The Krull dimension of the semiring of natural numbers is equal to 2

Let $R$ be a ring. Its Krull dimension is the supremum of the lengths $n$ of chains $P_0\subsetneq P_1 \subsetneq\dots\subsetneq P_n$ of prime ideals of $R$. When $R$ is a field, the null ideal is the only prime ideal, and it is a maximal ideal so that its Krull dimension is zero. When $R$ is a principal ideal domain which is not a field, there are two kinds of prime ideals: the null ideal is prime (as in any domain), and the other prime ideals are the maximal ideals of $R$, generated by a prime element $p$. In particular, the Krull dimension of the ring of integers is equal to $1$.

It is classic that these concepts can be defined for semirings as well.

A semiring $R$ is a set endowed with a commutative and associative addition with a neutral element $0$, an associative multiplication with a neutral element $1$, such that addition distributes over multiplication: $(a+b)c=ac+bc$ and $c(a+b)=ca+cb$. When its multiplication is commutative, the semiring is said to be commutative.

Semirings $R$ have ideals: these are nonempty subsets $I$ which are stable under addition ($a+b\in I$ for $a,b\in I$), and stable under multiplication by any element of $R$: for general semirings, one has to distinguish between left, right, and two-sided ideals; for commutative semirings, the notions coincide.

An ideal $P$ of a semiring $R$ is said to be prime if $R\setminus P$ is a multiplicative subset; explicitely, $P\neq R$, and if $ab\in P$, then $a\in P$ or $b\in P$.

An ideal $P$ of a semiring $R$ is said to be maximal if $P\neq R$ and if there is no ideal $I$ such that $P\subsetneq I\subsetneq R$. A semiring $R$ is said to be local if it admits exactly one maximal ideal; this means that the set of non-invertible elements of $R$ is an ideal.

The amusing part comes from the classification of prime and maximal ideals of the semiring $\mathbf N$ of natural numbers, which I learned of via a Lean formalization project led by Junyan Xu.

Theorem.

1. The semiring $\mathbf N$ is local; its maximal ideal is the set $\mathbf N\setminus\{1\}$.
2. The null ideal is a prime ideal.
3. The other prime ideals are the sets $p\mathbf N$, for all prime numbers $p$.

In particular, we have chains $\langle 0\rangle \subsetneq \langle p\rangle \subsetneq \mathbf N\setminus\{1\}$ of prime ideals, hence the result stated in the title of this post:

Corollary. — The Krull dimension of the semiring of natural numbers is equal to $2$.

Proof.

1. The element $1$ is the only unit of $\mathbf N$, and $\mathbf N\setminus\{1\}$ is obviously an ideal, necessarily its unique maximal ideal.
2. The null ideal is a prime ideal, as in any semiring which is a domain.
3. Let now $P$ be a nonzero prime ideal of $\mathbf N$ and let $p$ be the smallest nonzero element of $P$. Then $p\neq 1$ (otherwise, $P=\mathbf N$, which is not prime). The hypothesis that $P$ is prime implies that one of the prime factors of $p$ belongs to $p$; by the choice of $p$, this must be $p$ itself, so that $p$ is a prime number. Then $P$ contains the set $p\mathbf N $ of multiples of $p$, which is a prime ideal of $\mathbf N$. Let us assume that $p\mathbf N\subsetneq P$ and let $n\in P\setminus p\mathbf N$. By assumption, $p$ does not divide $n$, so that these two integers are coprime. By the following proposition, $P$ contains every integer at least equal to $(p-1)(n-1)$; in particular, it contains both a power of $2$ and a power of $3$; since $P$ is prime, it contains $2$ and $3$, and it contains any integer at least $(2-1)(3-1)=2$, hence $P=\mathbf N\setminus\{1\}$. This concludes the proof.

Proposition. — Let $a$ and $b$ be nonzero coprime natural numbers. For any integer $n\geq (a-1)(b-1)$, there are natural numbers $u$ and $v$ such that $n=au+bv$.

Proof. — Since $a$ and $b$ are coprime, there are integers $u$ and $v$ such that $n=au+bv$. Replacing $u$ by $u+b$ and $v$ by $v-a$, we may assume that $0\leq u$. Replacing $u$ by $u-b$ and $v$ by $v+a$, we may assume that $u < b$. Then \[ bv = n - au \geq (a-1)(b-1)-a(b-1)= -(b-1). \] This implies $v\geq0$, so that $u$ and $v$ are natural numbers.

