If someone could check my proof for the following question, I would appreciate it!
Let $A$ be a ring, and let $A[[x]]$ be the formal power series ring. Show that every prime ideal of $A$ is the contraction of a prime ideal in $A[[x]]$.
Let $\mathfrak{p}$ be a prime ideal in $A$. There is the canonical surjective quotient map: $$ \psi: A\to A/\mathfrak{p} $$ There is an isomorphism between $A[[x]]/(x)$ and $A$ induced by the following map $\phi$ which sends every power series to its constant term with kernel $(x)\subset A[[x]]$: $$ \phi: A[[x]] \to A $$ Therefore, the surjective map $\psi\circ \phi: A[[x]]\to A/\mathfrak{p}$ has kernel: $$ \ker(\psi\circ \phi) = (x) + \mathfrak{p}A[[x]] $$ ($\supset$ is clear. Suppose $f\in \ker(\psi\circ \phi)$ then $f$ has constant term in $\mathfrak{p}$ and the higher order terms can be anything. Thus, clearly $f\in (x) + \mathfrak{p}A[[x]]$.)
Therefore, $(x)+\mathfrak{p}A[[x]]$ is prime, since $A[[x]]/((x)+\mathfrak{p}A[[x]]) \cong A/\mathfrak{p}$ is an integral domain. Moreover, $((x)+\mathfrak{p}A[[x]])^c = \mathfrak{p}$ since the constant terms of $(x)+\mathfrak{p}A[[x]]$ are exactly the elements of $\mathfrak{p}\subset A$. This concludes the proof.
(Sidenote: Does the same proof show that every maximal ideal of $A$ is the contraction of a maximal ideal in $A[[x]]$?)
