$\mathfrak{q}$ is primary iff: given $a,b \in A$, $ab \in \mathfrak{q}$ and $a \not\in \mathfrak{q}$, then $b^n \in \mathfrak{q}$ for some $n \geq 1$

Question

I am studying from Lang's Algebra, and in Chapter X Noetherian Rings and Modules, $\S$3 Primary Decomposition, he makes the following definitions on page 421, third edition (assume that $A$ is a commutative ring and that $M$ is an $A$-module):

Let $M$ be a module. A submodule $Q$ of $M$ is said to be primary if $Q \neq M$, and if given $a \in A$, the homomorphism $a_{M/Q}$ is either injective or nilpotent. Viewing $A$ as a module over itself, we see that an ideal $\mathfrak{q}$ is primary if and only if it satisfies the following condition: $$ \textit{Given $a,b \in A$, $ab \in \mathfrak{q}$ and $a \not\in \mathfrak{q}$, then $b^n \in \mathfrak{q}$ for some $n \geq 1$.} $$

(Here, $a_{M/Q}$ is the homomorphism $M \to M$ given by $x \mapsto ax$.)

I am not able to deduce the equivalent condition given for the ideal $\mathfrak{q}$ to be primary. This is what I have so far:

1. If $a \in \mathfrak{q}$, then $a_{A/\mathfrak{q}}$ is nilpotent. So, to check whether or not $\mathfrak{q}$ is primary, we only need to see what happens when $a \not\in\mathfrak{q}$.

2. So, suppose that $a \not\in \mathfrak{q}$. Then, $a_{A/\mathfrak{q}}$ is injective if and only if for all $\bar{b} \in A/\mathfrak{q}$, $a_{A/\mathfrak{q}}(\bar{b}) = \bar{0} \implies \bar{b} = \bar{0}$, that is, if and only if for all $b \in A$, $ab \in \mathfrak{q} \implies b \in \mathfrak{q}$.

3. And, $a_{A/\mathfrak{q}}$ is nilpotent if and only if there exists $n \geq 1$ such that $(a_{A/\mathfrak{q}})^n(\bar{b}) = \bar{0}$ for all $\bar{b} \in A/\mathfrak{q}$, that is, if and only if $a^n b \in \mathfrak{q}$ for all $b \in A$.

So, what I have is that $\mathfrak{q}$ is primary if and only if for each $a \not\in \mathfrak{q}$, either $ab \in \mathfrak{q} \implies b \in \mathfrak{q}$ for all $b \in A$, or there exists $n \geq 1$ such that $a^n b \in \mathfrak{q}$ for all $b \in A$.

How do I go from here to the statement given in Lang?

Answer 1

Note that $\newcommand{\q}{\mathfrak{q}}a_{A/\q}$ is nilpotent if for some $n\ge1$, $a^n(A/\q)=0$, that is $a^n\in\q$. Thus $a_{A/\q}$ is nilpotent iff $a\in\sqrt\q$, the radical of $\q$.

If $\q$ is primary and $ab\in\q$ then either $b_{A/\q}$ is nilpotent or $b_{A/\q}$ is injective, but that implies $a$ is zero in $A/\q$, that is $a\in\q$. Therefore either $a\in\q$ or $b\in\sqrt\q$.

Now suppose that $\q$ satifies the alternative condition. If $b_{A/\q}$ is not injective, there is $a\notin\q$ with $ab\in\q$. Then $b\in\sqrt\q$, that is, $b_{A/\q}$ is nilpotent.

Answer 2

I realised how to do it a few moments after posting the question.

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Suppose that $\mathfrak{q}$ is primary. Let $a,b \in A$, $ab \in \mathfrak{q}$ and $a \not\in \mathfrak{q}$. We need to show that $b^n \in \mathfrak{q}$ for some $n \geq 1$.

Let $b_{A/\mathfrak{q}}$ be injective. The condition $ab \in \mathfrak{q}$ implies that $b_{A/\mathfrak{q}}(\bar{a}) = \bar{0}$, since $ab = ba$. Now, since $b_{A/\mathfrak{q}}$ is injective by assumption, this implies that $\bar{a} = \bar{0}$, that is, $a \in \mathfrak{q}$, a contradiction.

So, $b_{A/\mathfrak{q}}$ must be nilpotent. Hence, there exists $n \geq 1$ such that $b^n r \in \mathfrak{q}$ for all $r \in A$, as shown in point no. 3. In particular, this is true for $r = 1$, so we get that $b^n \in \mathfrak{q}$ for some $n \geq 1$.

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Conversely, suppose that the following condition is satisfied:

$$\textit{Given $a,b \in A$, $ab \in \mathfrak{q}$ and $a \not\in \mathfrak{q}$, then $b^n \in \mathfrak{q}$ for some $n \geq 1$.}\tag{$*$}$$

We need to show that for each $a \in A$, $a_{A/\mathfrak{q}}$ is either injective or nilpotent.

If $a \in \mathfrak{q}$, then $a_{A/\mathfrak{q}}$ is nilpotent, as shown in point no. 1.

So, let $a \not\in \mathfrak{q}$. If it so happens that whenever $ab \in \mathfrak{q}$ we have $b \in \mathfrak{q}$, then $a_{A/\mathfrak{q}}$ will be injective. That is, if it so happens given $a \not\in \mathfrak{q}$ we can take $n = 1$ in $(*)$ for any $b \in A$, then $a_{A/\mathfrak{q}}$ will be injective.

So, suppose that $a \not\in \mathfrak{q}$ and it is not the case that $n = 1$ works in $(*)$ for all $b \in A$. Then, there exists $b \in A$ such that $ab \in \mathfrak{q}$ but $b \not\in \mathfrak{q}$. Now, since $ab = ba$, by swapping the roles of $a$ and $b$ we see that the hypotheses of $(*)$ are again satisfied. Hence, $a^n \in \mathfrak{q}$ for some $n \geq 1$ (in fact, $n > 1$ since $a \not\in \mathfrak{q}$ by assumption). Hence, $a^n r \in \mathfrak{q}$ for all $r \in A$. Thus, $a_{A/\mathfrak{q}}$ is nilpotent.

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Hence, the two definitions of $\mathfrak{q}$ being primary are equivalent.