On the other hand, not all natural numbers $< (a-1)(b-1)$ can be written as such as sum. For example, $ab-a-b$ can't. Indeed, if $ab-a-b=au+bv$, hence $ab=a(u+1)+b(v+1)$, then $b$ divides $a(u+1)$, hence $b$ divides $u+1$, and similarly $a$ divides $v+1$. Then $ab$ is the sum of two nonzero multiples of $ab$, a contradiction. The precise distribution of the natural numbers $< (a-1)(b-1)$ which can be written as $au+bv$, for some natural numbers $u$ and $v$ is complicated, but at least one result is known : such an integer $n$ can be written in this form if and only if $ab-a-b-n$ cannot! One direction is clear, if $n$ and $ab-a-b-n$ can both be written in this form, then so can their sum, which is $ab-a-b$, a contradiction. On the other hand, let $n$ be an integer that cannot be written in this form, and write it as $n=au+bv$, for some integers $u$ and $v$, with $0\leq u< b$. By assumption, $v<0$, hence $v\leq -1$. Then \[ ab-a-b-n=ab-a-b-au-bv=a(b-1-u)+b(-v-1).\] We see that $b-1-u\geq 0$ and $-v-1\geq 0$, which shows that $ab-a-b-n$ can be written in the desired form.

Another open question of the style of the proposition had been raised by Frobenius: consider mutually coprime integers $a_1,\dots,a_r$; then any large enough integer $n$ can be written as $n=a_1u_1+\dots+a_ru_r$, for some natural numbers $u_1,\dots,u_r$, but when $r\geq 3$, there is no known formula for largest natural number that cannot be written in this form. The case $r=2$ that we discussed here was due to Sylvester (1884).

Flatness and projectivity: when is the localization of a ring a projective module?

Projective modules and flat modules are two important concepts in algebra, because they characterize those modules for which a general functorial construction (Hom module and tensor product, respectively) behave better than what is the case for general modules.

This blog post came out of reading a confusion on a student's exam: projective modules are flat, but not all flat modules are projective. Since localization gives flat modules, it is easy to obtain a an example of a flat module which is not projective (see below, $\mathbf Q$ works, as a $\mathbf Z$-module), but my question was to understand when the localization of a commutative ring is a projective module.

$\gdef\Hom{\operatorname{Hom}}\gdef\Spec{\operatorname{Spec}}\gdef\id{\mathrm{id}}$

Let me first recall the definitions. Let $R$ be a ring and let $M$ be a (right)$R$-module.

The $\Hom_R(M,\bullet)$-functor associates with a right $R$-module $X$ the abelian group $\Hom_R(M,X)$. By composition, any linear map $f\colon X\to Y$ induces an additive map $\Hom_R(M,f)\colon \Hom_R(M,X)\to \Hom_R(M,X)$: it maps $u\colon M\to X$ to $\phi\circ u$. When $R$ is commutative, these are even $R$-modules and morphisms of $R$-modules. If $f$ is injective, $\Hom_R(M,f)$ is injective as well, but if $f$ is surjective, it is not always the case that $\Hom_R(M,f)$ is surjective, and one says that the $R$-module $M$ is projective if $\Hom_R(M,f)$ is surjective for all surjective linear maps $f$.

The $\otimes_R$-functor associates with a left $R$-module $X$ the abelian group $M\otimes_R X$, and with any linear map $f\colon X\to Y$, the additive map $M\otimes_R X\to M\otimes_R Y$ that maps a split tensor $m\otimes x$ to $m\otimes f(x)$. When $R$ is commutative, these are even $R$-modules and morphisms of $R$-modules. If $f$ is surjective, then $M\otimes_R f$ is surjective, but if $f$ is injective, it is not always the case that $M\otimes_R f$ is injective. One says that $M$ is flat if $M\otimes_R f$ is injective for all injective linear maps $f$.

These notions are quite abstract, and the development of homological algebra made them prevalent in modern algebra.

Example. — Free modules are projective and flat.

Proposition. — An $R$-module $M$ is projective if and only if there exists an $R$-module $N$ such that $M\oplus N$ is free.
Indeed, taking a generating family of $M$, we construct a free module $L$ and a surjective linear map $u\colon L\to M$. Since $M$ is projective, the map $\Hom_R(M,u)$ is surjective and there exists $v\colon M\to L$ such that $u\circ v=\id_M$. Then $v$ is an isomorphism from $M$ to $u(M)$, and one can check that $L=u(M)\oplus \ker(v)$.

