On Mar 10, 10:45 pm, leiko <leikomats...@gmail.com> wrote:
> Is the completion of a noetherian domain at a a prime P also a domain? > If so, could someone give me a proof of this? > Thanks
What I understand is the "completion of R at the ideal J" is the inverse limit of the rings R/J^n, with n=1, 2, ... etc.
That is, the subring of the product Prod(R/J^n), n=1,2,3,..., consisting of all coherent sequences, (a_1, a_2, ..., a_n,...) where a_{n+1}=a_n (mod J^n); if we think of a_i as elements of R, then we have (a_1,a_2,...) = (b_1,b_2,...) if and only if b_i = a_i (mod P^i).
Assuming this is what you mean, (a_1,a_2,...)=0 if and only if a_i is in P^i for all i.
Now, the argument below is done thinking about the p-adic numbers. I am using:
Lemma: Let R be a commutative ring, let P be a prime ideal, and let m be a positive integer. Let a and b be elements of R. Suppose that a lies in P^m but not in P^{m+1} and b lies in P^n but not in P^{n+1}; then ab lies in P^{n+m} but not in P^{n+m+1}.
I think this holds in arbitrary rings (it is certainly true in the integers), but I could certainly be wrong. If I am, sorry for wasting your time.
Suppose that (a_1,a_2,...)(b_1,b_2,..) = 0. If a_1 is not in P, then a_i is not in P for all i, hence the fact that a_ib_i is in P^i forces b_i to be in P^i, so (b_1,b_2,...) = 0.
If a_1 is in P, but (a_1,a_2,...) =/=0, then there exists a least k, k>0, such that a_k is in P^k but a_{k+1} is not in P^{k+1}; in particular, a_{k+m} is in P^k but not in P^{k+1} for all m>0.
Let m>0. Since a_{k+m}b_{k+m} lies in P^m, but a_{k+m} lies in P^k but not in P^{k+1}, then b_{k+m} must lie in P^m. If b_{k+m} is not 0, let t be the largest positive integer such that b lies in P^{m+t}. I claim that t>=k. For if t<k, then for any s>0 we know that b_{k+m+s} = b_{k +m} (mod P^{k+m}), and therefore b_{k+m+s} lies in P^{m+t} but not in P^{m+t+1}. Thus, a_{k+m+s}b_{k+m+s} lies in P^{k}P^{m+t}=P^{k+m+t} but not in P^{k+m+t+1}. However, we know that a_{k+m+s}b_{k+m+s} lies in P^{k+m+s}, so picking s>t yields a contradiction. Thus, t>=k, hence b_{k+m} lies in P^{k+m} for all m>0, and this means that (b_1,b_2,...) = 0.
if b_{k+m} = 0, but some b_{k+m+t}, t>0, is nonzero, repeat the above argument there. (If all b_{k+m+t} = 0, there is nothing to do, of course).
Thus, if the product is 0, and (a_1,a_2,...) =/=0, then (b_1,b_2,...) = 0, so the completion is a domain. (Modulo the lemma)
> On Mar 10, 10:45 pm, leiko <leikomats...@gmail.com> wrote:
> > Is the completion of a noetherian domain at a a prime P also a domain? > > If so, could someone give me a proof of this? > > Thanks
> What I understand is the "completion of R at the ideal J" is the > inverse limit of the rings R/J^n, with n=1, 2, ... etc.
> That is, the subring of the product Prod(R/J^n), n=1,2,3,..., > consisting of all coherent sequences, (a_1, a_2, ..., a_n,...) where > a_{n+1}=a_n (mod J^n); if we think of a_i as elements of R, then we > have (a_1,a_2,...) = (b_1,b_2,...) if and only if b_i = a_i (mod P^i).
> Assuming this is what you mean, (a_1,a_2,...)=0 if and only if a_i is > in P^i for all i.
> Now, the argument below is done thinking about the p-adic numbers. I > am using:
> Lemma: Let R be a commutative ring, let P be a prime ideal, and let m > be a positive integer. Let a and b be elements of R. Suppose that a > lies in P^m but not in P^{m+1} and b lies in P^n but not in P^{n+1}; > then ab lies in P^{n+m} but not in P^{n+m+1}.
