal"y") : ideal"x,y"

o11 = ideal (y, x)

o11 : Ideal of R

i12 : ann((I + ideal"x")/(I + ideal"y"))

o12 = ideal (y, x)

o12 : Ideal of R

This gives us

\begin{equation*} M_0 = 0 \underset{R/(x,z)}{\subseteq} M_1 = \frac{I + (xy)}{I} \underset{R/(x,z)}{\subseteq} M_2 = \frac{I + (y)}{I} \underset{R/(x,y)}{\subseteq} M_3 = \frac{I + (x,y)}{I}. \end{equation*}

Next, we take \(M/M_3 \cong R/(I + (x,y))\) and find that

i13 : associatedPrimes (R^1/(I + ideal"x,y"))

o13 = {ideal (y, x)}

o13 : List

i14 : (I + ideal"x,y") : ideal"x,y"

o14 = ideal 1

o14 : Ideal of R

This last computation actually says we are done: since \((x,y)\) kills everything inside \(M/M_3\text{,}\) we can now complete our prime filtration with

\begin{equation*} 0 \underset{R/(x,z)}{\subseteq} M_1 = \frac{I + (xy)}{I} \underset{R/(x,z)}{\subseteq} M_2 = \frac{I + (y)}{I} \underset{R/(x,y)}{\subseteq} M_3 = \frac{I + (x,y)}{I} \underset{R/(x,y)}{\subseteq} M_4 = R/I. \end{equation*}

The prime ideals that appear in this filtration are \((x,y)\) and \((x,z)\text{.}\) From the computation in the beginning, these are precisely the associated primes of \(M\text{.}\)

Corollary 12.37. Ass Finite.

If \(R\) is a noetherian ring, and \(M\) is a finitely generated module, and

\begin{equation*} M=M_t \supsetneq M_{t-1} \supsetneq M_{t-2} \supsetneq \cdots \supsetneq M_1 \supsetneq M_0 = 0 \end{equation*}

is a prime filtration of \(M\) with \(M_i/M_{i-1}\cong R/P_i\text{,}\) then

\begin{equation*} \Ass_R(M)\subseteq \{ P_1, \dots, P_t \}. \end{equation*}

Therefore, \(\Ass_R(M)\) is finite. Moreover, if \(M\) is graded, then \(\Ass_R(M)\) is a finite set of homogeneous primes.

Proof.

For each \(i\text{,}\) we have a short exact sequence

\begin{equation*} \begin{CD} 0@>>>M_{i-1}@>>>M_{i}@>>>M_{i}/M_{i-1}@>>>0 \end{CD} \end{equation*}

By , \(\Ass(M_i) \subseteq \Ass(M_{{i-1}})\cup \Ass(M_i/M_{i-1}) = \Ass(M_{{i-1}}) \cup \{P_i\}\text{.}\) Inductively, we have \(\Ass(M_i)\subseteq\{P_1,\dots,P_i\}\text{,}\) and \(\Ass_R(M) = \Ass_R(M_t) \subseteq \{ P_1, \dots, P_t \}\text{.}\) This immediately implies that \(\Ass(M)\) is finite. In the graded case, gives us a filtration where all the \(P_i\) are homogeneous primes, and those include all the associated primes.

Example 12.38. Prime Ideals are Ass(M) for some M.

Any subset \(X\subseteq\Spec(R)\) (for any \(R\)) can be realized as \(\Ass(M)\) for some module \(M\text{:}\) take \(M=\displaystyle\bigoplus_{P \in X} R/P\text{.}\) However, \(M\) is not finitely generated when \(X\) is infinite.

Solution.

Let \(X\) be a subset of \(\Spec(R)\text{.}\) We construct an \(R\)-module \(M\) such that \(\Ass(M) = X\text{.}\)

Let \(I = \bigcap_{\fp \in X} \fp\) and consider the \(R\)-module \(M = R/I\text{.}\) Then for any prime ideal \(\fp \in \Spec(R)\text{,}\) we have \(\fp \in \Ass(M)\) if and only if \(\fp \supseteq I\text{.}\) Thus, we have \(\Ass(M) = { \fp \in \Spec(R) \mid \fp \supseteq I }\text{.}\)