Corollary. — Projective modules are flat.

Theorem (Kaplansky). — If $R$ is a local ring, then a projective $R$-module is free.

The theorem has a reasonably easy proof for a finitely generated $R$-module $M$ over a commutative local ring. Let $J$ be the maximal ideal of $R$ and let $k=R/J$ be the residue field. Then $M/JM$ is a finite dimensional $k$-vector space; let us consider a family $(e_1,\dots,e_n)$ in $M$ whose images form a basis of $M/JM$. Now, one has $\langle e_1,\dots,e_n\rangle + J M = M$, hence Nakayama's lemma implies that $M=\langle e_1,\dots,e_n\rangle$. Let then $u\colon R^n\to M$ be the morphism given by $u(a_1,\dots,a_n)=\sum a_i e_i$; by what precedes, it is surjective, and we let $N$ be its kernel. Since $M$ is projective, the morphism $\Hom_R(M,u)$ is surjective, and there exists $v\colon M\to R^n$ such that $u\circ v=\id_M$. We then have an isomorphism $M\oplus N\simeq R^n$, where $N=\ker(v)$. Moding out by $J$, we get $M/JM \oplus N/JN \simeq k^n$. Necessarily, $N/JN=0$, hence $N=JN$; since $N$ is a direct summand of $R^n$, it is finitely generated, and Nakayama's lemma implies that $N=0$.

Example. — Let $R$ be a commutative ring and let $S$ be a multiplicative subset of $R$. Then the fraction ring $S^{-1}R$ is a flat $R$-module.
Let $u\colon X\to Y$ be an injective morphism of $R$-modules. First of all, one identifies the morphism $S^{-1}R\otimes_R u\colon S^{-1}R\otimes_R X\to S^{-1}R\otimes_R Y$ to the morphism $S^{-1}u\colon S^{-1}X\to S^{-1}Y$ induced by $u$ on fraction modules. Then, it is easy to see that $S^{-1}u$ is injective. Let indeed $x/s\in S^{-1}X$ be an element that maps to $0$; one then has $u(x)/s=0$, hence there exists $t\in S$ such that $tu(x)=0$. Consequently, $u(tx)=0$, hence $tx=0$ because $u$ is injective. This implies $x/s=0$.

Theorem. — Let $R$ be a commutative ring. If $M$ is a finitely presented $R$-module, then $M$ is locally free: there exists a finite family $(f_1,\dots,f_n)$ in $R$ such that $R=\langle f_1,\dots,f_n\rangle$ and such that for every $i$, $M_{f_i}$ is a free $R_{f_i}$-module.
The proof is a variant of the case of local rings. Starting from a point $p\in\Spec(R)$, we know that $M_p$ is a finitely presented flat $R_p$-module. As above, we get a surjective morphism $u\colon R^n\to M$ which induces an isomorphism $\kappa(p)^n\to \kappa(p)\otimes M$, and we let $N$ be its kernel. By flatness of $M$ (and an argument involving the snake lemma), the exact sequence $0\to N\to R_p\to M\to 0$ induces an exact sequence $0\to \kappa(p)\otimes N\to \kappa(p)^n\to \kappa(p)\otimes M\to 0$. And since the last sequence is an isomorphism, we have $\kappa(p)\otimes N$. Since $M$ is finitely presented, the module $N$ is finitely generated, and Nakayama's lemma implies that $N_p=0$; moreover, there exists $f\not\in p$ such that $N_f=0$, so that $u_f\colon R_f^n\to M_f$ is an isomorphism. One concludes by using the quasicompactness of $\Spec(R)$.

However, not all flat modules are projective. The most basic example is the following one.