> I think this holds in arbitrary rings (it is certainly true in the > integers), but I could certainly be wrong. If I am, sorry for wasting > your time.
> Suppose that (a_1,a_2,...)(b_1,b_2,..) = 0. If a_1 is not in P, then > a_i is not in P for all i, hence the fact that a_ib_i is in P^i forces > b_i to be in P^i, so (b_1,b_2,...) = 0.
> If a_1 is in P, but (a_1,a_2,...) =/=0, then there exists a least k, > k>0, such that a_k is in P^k but a_{k+1} is not in P^{k+1}; in > particular, a_{k+m} is in P^k but not in P^{k+1} for all m>0.
> Let m>0. Since a_{k+m}b_{k+m} lies in P^m, but a_{k+m} lies in P^k but > not in P^{k+1}, then b_{k+m} must lie in P^m. If b_{k+m} is not 0, let > t be the largest positive integer such that b lies in P^{m+t}. I claim > that t>=k. For if t<k, then for any s>0 we know that b_{k+m+s} = b_{k > +m} (mod P^{k+m}), and therefore b_{k+m+s} lies in P^{m+t} but not in > P^{m+t+1}. Thus, a_{k+m+s}b_{k+m+s} lies in P^{k}P^{m+t}=P^{k+m+t} but > not in P^{k+m+t+1}. However, we know that a_{k+m+s}b_{k+m+s} lies in > P^{k+m+s}, so picking s>t yields a contradiction. Thus, t>=k, hence > b_{k+m} lies in P^{k+m} for all m>0, and this means that (b_1,b_2,...) > = 0.
> if b_{k+m} = 0, but some b_{k+m+t}, t>0, is nonzero, repeat the above > argument there. (If all b_{k+m+t} = 0, there is nothing to do, of > course).
> Thus, if the product is 0, and (a_1,a_2,...) =/=0, then (b_1,b_2,...) > = 0, so the completion is a domain. (Modulo the lemma)
> -- > Arturo Magidin
Thanks a lot. I believe the lemma; however I could be even more wrong than you. I was also concerned about this other fact which I am not sure of: In a Noetherian ring, is the infinite intersection of powers P^n of a prime P equal to 0? I mgiht make a different psot with these two questions and maybe someone will correct these facts if they are wrong or confirm that they are right.
Thanks for the proof again..I had managed to show that x^ in the completion is not a zero divisor for any x, but that was not enough to conclude that the completion is not a domain, since Eisenbud has a counterexample: A completion of a domain at a nonprime ideal which is a nondomain (pages 187-188).
On Mar 11, 12:12 pm, leiko <leikomats...@gmail.com> wrote:
> On Mar 11, 11:52 am, Arturo Magidin <magi...@member.ams.org> wrote:
Alas:
> > Lemma: Let R be a commutative ring, let P be a prime ideal, and let m > > be a positive integer. Let a and b be elements of R. Suppose that a > > lies in P^m but not in P^{m+1} and b lies in P^n but not in P^{n+1}; > > then ab lies in P^{n+m} but not in P^{n+m+1}.
is false as stated; it's true in a PID, but false even in Principal Ideal Rings (take R = Z/36Z, P=(2), and a=b=2; then a,b are in P-P^2, but ab is in P^3 because P^3=P^2). I don't know what happens if P^n+1 is properly contained in P^n for all n, or if we have a domain.
On Mar 11, 3:40 pm, Arturo Magidin <magi...@member.ams.org> wrote:
> On Mar 11, 12:12 pm, leiko <leikomats...@gmail.com> wrote:
> > On Mar 11, 11:52 am, Arturo Magidin <magi...@member.ams.org> wrote:
> Alas:
> > > Lemma: Let R be a commutative ring, let P be a prime ideal, and let m > > > be a positive integer. Let a and b be elements of R. Suppose that a > > > lies in P^m but not in P^{m+1} and b lies in P^n but not in P^{n+1}; > > > then ab lies in P^{n+m} but not in P^{n+m+1}.