It remains to show that \(I\) is an ideal in \(R\text{.}\) Since \(X\) is a subset of \(\Spec(R)\text{,}\) we have \(R \setminus X \subseteq \bigcup_{\fp \in X} \fp^c\text{,}\) where \(\fp^c\) denotes the complement of \(\fp\) in \(R\text{.}\) Taking the intersection of both sides with \(X\text{,}\) we have \(X \subseteq \bigcup_{\fp \in X} \fp^c \cap X\text{.}\) Since \(\fp^c \cap X\) is closed under finite intersection and contains the empty set, it is a closed set of \(\Spec(R)\text{.}\) Hence, we can apply the prime avoidance lemma to get \(I = \bigcap_{\fp \in X} \fp \subseteq \bigcap_{\fp \in X} \fp^c \cap X = \emptyset\text{.}\) Thus, \(I\) is an ideal in \(R\text{.}\)

Example 12.39. Module with no Associated Primes.

If \(R\) is not noetherian, then there may be modules (or ideals even) with no associated primes. Let \(R=\bigcup_{n\in \N} \C[[x^{1/n}]]\) be the ring of nonnegatively-valued Puiseux series. We claim that \(R/(x)\) is a cyclic module with no associated primes, i.e., the ideal \((x)\) has no associated primes.

Solution.

First, observe that any element of \(R\) can be written as a unit times \(x^{m/n}\) for some \(m,n\text{,}\) so any associated prime of \(R/(x)\) must be the annihilator of \(x^{m/n}+(x)\) for some \(m \leq n\text{.}\) However, we claim that these are never prime. Indeed, we have \(\ann(x^{m/n}+(x))=(x^{1-m/n})\text{,}\) which is not prime since \((x^{1/2-m/2n})^2 \in (x^{1-m/n})\) but \(x^{1/2-m/2n} \notin (x^{1-m/n})\text{.}\)

Theorem 12.40. Associated Primes Localize in Noetherian Rings.

Let \(R\) be a noetherian ring, \(W\) a multiplicative set, and \(M\) a module. Then

\begin{equation*} \Ass_{W^{-1}R}(W^{-1}M)=\{W^{-1}P \ | \ P \in \Ass_R(M), \, P \cap W = \varnothing\}. \end{equation*}

Proof.

Given \(P \in \Ass_R(M)\) such that \(P \cap W= \varnothing\text{,}\) [provisional cross-reference: cite] [[Mathematics/Commutative Algebra/Results/Lemma - Localization Ideals|Lemma]] says that \(W^{-1}P\) is a prime in \(W^{-1}R\text{.}\) Then \(W^{-1}R /W^{-1}P \cong W^{-1}(R/P) \hookrightarrow W^{-1}M\) by exactness, so \(W^{-1}P\) is an associated prime of \(W^{-1} M\text{.}\)

Now suppose that \(Q \in \Spec(W^{-1}R)\) is associated to \(W^{-1}M\text{.}\) By [provisional cross-reference: cite] [[Mathematics/Commutative Algebra/Results/Lemma - Localization Ideals|Lemma]], we know this is of the form \(W^{-1}P\) for some prime \(P\) in \(R\) such that \(P \cap W = \varnothing\text{.}\) Since \(R\) is noetherian, \(P\) is finitely generated, say \(P = (f_1, \ldots, f_n)\) in \(R\text{,}\) and so \(Q = \left( \frac{f_1}{1}, \ldots, \frac{f_n}{1} \right)\text{.}\)

By assumption, \(Q = \ann(\frac{m}{w})\) for some \(m \in M\text{,}\) \(w \in W\text{.}\) Since \(w\) is a unit in \(W^{-1}R\text{,}\) we can also write \(Q = \ann(\frac{m}{1})\text{.}\) By definition, this means that for each \(i\)

\begin{equation*} \frac{f_i}{1} \frac{m}{1} = \frac{0}{1} \, \Longleftrightarrow \, u_i f_i m = 0 \textrm{ for some } u_i \in W. \end{equation*}

Let \(u = u_1 \cdots u_n \in W\text{.}\) Then \(u f_i m = 0\) for all \(i\text{,}\) and thus \(P u m = 0\text{.}\) We claim that in fact \(P = \ann(um)\) in \(R\text{.}\) Consider \(v \in \ann(um)\text{.}\) Then \(u (vm) = 0\text{,}\) and since \(u \in W\text{,}\) this implies that \(\frac{vm}{1} = 0\text{.}\) Therefore, \(\frac{v}{1} \in \ann(\frac{m}{1}) = W^{-1}P\text{,}\) and \(vw \in P\) for some \(w \in W\text{.}\) But \(P \cap W = \varnothing\text{,}\) and thus \(v \in P\text{.}\) Thus \(P \in \Ass(M)\text{.}\)

Theorem 12.41. Support and Associated Primes.