Example. — The $\mathbf Z$-module $\mathbf Q$ is flat, but is not projective.
It is flat because it is the total fraction ring of $\mathbf Z$. To show that it is not projective, we consider the free module $L={\mathbf Z}^{(\mathbf N)}$ with basis $(e_n)$ and the morphism $u\colon L\to\mathbf Q$ that maps $e_n$ to $1/n$ (if $n>0$, say). This morphism is surjective. If $\mathbf Q$ were projective, there would exist a morphism $v\colon \mathbf Q\to L$ such that $u\circ v=\id_{\mathbf Q}$. Consider a fraction $a/b\in\mathbf Q$; one has $b\cdot 1/b=1$, hence $b v(1/b)=v(1)$. We thus see that all coeffiencients of $v(1)$ are divisible by $b$, for any integer $b$; they must be zero, hence $v(1)=0$ and $1=u(v(1))=0$, a contradiction.
The proof generalizes. For example, if $R$ is a domain and $S$ does not consist of units, and does not contain $0$, then $S^{-1}R$ is not projective. (With analogous notation, take a nonzero coefficient $a$ of $v(1)$ and set $b=as$, where $s\in S$ is not $0$; then $as$ divides $a$, hence $s$ divides $1$ and $s$ is a unit.)

These recollections are meant to motivate the forthcoming question: When is it the case that a localization $S^{-1}R$ is a projective $R$-module?

Example. — Let $e$ be an idempotent of $R$, so that the ring $R$ decomposes as a product ot two rings $R\simeq eR \times (1-e)R$, and both factors are projective submodules of $R$ since their direct sum is the free $R$-module $R$. Now, one can observe that $R_e= eR$. Consequently, $R_e$ is projective. Geometrically, $\Spec(R)$ decomposes as a disjoint union of two closed subsets $\mathrm V(e)$ and $\mathrm V(1-e)$; the first one can be viewed as the open subset $\Spec(R_{1-e})$ and the second one as the open subset $\Spec(R_e)$.

The question was to decide whether this geometric condition furnishes the basic conditions for a localization $S^{-1}R$ to be projective. With the above notation, we recall that $\Spec(S^{-1}R)$ is homeomorphic to a the subset of $\Spec(R)$ consisting of prime ideals $p$ such that $p\cap S=\emptyset$. The preceding example corresponds to the case where $\Spec(S^{-1}R)$ is open and closed in $\Spec(R)$. In this case, we view $S^{-1}R$ as a quasicoherent sheaf on $\Spec(R)$, it is free of rank one on the open subset $\Spec(S^{-1}R)$, and zero on the complementary open subset. It is therefore locally free, hence the $R$-module $S^{-1}R$ is projective.

Observation. — The set $\Spec(S^{-1}R)$ is stable under generization. If $S^{-1}R$ is a projective $R$-module, then it is open.
The first part is obvious: if $p$ and $q$ are prime ideals of $R$ such that $p\subseteq q$ and $q\cap S=\emptyset$, then $p\cap S=\emptyset$. The second part follows from the observation that the support of $S^{-1}R$ is exactly $\Spec(S^{-1}R)$, combined with the following proposition.

Proposition. — The support of a projective module is open.
I learnt this result in the paper by Vasconcelos (1969), “On Projective Modules of Finite Rank” (Proceedings of the American Mathematical Society 22 (2): 430‑33). The proof relies on the trace ideal $\tau_R(M)$ of a module: this is the image of the canonical morphism $t\colon M^\vee \otimes_R M\to R$. (It is called the trace ideal, because when $M$ is free, $M^\vee\otimes_R M$ can also be identified with the module of endomorphisms of finite rank of $M$, a split tensor $\phi\otimes m$ corresponds with the endomorhism $x\mapsto \phi(x)m$, and then $t(\phi \otimes m)=\phi(m)$ is its trace.) Now, if $p$ belongs to the support of $M$, then $\tau_R(M)_p=R_p$, while if $p$ does not belong to the support of $M$, one has $M_p=0$, hence $\tau_R(M)_p=0$. In other words, the support of $M$ is the complement of the closed locus $\mathrm V(\tau_R(M))$ of $\Spec(R)$.

On the other hand, one should remember the following basic property of the support of a module.

Proposition. — The support of a module is stable under specialization. The support of a finitely generated module is closed.
Indeed, for every $m\in M$ and $p\in \Spec(R)$, saying that $m=0$ in $M_p$ means that there exist $s\in R$ such that $s\notin p$ with $sm=0$. In other words, this set is $\mathrm V(\mathrm{ann}_R(m))$. This shows that the support of $M$ is the union of the closed subsets $\mathrm V(\mathrm{ann}_R(m))$; it is in particular stable under specialization. If $M$ is finitely generated, this also shows its support is $\mathrm V(\mathrm{ann}_R(M))$, hence is closed.

