UFDs and reduced quotients

Question

Let $A$ be a UFD and let $x\in A$ an element. I don't understand why the following claim is true:

The quotient $A/(x)$ is reduced if and only if $x$ is a product of distinct primes.

Can you suggest a proof?

Answer 1

Write $x = p_1 ^ {\alpha_1} \cdots p_n ^ {\alpha_n}$.

Take $y \in A$ and write $y = p_1 ^ {\beta_1} \cdots p_n ^ {\beta_n} z$ with $z$ coprime with $x$.

Then $y^m \in (x)$ iff $m \beta_k \ge \alpha_k$ for all $k=1,\dots,n$.

Since $\alpha_k >0$, we must have $\beta_k >0$.

Therefore, if $y^m \in (x)$ then $\beta_k >0$ for all $k$, that is, all primes that divide $x$ also divide $y$.

If $\alpha_k=1$ for all $k$, this implies that $y$ is a multiple of $x$ and therefore $A/(x)$ is has no nilpotent elements.

In general, the radical of $x$ is $(p_1\cdots p_n)$.