# Prime Ideal & Noetherian Integral Domain

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1. Nov 17, 2014

### A.Magnus

I am reading a graduate-level Abstract Algebra lemma on noetherian integral domain, I am bring it up here hoping for pointers. The original passage is in one big-fat paragraph but I broke it down here for your easy reading. Let me know if I forget to include any underlying lemmas, thank you for your time and help. (I posted this in homework forum but apparently it was not the right place since I did not get any response at all.)

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LEMMA:
Let M be an R-module. Let T be maximal among the ideals of R such that M possesses a submodule L for which L/LT is not noetherian. Then T is a prime ideal of R.

PROOF:
(1) We are assuming that M possesses a submodule L for which L/LT is not noetherian. Thus, as L/LR = L/L is noetherian, T≠R.

(2) Let us assume, by way of contradiction, that T is not prime. Then R possesses ideals U and V such that T ⊂ U,T ⊂ V , and UV ⊆ T.

(3) The (maximal) choice of T forces L/LU and LU/LUV to be noetherian. [QUESTION: I understand that while T is maximal but U and V are strictly larger than T, and the only way to resolve this paradox is to take U and V as structures different from T. But I am lost on how all these "force L/LU and LU/LUVto be noetherian."]

(4) Thus, by Lemma below, L/LUV is noetherian. [QUESTION: Does it mean that since L/LU and LU/LUV are noetherian from above, therefore L, LU andLUV are noetherian, and therefore L/LUV is noetherian?]

(5) On the other hand, as UV ⊆ T, LUV ⊆ LT. Thus, L/LT is a factor module of L/LUV. [QUESTION: Here, I am begging explanation on how L/LT is a factor module of L/LUV, step-by-step if possible.]

(6) Thus, as L/LUV is noetherian, L/LT is noetherian; cf. Lemma below. This contradiction finishes the proof.

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This is the lemma quoted above: Let M be an R-module, and let L be a submodule of M. Then M is noetherian if and only if L and M/L are noetherian.

2. Nov 18, 2014

### mathwonk

well thisn is the most confusing pile of %%% I ever saw, but it is technically correct, as far as i have gotten.

I.e. T is an ideal such that M does have a submodule L with L/LT non mnoetherian. However if U,V are larger ideals than T, then

every sub module K of M has both K/UK and K/VK noetherian.

Since L is a submodule of M, thus L/UL is noetherian, and since LU is also a submodule of M, LU/LUV is noetherian.

more later...

3. Nov 18, 2014

### A.Magnus

Thanks! I share your pain, that's the reason I bring it up here. Do please come back again.

4. Nov 18, 2014

### mathwonk

Ok the basic trivial lemma is that a module A is noetherian if it has a submodule B such that both B and A/B are noetherian.

well, LU/LUV is a submodule of L/LUV, and we have shown both LU/LUV and (L/LUV)/(LU/LUV) ≈ L/LU are noetherian, hence so is

L/LUV...... I'm tired of this., but it looks ok. well, anytime A is in B is in C, then A A/B is ISOMORPHIC TO (not equal to) a factor of A/C, namely A/B ≈ (A/C)/(B/C).

sorry for shouting. not shouting at you but at the lousy book.

5. Nov 18, 2014

### mathwonk

ok, wind up, since UV is in T, thus UVL is in TL is in L, so L/(TL) is isomorphic to a factor of L/(UVL), hence is noetherian. blthhhh!!!

6. Nov 18, 2014

### mathwonk

may i know the name of the book? i want to make other suggestions for books.

and where is this boring tedious lemma leading us? to what hopefully interesting theorem?

7. Nov 18, 2014

### A.Magnus

Thanks again! I will read closely your pointers tomorrow first thing in the morning, since it is already late evening. To answer your last question: This lemma does not come from a textbook but from a class note we are supposed to read and understand. Thanks again and again!

8. Nov 19, 2014

### mathwonk

I think I would understand it better if I knew what good it is. Where is this going? How was this lemma used?