At this point, one can go either following Vasconcelos (1969) who shows that a projective module $M$ of the form $S^{-1}R$ is finitely generated if and only if its trace ideal is. In particular, if $R$ is noetherian and $S^{-1}R$ is a projective $R$-module, then $\Spec(S^{-1}R)$ is closed. It is thus open and closed, and we are in the situation of the basic example above.

One can also use a topological argument explained to me by Daniel Ferrand: a minimal prime ideal of $R$ that meets $\Spec(S^{-1}R)$ is disjoint from $S$, hence belongs to $\Spec(S^{-1}R)$. Consequently, $\Spec(S^{-1}R)$ is the union of the irreducible components of $\Spec(R)$ that it meets. If this set of irreducible components is finite (or locally finite), for example if $\Spec(R)$ is noetherian, for example if $R$ is a noetherian ring, then $\Spec(S^{-1}R)$ is closed.

I did not find the time to think more about this question, and it would be nice to have an example of a projective localization which does not come from this situation.

Associated prime ideals

$\gdef\ann{\mathop{\mathrm{ann}}} \gdef\Ass{\mathop{\mathrm{Ass}}}\gdef\Spec{\mathop{\mathrm{Spec}}}$

I would like to go back to a quite delicate question of commutative algebra, that of associated prime ideals of modules. In most textbooks (Bourbaki, Matsumura…), this concept is considered for modules over a noetherian ring, while it is also necessary to consider it in a greater generality for some applications in algebraic geometry. For my book, (Mostly) commutative algebra (Springer Nature, 2021), I preferred to introduce the general concept (§6.5), because I observed that the initial proofs are in fact easier. In yesterday's class (Cohomology of coherent sheaves, 2nd year of Master course at Université Paris Cité), some remarks of a student, Elias Caeiro, helped me simplify two steps of the treatment I proposed in my book.

Definition. — Let $A$ be a ring and let $M$ be an $A$-module. Say that a prime ideal $P$ of $A$ is associated to $M$ if there exists an element $m\in M$ such that $P$ is minimal among all prime ideals containing $\ann_A(m)$.
We write $\Ass_A(M)$ (sometimes spelt out as “assassin”) for the set of all associated prime ideals of $M$.
(Here, $\ann_A(m)$ is the annihilator of $m$, the ideal of all $a\in A$ such that $am=0$.)

There is a geometric way to intepret this definition: it means that in the spectrum $\Spec(A)$, the irreducible closed set $V(P)$ (of which $P$ is the generic point) is an irreducible component of $V(\ann_A(m))$. Thanks to this remark, associated prime ideals are compatible with localisation: \[ \Ass_{S^{-1}A}(S^{-1}A) = \Ass_A(M) \cap \Spec(S^{-1}A), \] where $\Spec(S^{-1}A)$ is identified as the subset of $\Spec(A)$ consisting of prime ideals $P$ which are disjoint from $S$. In particular, $P$ is associated to $M$ if and only if the maximal ideal $PA_P$ of the local ring $A_P$ is associated to the module $M_P$.

Here is what the associated prime ideals mean, from the point view of module theory.
Proposition. — Let $a\in A$.
a) The multiplication by $a$ is injective in $M$ if and only if $a$ does not belong to any associated prime ideal of $M$.
b) The localized module $M_a$ is zero if and only if $a$ belongs to all associated prime ideals of $M$.
c) In particular, $M=0$ if and only if $\Ass_A(M)=\emptyset$.

Proof. — a) If $a$ belongs to the associated prime ideal $P$, then $a$ belongs to the associated prime ideal $PA_P$ of $M_P$, which means that there exists $m\in M$ such that $PA_P$ is the only prime ideal containing $\ann_{A_P}(m)$. Consequently, $a$ is nilpotent modulo $\ann_{A_P}(m)$ and there exists $n\geq 0$ and $b\in A\setminus P$ such that $a^nb\in\ann_A(m)$. Take a minimal such $n$. Since $b\notin P$, one has $n\geq 1$; then $a^{n-1}b m\neq0$, while $a\cdot a^{n-1}bm=0$ and the homothety $(a)_M$ is not injective. Conversely, if $(a)_M$ is not injective, take $m\neq0$ in $M$ such that $am=0$; the annihilator $\ann_A(m)$ is not equal to $A$, hence $V(\ann_A(m))\neq \emptyset$; take an irreducible component of this closed subset — equivalently a minimal prime ideal $P$ among those containing $\ann_A(m)$; one has $a\in \ann_A(m)$, hence $a\in P$.
b) follows from c), with $a=1$.
c) The module $M$ is zero if and only if the multiplication by $0$ is injective on $M$. By a), this is equivalent to the fact that $\Ass_A(M)$ is empty.