Let \(R\) be noetherian and \(M\) be an \(R\)-module.

\begin{equation*} \Supp(M) \,\, = \displaystyle\bigcup_{P \in \Ass(M)} V(P). \end{equation*}

Proof.

Let \(P \in \Ass_R(M)\text{,}\) and fix \(m \in M\) such that \(P = \ann_R(m)\text{.}\) Let \(Q \in V(P)\text{.}\) By [provisional cross-reference: cite] [[Mathematics/Commutative Algebra/Results/Proposition - Supp(M) = Ann(M)|Proposition]], \(Q \in \Supp(R/P)\text{.}\) Since \(0 \to R/P \xrightarrow{m} M\) is exact and localization is exact, \(0 \to (R/P)_Q \to M_{Q}\) is also exact. Since \((R/P)_Q \neq 0\text{,}\) we must also have \(M_{Q} \neq 0\text{,}\) and thus \(Q \in \Supp(M)\text{.}\) This shows \(\Supp(M) \, \supseteq \displaystyle\bigcup_{P \in \Ass(M)} V(P)\text{.}\) Now let \(Q\) be a prime ideal and suppose that

\begin{equation*} Q \notin \bigcup_{P\in \Ass(M)} V(P). \end{equation*}

In particular, \(Q\) does not contain any associated prime of \(M\text{.}\) Then there is no associated prime of \(M\) that does not intersect \(R \setminus Q\text{,}\) so \(\Ass_{R_{Q}}(M_{Q})=\varnothing\text{.}\) By [provisional cross-reference: cite] [[Mathematics/Commutative Algebra/Results/Lemma - Nonzero Element Localizes|Lemma]], \(M_{Q}=0\text{.}\)

Theorem 12.42. Min Contained in Ass.

Let \(R\) be noetherian and \(M\) be an \(R\)-module. If \(M\) is a finitely generated \(R\)-module, then \(\Min(\ann_R(M))\subseteq \Ass_R(M)\text{.}\) In particular, \(\Min(I)\subseteq \Ass_R(R/I)\text{.}\)

Proof.

By [provisional cross-reference: cite] [[Mathematics/Commutative Algebra/Results/Theorem - Support and Associated Primes|Theorem]],

\begin{equation*} V(\ann_R(M))=\Supp_R(M)=\bigcup_{P \in \Ass(M)} V(P), \end{equation*}

so the minimal elements of both sets agree. In particular, the right hand side has the minimal primes of \(\ann_R(M)\) as minimal elements, and they must be associated primes of \(M\text{,}\) or else this would contradict minimality.[^3

Definition 12.43. Embedded Prime.

If \(I\) is an ideal, then an associated prime of \(I\) that is not a minimal prime of \(I\) is called an embedded prime of \(I\text{.}\)

Theorem 12.44. Ideal of Zerodivisors.

Let \(I\) be an ideal and \(M\) a finitely generated module over a noetherian ring \(R\text{.}\) If \(I\) consists of zerodivisors on \(M\text{,}\) then \(Im=0\) for some nonzero \(m\in M\text{.}\)

Proof.

The assumption says that

\begin{equation*} I\subseteq \displaystyle\bigcup_{P \in \Ass(M)}(P). \end{equation*}

By the assumptions, [provisional cross-reference: cite] [[Mathematics/Commutative Algebra/Results/Corollary - Ass Finite|Corollary]] applies, and it guarantees that this is a finite set of primes. By [provisional cross-reference: cite] [[Mathematics/Commutative Algebra/Results/Lemma - Prime Avoidance|prime avoidance]], \(I \subseteq P\) for some \(P \in \Ass(M)\text{.}\) Equivalently, \(I\subseteq \ann_R(m)\) for some nonzero \(m\in M\text{.}\)

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