TITLE: If $A[[x]]$ is Noetherian, will $A$ be Noetherian?
QUESTION [10 upvotes]: Let $A$ be a commutative ring with unit and $A[x]$ be the polynomial ring with coefficients in $A$. Then $A$ is Noetherian if and only if $A[x]$ is Noetherian. This is obtained by Hilbert Basis Theorem and that any quotient ring of a Noetherian ring is Noetherian. 
Let $A[[x]]$ be the formal power series ring with coefficients in $A$. It can be proved that if $A$ is Noetherian, then $A[[x]]$ is Noetherian. How about the converse? If $A[[x]]$ is Noetherian, will $A$ be Noetherian? Note that $A$ does not equal to $A[[x]]/(x)$ so the argument in polynomial case does not hold. Could you give a counterexample? Thanks a lot.

REPLY [21 votes]: Despite your claim, I think that $A=A[[x]]/(x)$.  Take the map taking a general element $\sum_{n=0}^\infty a_nx^n \to a_0$. It is clearly a homomorphism from $A[[x]] \to A$. Then, the kernel contains every element with constant term zero. Isn't this equal to the ideal $(x)$? By the "quotients of noetherian rings are noetherian" argument you cite, this proves the result.
Let me know if there is a mistake in my logic.