Corollary. — A prime ideal $P$ is in the support of $M$ if and only if it contains some associated prime ideal.
The prime ideal $P$ belongs to the support of $M$ if and only if $M_P\neq0$, if and only if $\Ass_{A_P}(M_P)$ is not empty, if and only if there exists an associated prime ideal of $M$ which belongs to $\Spec(A_P)$, that is, is contained in $P$.

For noetherian rings, one has the following characterization of associated prime ideals, which is usually taken at their definition.

Theorem. — Let $A$ be a noetherian ring and $M$ be an $A$-module. A prime ideal $P$ of $A$ is associated to $M$ if and only if there exists $m\in M$ such that $P=\ann_A(m)$.
If $P=\ann_A(m)$, then $P$ is associated to $M$. Conversely, let $m\in M$ and let $P$ be a minimal prime ideal of $A$ among those containing $\ann_A(m)$. We first assume that $A$ is local with maximal ideal $P$; then $P$ is the only prime ideal of $A$ that contains $\ann_A(m)$, which implies that any element of $P$ is nilpotent modulo $\ann_A(m)$. Since $P$ is finitely generated (because $A$ is noetherian), there exists an integer $n$ such that $P^n\subseteq \ann_A(m)$. Take a minimal such $n$. Since $\ann_A(m)\subseteq P$, one has $n\geq 1$; then $P^{n-1}\not\subseteq\ann_A(m)$ so that there exists $b\in P^{n-1}$ such that $bm\neq0$. Then $ab\in P^n$ for every $a\in P$, so that $P\subseteq \ann_A(bm)$, and $\ann_A(bm)\subseteq P$ because $bm\neq0$. Consequently, $P=\ann_A(bm)$. In the general case, we use the case of a local ring to obtain $m\in M$ such that $\ann_{A_P}(m/1)=PA_P$. Consequently, $\ann_A(m)\subseteq P$, and for every $a\in P$, there exists $b\notin P$ such that $abm=0$. Using that $P$ is finitely generated, one finds $b\notin P$ such that $abm=0$ for every $a\in P$; then $\ann_A(bm)=P$, as was to be shown.

From that point on, both presentations converge. One deduces from the preceding theorem that if $A$ is noetherian and $M$ is finitely generated, there exists a composition series $0=M_0\subseteq M_1 \subseteq \dots \subseteq M_n=M$, with successive quotients $M_k/M_{k-1}$ of the form $A/P_k$, for some prime ideals $P_k$ of $A$, and then $\Ass_A(M)$ is contained in $\{P_1,\dots,P_n\}$, in view of the following lemma. In particular, $\Ass_A(M)$ is finite.

Lemma. — Let $M$ be an $A$-module and let $N$ be a submodule of $M$; then $ \Ass_A(N)\subseteq \Ass_A(M)\subseteq \Ass_A(N)\cup \Ass_A(M/N)$.
The first inclusion $\Ass_A(N)\subseteq \Ass_A(M)$ follows from the definition. Let us prove the second one. Let $P\in\Ass_A(M)$ and let $m\in M$ be such that $P$ is a minimal prime ideal of $A$ among those containing $\ann_A(m)$. Let $m'$ be the image of $M$ in $M/N$. If $P$ contains $\ann_A(m')$, then $P$ is also minimal among such prime ideals, hence $P\in\Ass_A(M/N)$. Otherwise, there exists $b\in \ann_A(m')$ such that $b\notin P$. Let us prove that $P$ is minimal among the prime ideals containing $\ann_A(bm)$. First of all, let $a\in\ann_A(bm)$; then $abm=0$, hence $ab\in P$, hence $a\in P$ since $b\notin P$. Since $\ann_A(m)\subseteq\ann_A(bm)$, it also follows that $P$ is minimal among the prime ideals containing $\ann_A(bm)$. Since $b\in\ann_A(m')$, one has $bm'=0$, hence $bm\in N$ and $P\in\Ass_A(N)$.