If $\mathfrak{p}$ is a prime such that $M_\mathfrak{p} \neq 0$, then $\mathfrak{p}$ contains an associated prime of $M$

Question

I am studying from Serge Lang's Algebra (3rd edition), and in Chapter X Noetherian Rings and Modules, $\S2$ Associated Primes, we have the following proposition:

Proposition 2.10. Let $A$ be Noetherian, and let $M$ be a module. Let $a \in A$. The following conditions are equivalent:

1. $a_M$ is locally nilpotent.
2. $a$ lies in every associated prime of $M$.
3. $a$ lies in every prime $\mathfrak{p}$ such that $M_\mathfrak{p} \neq 0$.

If $\mathfrak{p}$ is a prime such that $M_\mathfrak{p} \neq 0$, then $\mathfrak{p}$ contains an associated prime of $M$.

I don't follow one part of the proof, namely the proof of the last statement of this proposition. It goes as follows in the book:

Proof (of last statement). Let $\mathfrak{p}$ be a prime such that $M_\mathfrak{p} \neq 0$. Then there exists $x \in M$ such that $(Ax)_\mathfrak{p} \neq 0$. By Corollary 2.7, there exists an associated prime $\mathfrak{q}$ of $(Ax)_\mathfrak{p}$ in $A$. Hence there exists an element $y/s$ of $(Ax)_\mathfrak{p}$, with $y \in Ax$, $s \not\in \mathfrak{p}$, and $y/s \neq 0$, such that $\mathfrak{q}$ is the annihilator of $y/s$.

It follows that $\mathfrak{q} \subset \mathfrak{p}$, for otherwise, there exists $b \in \mathfrak{q}$, $b \not\in \mathfrak{p}$, and $0 = by/s$, whence $y/s = 0$, contradiction.

Let $b_1,\dots,b_n$ be generators for $\mathfrak{q}$. For each $i$, there exists $s_i \in A$, $s_i \not\in \mathfrak{p}$, such that $s_i b_i y = 0$ because $b_i y / s = 0$. Let $t = s_1 \cdots s_n$. Then $\mathfrak{q}$ is the annihilator of $ty$ in $A$. Hence $\mathfrak{q} \subset \mathfrak{p}$, as desired.

My questions are:

1. How does the existence of $b \in \mathfrak{q}$ such that $b \not \in \mathfrak{p}$ and $0 = by/s$ imply that $y/s = 0$?

2. I see that $\mathfrak{q}$ is contained in the annihilator of $ty$, because $b_i(ty) = 0$ for each $i$. How can I show the reverse containment to deduce that $\mathfrak{q}$ is the annihilator of $ty$?

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For the sake of completeness, here are some of the definitions and results used in the above discussion:

• $a_M$ denotes the homomorphism $M \to M$ given by $x \mapsto ax$.

• $a_M$ is said to be locally nilpotent if for each $x \in M$ there exists an integer $n(x) \geq 1$ such that $a^{n(x)}x = 0$.

• A prime ideal $\mathfrak{p}$ of $A$ is called an associated prime of $M$ if there exists $x \in M$ such that $\mathfrak{p}$ is the annihilator of $x$.

• Corollary 2.7. If $A$ is Noetherian and $M$ is a module $\neq 0$, then there exists a prime associated with $M$.

Answer 1

1. View $(Ax)_\mathfrak{p}$ as a module over $A_\mathfrak{p}$. Then, $$\frac {by}s = \frac b1 \cdot \frac ys.$$ Since $b \not\in \mathfrak{p}$, $\frac b1$ is a unit in $A_\mathfrak{p}$ as noted by @Bernard in the comments above. Thus, $$\frac b1 \cdot \frac ys = 0 \implies \frac ys = 0.$$

2. Note that $t \not\in \mathfrak{q}$ because $s_i \not\in \mathfrak{p}$ for all $i \implies$ $t = s_1 \cdots s_n \not\in \mathfrak{p}$ as $\mathfrak{p}$ is a prime ideal, and since $\mathfrak{q} \subset \mathfrak{p}$, $t \not\in \mathfrak{q}$. Now, let $r$ be in the annihilator of $ty$. So, $$r(ty)=0 \implies (rt)(y/s) = 0 \implies rt \in \mathfrak{q}.$$ Recall that $\mathfrak{q}$ is an associated prime of $(Ax)_\mathfrak{p}$, so it is a prime ideal of $A$. Hence, $rt \in \mathfrak{q}$ and $t \not\in \mathfrak{q}$ imply that $r \in \mathfrak{q}$. Thus, $\mathfrak{q}$ contains the annihilator of $ty$.

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Also, it is a bit confusing that Lang concludes twice that $\mathfrak{q} \subset \mathfrak{p}$. Instead, he should say the second time that we have proved that $\mathfrak{q}$, which was a priori an associated prime of $(Ax)_\mathfrak{p}$, is actually an associated prime of $M$.