Prime ideals in formal power series

Question

Let $A$ be a commutative ring with unit. If $\mathfrak{p} \subset A $ is a prime ideal, then $\mathfrak{p}$ is the contraction of a prime ideal of $A[[x]]$, the ring of formal power series.

Why is this true ?

Answer 1

Hint: consider the subset $\tilde{\mathbb{p}}$ of $A[[x]]$ consisting of power series $$ \sum_{n\ge 0}a_n x^n $$ where $a_n\in\mathbb{p}$. Then:

1. $\tilde{\mathbb{p}}$ is an ideal of $A[[x]]$

2. $\tilde{\mathbb{p}}$ is a prime ideal of $A[[x]]$

3. The contraction of $\tilde{\mathbb{p}}$ is …

You can consider the map $e\colon A[[x]]\to A$, where $e\left(\sum_{n\ge0}a_nx^n\right)=a_0$. This is a ring homomorphism.