> is false as stated; it's true in a PID, but false even in Principal > Ideal Rings (take R = Z/36Z, P=(2), and a=b=2; then a,b are in P-P^2, > but ab is in P^3 because P^3=P^2). I don't know what happens if P^n+1 > is properly contained in P^n for all n, or if we have a domain.
Also false in Noetherian domains, as I posted elsewhere: R = k[x^2,x^3], P=(x^2,x^3), a=b=x^3.
I think I can fix the proof; but:
(i) I'm about to go proctor, so I can't do it now; and (ii) My laptop decided to stop booting up yesterday. I think it's just the hard drive dying, in which case I hope I can get it back up in short order, but I may not be able to post until the weekend is over.
> On Mar 11, 3:40 pm, Arturo Magidin <magi...@member.ams.org> wrote:
> > On Mar 11, 12:12 pm, leiko <leikomats...@gmail.com> wrote:
> > > On Mar 11, 11:52 am, Arturo Magidin <magi...@member.ams.org> wrote:
> > Alas:
> > > > Lemma: Let R be a commutative ring, let P be a prime ideal, and let m > > > > be a positive integer. Let a and b be elements of R. Suppose that a > > > > lies in P^m but not in P^{m+1} and b lies in P^n but not in P^{n+1}; > > > > then ab lies in P^{n+m} but not in P^{n+m+1}.
> > is false as stated; it's true in a PID, but false even in Principal > > Ideal Rings (take R = Z/36Z, P=(2), and a=b=2; then a,b are in P-P^2, > > but ab is in P^3 because P^3=P^2). I don't know what happens if P^n+1 > > is properly contained in P^n for all n, or if we have a domain.
> Also false in Noetherian domains, as I posted elsewhere: R = > k[x^2,x^3], P=(x^2,x^3), a=b=x^3.
> I think I can fix the proof; but:
> (i) I'm about to go proctor, so I can't do it now; and > (ii) My laptop decided to stop booting up yesterday. I think it's just > the hard drive dying, in which case I hope I can get it back up in > short order, but I may not be able to post until the weekend is over.
> -- > Arturo Magidin
Thank you a lot! I am looking forward to your post then, whenever it is... I'll think about it some more also meanwhile.
On Mar 12, 2:51 pm, leiko <leikomats...@gmail.com> wrote:
> Thank you a lot! I am looking forward to your post then, whenever it > is... I'll think about it some more also meanwhile.
Unfortunately, I can't quite seem to get the argument to work. We can localize at P so that we are working in a local Noetherian ring and we are localizing at the maximal ideal (here we are implicitly using that P is a prime ideal and not just any old ideal, in order to get R to embed in its localization; this takes care of the Eisenbud counterexample)
And in fact, as I look at Eisenbud's "Commutative Algebra with a View Toward Algebraic Geometry", I see why I can't quite get it to work. It's false!
Specifically, in page 191, it says:
"In fact, a theorem of Larfeldt and Lech [Larfeldt, T. and C. Lech (1981). Analytic ramifications and flat couples of local rings. Acta Math. 146, pp. 201-208, says that if A is any finite-dimensional algebra over a field k (for example, k[x]/(x^2) ), then there is a Noetherian local integral domain R with maximal ideal M such that [the completion of R at M is isomorphic to] A[[x_1,...,x_n]] for some n."
Since A[[x_1,...,x_n]] is not a domain if A has zero divisors (as is the case in the example given), the completion may very well fail to be a domain even when R is a Noetherian local integral domain.
Eisenbud also says: "there have been attempts to define a more special class of rings that would not only be Noetherian, but would also share other good properties of affine rings", such as the completion with respect to a maximal ideal having no nilpotent elements. He mentiones Grothendieck's "excellent rings".
So, that's why I can't get it to work. Sorry for thinking I could.
> On Mar 12, 2:51 pm, leiko <leikomats...@gmail.com> wrote:
> > Thank you a lot! I am looking forward to your post then, whenever it > > is... I'll think about it some more also meanwhile.
> Unfortunately, I can't quite seem to get the argument to work. We can > localize at P so that we are working in a local Noetherian ring and we > are localizing at the maximal ideal (here we are implicitly using that > P is a prime ideal and not just any old ideal, in order to get R to > embed in its localization; this takes care of the Eisenbud > counterexample)
> And in fact, as I look at Eisenbud's "Commutative Algebra with a View > Toward Algebraic Geometry", I see why I can't quite get it to work. > It's false!