9. Nov 19, 2014

### A.Magnus

The lemma was quoted from a heading titled: Noetherian Integral Domains, it lists about 9 lemmas but at the end of section they do not culminate to any big theorem what so ever. I would love to understand them because I like the subject. But anyway, I read your pointers early in the morning and I think now they are beginning to make some sense. Now I fully understand #3 and #4, although I am still pondering about #5. Having said that, I would like to ask you another question, this time around is a general question. See my next reply. Thanks again and again for your help!

10. Nov 19, 2014

### A.Magnus

Hi Mathwonk, this morning I came across your reply to an old posting posted back in 2008, the link is here:

In responding to this posting, you wrote the following:
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a prime ideal of a cone is (Z^2 -X^2-Y^2).
a prime ideal of a plane is (Y).
the sum of these ideals (Z^2 -X^2-Y^2, Y) = (Z^2 -X^2, Y), is the ideal of the intersection, which is the two lines Z=X, Z=-X, in the X,Z plane.
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It seems to me that you are using some kind of mental aids for visualizing abstract algebra, at this posting it looks like you were using cone (3D object) and plane (2D) to visualize prime ideal. I am wondering if you could elaborate more about using such visualization aid. In that posting, Hurkyl (who responded just before you did) also used R[x, y] and parabola as visualization, but since he has been long inactive (since June 2013) therefore I am not sure I can contact him.

As a matter of fact, I have been using Venn diagram myself to visualize abstract algebra, although it is not very effective. Therefore I would love to hear any other visualization aids from seasoned mathematicians like you. Thank you for your time and help, again.

PS. If you would like me to post a new thread since it is an entirely new topic, I would be most happy to do so. Thanks.

11. Nov 19, 2014

### mathwonk

this is algebraic geometry, the geometry of polynomial rings and generalizations.

To visualize an ideal of polynomials, you just graph their common zero locus.

the common zero locus of a family of polynomials equals the common zero locus of the ideal they generate.

If you use complex numbers as coefficients, in fact there is a one one correspondence between subsets of affine n-space that are define as common zero loci of some polynomials, and the set of ideals in the polynomial ring of n variables which are equal to their own radical.

More simply there is a one one correspondence between points and maximal ideals, and a one one correspondence between algebraic sets that consist of only one piece and prime ideals.

Thus abstractly one can start from any commutative ring and construct a geometric object from it, by taking the closed points to be the maximal ideals, and the irreducible sets to be the prime ideals. Sometimes one also calls these latter "points",

Here is a link to some algebra notes of mine. there is a bit of "geometry of rings" starting at the bottom of page 32. The correspondence between points and maximal ideals in the case of complex coefficients is proved on page 33.

http://www.math.uga.edu/%7Eroy/80006b.pdf [Broken]

Last edited by a moderator: May 7, 2017
12. Nov 19, 2014

### A.Magnus

Oh, I did not know that you were discussing about algebraic geometry, I thought you were talking about abstract algebra. But anyway thank you very much for your response, now I learned something. Thanks also for the link to your class note.

13. Nov 20, 2014

### mathwonk

well today abstract algebraic geometry is a way to visualize ring theory. You can start from any commutative ring with identity and form the spectrum, the set of all prime ideals, and then put a spectral topology on it, by saying the closure of a set of prime ideals equals all those primes that contain the intersection of the given set.....This is the main area in which noetherian rings are studied today, i.e. commutative algebra is almost just a sub subject of abstract algebraic geometry today.

14. Nov 20, 2014

### A.Magnus

Thanks again. I sent the note to your school email address. Did you get it? Thanks.

15. Nov 22, 2014

### A.Magnus

Many many thanks to Mathwonk, now I am able to piece together the logic and break down the proof into dumb line-by-line, step-by-step:

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LEMMA:
Let M be an R-module. Let T be maximal among the ideals of R such that M possesses a submodule L for which L/LT is not noetherian. Then T is a prime ideal of R.

