Specifically, is it possible for a non-Noetherian ring $R$ to have $R[x]$ Noetherian? Every reference I've seen for the Hilbert basis theorem only states the direction "$R$ Noetherian $\Rightarrow$ $R[x]$ Noetherian", which would certainly seem to imply that the converse is false. Unfortunately, it's tough to think about non-Noetherian rings, and what I'm sure is most people's favorite example of one, $K[x_1,x_2,\ldots]$ for a field $K$, is obviously not going to help us here.

  • 16
    $\begingroup$ Isn't there a quotient HM from $R[X]$ onto $R$, given by sending $X$ to $0$? and doesn't Noetherian-ity pass to quotient rings? $\endgroup$ – Yemon Choi Feb 13 '10 at 2:27
  • 1
    $\begingroup$ Another excellent way of seeing this result! Now I see that this is quite clear - it just seemed odd that no one states that it's an "if and only if". $\endgroup$ – Zev Chonoles Feb 13 '10 at 2:38

If $A$ is an ideal of $R$, then $A[X]$ is an ideal of $R[X]$, right? So an ascending chain of ideals in $R$ which does not stabilize gives you an ascending chain of ideals in $R[X]$ which doesn't stabilize either?

  • 1
    $\begingroup$ Ah, of course - thanks for the answer. Still wish books / professors would mention this though... $\endgroup$ – Zev Chonoles Feb 13 '10 at 2:30
  • 2
    $\begingroup$ You hardly see the other direction mentioned since the usual direction is by far the more important one. Consider the case of the Chinese remainder theorem, which says the natural ring homomorphism Z/(mn) ---> Z/(m) x Z/(n) is an isomorphism when (m,n) = 1. This map makes sense even if (m,n) is not 1, but in that case we don't get an isomorphism. So should you be annoyed that the CRT is not usually stated in the form that this natural map is an isomorphism if and only if (m,n) = 1, rather than just if (m,n) = 1? I don't think so. It's hardly ever useful. Left to the reader! $\endgroup$ – KConrad Feb 13 '10 at 3:12
  • 1
    $\begingroup$ I suppose that's the idea. Either way, I probably should have thought about the problem a bit more before posting it. $\endgroup$ – Zev Chonoles Feb 13 '10 at 3:20
  • $\begingroup$ Also, does a book / professor mention that R is a UFD iff R[x] is so. :) $\endgroup$ – Abhishek Parab Feb 13 '10 at 6:08

Dear Zev,

There are some sources which give the converse. See e.g. pp. 64-65 of


For that matter, see also pp. 32-33 of loc. cit. for the Chinese Remainder Theorem and its converse. (And I am not the only one to do this...)

Note that in both cases the converse is left as an exercise. I think (evidently) that this is the right way to go: it may not be so easy for the journeyman mathematician to come up with the statement of the converse, but having seen the statement it is a very valuable exercise to come up with the proof. (In particular, I believe that in a math text or course at the advanced undergraduate level and beyond, most exercises should indeed be things that one could find useful later on, and not just things which are challenging to prove but deservedly forgettable.)

Finally, by coincidence, just yesterday in my graduate course on local fields I got to the proof of the "tensor product theorem" on the classification of norms in a finite-dimensional field extension (which came up in a previous MO answer). The key idea of the proof -- which I found somewhat challenging to write; I certainly admit to the possibility of improvements in the exposition -- seems to be suspiciously close to the valuation-theoretic analogue of the converse of the Chinese Remainder Theorem! See pp. 18-19 of


if you're interested.

  • $\begingroup$ Well, I guess that's just a testament to how great your commutative algebra notes (pre-book?) are :) Also, I very much agree with your philosophy on exercises. $\endgroup$ – Zev Chonoles Feb 13 '10 at 17:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.