Let $A$ be a commutative ring (with unit) such that $A[[T]]$ is Noetherian. Can we conclude that $A[T]$ is Noetherian without invoking Hilbert's Basis Theorem (or a similarly heavy argument)? In other words, if we know Noetherianity of the power series ring, do we get Noetherianity of the polynomial ring "for almost free"?
The canonical map $A[[T]]\twoheadrightarrow A$ gives us the Noetherian property for $A$, which then implies Noetherianity of $A[T]$ by the usual polynomial Hilbert Basis Theorem. However, if we wish to skip HBT, the implication is non-obvious (at least to me) since Noetherianity is not "inclusion-borne".
Also notice that the opposite direction does not require HBT as $A[[T]]$ is the $(T)$-adic completion of $A[T]$.
All the standard texts on Commutative Algebra seem to give separate proofs one way or another.
PS: Noticing that nearly same proof works does not count, sorry :-)