I think that, in thiscase, A is necessarily noetherian:

if any radical ideal in A is finitely generated, in particular, any prime ideal is a finitely generated A-module. As we know, this implies that A is noetherian. 

By the bijection between the prime ideals of A and the localization A_S, we conclude that A_S is noetherian and so every ideal is finitely generated.

Hello Lucas, I am not convinced at all : reading Ritt or Kolchin works, they mention a non reduced differential ideal which is not finitely generated, while its radical differential ideal is.The base ring is K{x,y_1, y_2, ..., y_n}  (differential polynomials).

Thank you for this first answer,

 Best Regards,

Gabriel

Dear Lucas, 

A ring R is noetherian iff every ideal in R is finitely generated R-module.  Now if we know that prime ideals fulfill this condition, can we deduce that every ideal fulfill it?

I think it is not right.

It is true that, if all the prime ideals in a (commutative unitary) ring are finitely generated, then the ring is Noetherian.  The idea of the proof is that an ideal that is maximal among the non-finitely generated ideals is prime.  The proof appears, for example, in Larsen and McCarthy's book \textit{Multiplicative Theory of Ideals.}

Dear David,

First of all, thank you very much for the reference. I downloaded it, but I didn't find what you said. I will be thankful if you have the page.

Best 

Hello all and thanks for the indications.

I have to precise my post :

- the base ring isn't noetherian for all ideals, it's only for radical ones.

- the meaning of "finitely generated" is not the same in differential algebra : a finite set  f1, f2, ..., fN generates an ideal in a diffierential sense when any element of this ideal is a finite combination of the fi and some of their derivatives. Thus, an infinite number of symbols is necessary (take e.g. the differential ideal [y] generated by the basic differential polynomial y).

Gabriel

the first place where the result mentioned by David Lantz appears is the following article by Cohen: "Commutative rings with restricted minimum condition", Duke Math. J.Volume 17, Number 1 (1950), 27-42. link: http://projecteuclid.org/euclid.dmj/1077475897. Gabriel, is your ring commutative and unitary? I would like to know whether Cohen's theorem holds or not in the non-commutative case, and if not an example of a non-commutative ring in which every prime ideal is finitely generated but the ring is not noetherian.

Please dear Giulio, if you have the text of the given link, 

I tried to access, but  I recievd: Full-text: Access denied

Hi Hanifa, see here: http://www.filedropper.com/commutativeringswithrestrictedminimumcondition

Theorem 2.

Hello,

 the ring is commutative and unitary.

Happy New Year to all!

Dear Giulio,

Thank you very much for the link, I downloaded it and I read Theorem 2, it is for prime ideals with a finite basis

My question:  Is finite generated module a free module ( means has a basis)?

In my knowledge, Every free module of finite rank is finite generated, but the reciprocal is not necessary. is it right for prime ideals in a unitary commutative ring (just)?

no, an ideal, or more generally a module over a commutative ring which is finitely generated is not necessarily free. For example, consider a Dedekind domain with non-trivial class group: every ideal is projective, but not necessarily free (unless it is a principal ideal). However, the rank of every ideal (defined as the dimension over the quotient field K of its extension by scalars of K) is 1.

I don't  understand your last question.... are you asking if a finitely generated prime ideal is free as a module? in this case the answer is no for the same reason as above.

@Thomas: I am not aware of the definition you mention of "finitely generated" in the context of differential algebra: can you give me some reference (book, for example) for it?

Cohen's Theorem also appears in Matsumura's "Commutative Ring Theory" (Theorem 3.4)

Ok dear Giulio.

I also see this link: http://stacks.math.columbia.edu/tag/05KG#code

We have the following:

If every prime ideal of R is finitely generated (not necessary finite basis as mentioned theorem 2 in Cohn's paper), then every ideal of R is finitely generated.

Consequently, R is noetherian.

yes, Hanifa, the link you copied is precisely Cohen's Theorem, which can be now found in many books of commutative algebra (in MAtsumura's book, as Alexis pointed out).  Beware that "finite basis" in Cohen's article nowadays does not mean that the ideal has a basis as a free module, but just that it is finitely generated.

Dear Waldemar, I eventually got your idea. Radical ideals was the point, not noetherian. Best Regards!

If each radical ideal of A is finitely generated then each prime ideal of A is finitely generated too.  So, by Cohn's  theorem we conclude that A is Noetherian.

Bonjour et merci François. Il s'agit bien de Cohn, non de Cohen? Y a-t-il une référence facile à consulter pour ce théorème?

Many Thanks François!

I am sure that it is Irvin S. Cohen and not Cohn (??), who gave that result in the following article, as I wrote:

"Commutative rings with restricted minimum condition", Duke Math. J.Volume 17, Number 1 (1950), 27-42. 

https://en.wikipedia.org/wiki/Irvin_Cohen

Giulo is right. This result is due to I. Cohen (not to Cohn).