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First of all, I apologize if this is too elementary for MO. I am reposting this from MSE where I asked this question two days ago (link to MSE post).

I've been wondering about the following thing. 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 similar argument)? In other words, if we know Noetherianity of the power series ring, do we get Noetherianity of the polynomial ring "for almost free"?

Notice that the canonical epimorphism $A[[T]]\twoheadrightarrow A$ gives us the Noetherian property for $A$, which then implies Noetherianity of $A[T]$ by the usual polynomial Hilbert Basis Theorem (HBT for short). However, if we wish to skip HBT, the implication is non-obvious (at least to me) since Noetherianity is not "inclusion-borne". Moreover, 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 (AM, Kaplansky, Matsumura, Zariski-Samuel) appear to give separate proofs one way or another. Even though the proofs of both cases are deceivingly similar and polynomials are a special case of power series, both theorems seem content-wise quite distant from each other.

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  • $\begingroup$ How would one be in the situation of having proved $A[\![T]\!]$ is noetherian while regarding the Hilbert Basis Theorem to be a "heavy" tool to be avoided? Very little can be done in the theory of noetherian rings with using the Hilbert Basis Theorem, so the rationale to want to avoid it is unclear. Moreover, the preservation of the noetherian property under completion is a harder fact than the Hilbert Basis Theorem (and its proof uses much more). So it would help to hear more about the rationale. $\endgroup$
    – nfdc23
    Commented Feb 11, 2017 at 15:26
  • $\begingroup$ @nfdc23: No, the point is not the avoid some version of Hilbert's Basis Theorem altogether. I guess the rationale is one of exposition: prove the power series version of HIB as the (seemingly) "more general result" and then give some (short) argument why it subsumes the polynomial version. The only such "argument" I am aware of goes by saying that the proof is nearly the same except we consider the highest non-zero coefficient and so on. On the other hand, we do have a proper argument if we go from $A[T]$ to $A[[T]]$, namely completion. $\endgroup$
    – M.G.
    Commented Feb 11, 2017 at 15:44
  • $\begingroup$ Also, I removed "heavy" as it is somewhat subjective. $\endgroup$
    – M.G.
    Commented Feb 11, 2017 at 16:07
  • $\begingroup$ If you can show that $A[[T]]$ is faithfully flat over $A[T]$ without assuming that $A[T]$ is Noetherian, then you can descend from one to the other. $\endgroup$
    – Phil Tosteson
    Commented Feb 11, 2017 at 18:12
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    $\begingroup$ @Phil: $A[[T]]$ is never faithfully flat over $A[T]$ (for $A \neq 0$), e.g. $1+T$ generates the unit ideal in $A[[T]]$, but not in $A[T]$. $\endgroup$
    – js21
    Commented Feb 13, 2017 at 9:34

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