Why is this sequence of ideals an ascending chain? In proof of irreducible ideals are primary.

Question

Let $P$ be an irreducible ideal in commutative ring $R$. Suppose $ab \in P$, $a \notin P$, and define $A_n = \{b^n x : x \in R\} \cap P$. Then "clearly $A_n \subset A_{n+1}$" says D&F. But CLEARLY it's a descending chain as $bA_n = A_{n+1}$ and $bA_n \subset A_n$. Why do they say CLEARLY!!!!!! AHHHHH

Thanks.

Picture of Proof:

Answer 1

Well, $A_n=\{b^nx:x∈R\}∩P\supseteq \{b^n(bx):x∈R\}∩P=A_{n+1}$, since $bx\in R$ whenever $x\in R$.

The only unclear thing is, why can't they be equal. And this is equivalent to saying, we cannot write $1 = bc$ for some $c\in R$. And this follows since $b\in P, 1\notin P$; since $P$ is an ideal, then $\forall c\in R, P\ni bc \neq 1$.

Answer 2

The text says that $$ A_n=\{x\in R: a^nx\in Q\} $$

With your different notation, $$ A_n=\{x\in R: b^nx\in P\} $$ because you have $P$ in place of $Q$ and $a\notin P$ instead of $b\notin Q$.

Let $x\in A_n$: then $b^nx\in P$, so $bb^nx=b^{n+1}x\in P$. Hence $x\in A_{n+1}$.

You're interpreting the definition of $A_n$ in a wrong fashion.