Exercise 6.3 of Matsumura about associated prime ideals

Question

My question is about this exercise of Matsumura:

In the proof hint he uses $xA/x^nA\simeq A/x^{n-1}A$. Is this obvious or should I define an isomorphism?

Answer 1

It is almost obvious: intuitively, since $x$ is not a zero divisor, we can cancel it from numerator and denominator. Formally, define a map $A \rightarrow xA \rightarrow xA/x^nA$, in which the first arrow is multiplication by $x$ and second is just the canonical projection. Certainly this map is surjective. Let $a$ be in the kernel. This means that $xa \in x^n A$ and so $xa = x^n b$ for some $b \in A$. Since $x$ is not a zero divisor this gives $a = x^{n-1} b$. This proves the claimed isomorphism.