PROOF:
(1) Here we are given M possesses a submodule L for which L/LT is not noetherian. Since L is submodule of M and M is R-module, L = LR. Thus L/LR = L/L = {0}. Because singleton zero is noetherian, L/LR is noetherian.
(2) Since
L/LT is not noetherian while L/LR is noetherian, we conclude that T $\neq$ R.
(3) Let’s assume, by way of contradiction, that
T is not prime.
(4) Since
T is not a prime, then R must possess ideals U and V such that T $\subset$ U, T $\subset$ V, and UV $\subseteq$ T.
(5) We are given that
T is maximal, but at the same time from #(4) above we have concluded that there are ideal U and ideal V which are strictly larger than T.
(6) The only way to reconcile the above paradox is to conclude that
U and V must be of different structure than T. Thus as L/LT is not noetherian, L/LU must be noetherian.
(7) Since
L/LU is noetherian, L and LU must be noetherian; cf. Lemma above in the original posting.
(8) Notice here that
L is submodule of M, LU $\subset$ L, and that L/LU is noetherian. At the same time, notice that LU is submodule of M, and that LUV $\subset$ LU.
(9) The above two facts then lead us to conclude that
LU/LUV is noetherian too, as in the case of L/LU.
(10) Since
LU/LUV is noetherian, thus by Lemma above both LU and LUV are noetherian too.
(11) From #(7) we know that L is noetherian, and from #(10) we know that
LUV is noetherian. Thus L/LUV is noetherian by the Lemma above.
(12) On the other hand, #(4) gives us that
UV $\subseteq$ T. Since L is submodule to both UV and T, therefore LUV $\subset$ LT .
(13) In
L/LT from #(2), we notice that LT $\subset$ L, where L is a submodule of M and T is an ideal of R. In L/LUV from #(11), we also notice that LUV $\subset$ L, where L is a submodule of M and UV is an ideal of R.
(14) From #(13) above, therefore we conclude that
L/LT is isomorphic to L/LUV.
(15) The above isomorphism leads us further to conclude that since
L/LUV is noetherian, L/LT must be noetherian too.
(16) But this conclusion contradicts the given fact from #(1) that
L/LT is not noetherian.
(17) Because of this contradiction, we conclude that the assumption we made on #(1), that
T is not prime, is false. Therefore T is prime. $\blacksquare$

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Thank you again to Mathwonk, all the credits should go to him.

16. Nov 22, 2014

### mathwonk

unfortunately statement (7) is not correct. certainly for any non noetherian module M, M/M is noetherian, but M is not. fortunately (7) is niot needed for (9), since that follows by the "maximality" of T. I.e. V is larger than T so any module of form K/VK is noetherian, e.g. LU/LUV has this form, hence isn noetherian.

again i urge you to not focus just on the technicalities of the proof, but to try to get a sense of what all this is good for, i.e. where is it going? maybe looking at all the lemmas together will allow one to get some sense of this.

17. Nov 23, 2014

### A.Magnus

@mathwonk : You have brought up two very good points here:

(1) You urged me not to focus too much on small technicalities. However, the reason I am bean-counting is that because I was told that reading proofs, understanding proofs and doing proofs are the best way to accelerate your math maturity. (Please hold on to your fire for the time being. I know this elusive (but much coveted) math maturity is an exciting subject, I plan start a new posting so that other members can benefit too from your wisdom.)

(2) You urged me to focus instead on the end result, on the big picture. Again, I have an personal issue in this respect. Seeing the beauty of lower math theorems (at least up to the second year of undergraduate) is easy, but most of times I see only convoluted pictures in higher math. (Again, please hold on to your fire, I plan to post the whole story within couple of days, for me it is a burning question.)

Thank you.

18. Jan 5, 2015

### A.Magnus

Hi Mathwonk, I just posted a problem on the radical of finite group here. If you don't mind, I would love to get help from you. Thank you for your time.