> Specifically, in page 191, it says:
> "In fact, a theorem of Larfeldt and Lech [Larfeldt, T. and C. Lech > (1981). Analytic ramifications and flat couples of local rings. Acta > Math. 146, pp. 201-208, says that if A is any finite-dimensional > algebra over a field k (for example, k[x]/(x^2) ), then there is a > Noetherian local integral domain R with maximal ideal M such that [the > completion of R at M is isomorphic to] A[[x_1,...,x_n]] for some n."
> Since A[[x_1,...,x_n]] is not a domain if A has zero divisors (as is > the case in the example given), the completion may very well fail to > be a domain even when R is a Noetherian local integral domain.
> Eisenbud also says: "there have been attempts to define a more special > class of rings that would not only be Noetherian, but would also share > other good properties of affine rings", such as the completion with > respect to a maximal ideal having no nilpotent elements. He mentiones > Grothendieck's "excellent rings".
> So, that's why I can't get it to work. Sorry for thinking I could.
> -- > Arturo Magidin
Thank you a lot for thinking about it. So I did show that if x is not a zero divisor, then its image in the completion is not a zero divisor either. I know that this is not enough, since we have the counterexample in Eisenbud, to conclude that if A is a domain, the completion at any ideal is also a domain. But I want to show that the completion at a prime ideal is also a domain if the initial ring is a domain... Are you saying that this is not true? That's what I am understanding from what you wrote, but I am very surprised....
Well, what about the completion of a Noetherian domain at a prime ideal which is generated by a regular sequence of elements in the ring...would that be a domain?
> On Mar 14, 4:16 pm, Arturo Magidin <magi...@member.ams.org> wrote:
> > On Mar 12, 2:51 pm, leiko <leikomats...@gmail.com> wrote:
> > > Thank you a lot! I am looking forward to your post then, whenever it > > > is... I'll think about it some more also meanwhile.
> > Unfortunately, I can't quite seem to get the argument to work. We can > > localize at P so that we are working in a local Noetherian ring and we > > are localizing at the maximal ideal (here we are implicitly using that > > P is a prime ideal and not just any old ideal, in order to get R to > > embed in its localization; this takes care of the Eisenbud > > counterexample)
> > And in fact, as I look at Eisenbud's "Commutative Algebra with a View > > Toward Algebraic Geometry", I see why I can't quite get it to work. > > It's false!
> > Specifically, in page 191, it says:
> > "In fact, a theorem of Larfeldt and Lech [Larfeldt, T. and C. Lech > > (1981). Analytic ramifications and flat couples of local rings. Acta > > Math. 146, pp. 201-208, says that if A is any finite-dimensional > > algebra over a field k (for example, k[x]/(x^2) ), then there is a > > Noetherian local integral domain R with maximal ideal M such that [the > > completion of R at M is isomorphic to] A[[x_1,...,x_n]] for some n."
> > Since A[[x_1,...,x_n]] is not a domain if A has zero divisors (as is > > the case in the example given), the completion may very well fail to > > be a domain even when R is a Noetherian local integral domain.
> > Eisenbud also says: "there have been attempts to define a more special > > class of rings that would not only be Noetherian, but would also share > > other good properties of affine rings", such as the completion with > > respect to a maximal ideal having no nilpotent elements. He mentiones > > Grothendieck's "excellent rings".
> > So, that's why I can't get it to work. Sorry for thinking I could. > Thank you a lot for thinking about it. > So I did show that if x is not a zero divisor, then its image in the > completion is not a zero divisor either. > I know that this is not enough, since we have the counterexample in > Eisenbud, to conclude that if A is a domain, the completion at any > ideal is also a domain. > But I want to show that the completion at a prime ideal is also a > domain if the initial ring is a domain... Are you saying that this is > not true?
I am saying that what Eisenbud says shows that this is not always the case: pick your favorite finite dimensional algebra over a field that has nilpotent elements or zero divisors: for example, take k=Q[x]/ (x^2), with Q the rational numbers. The theorem quoted says that there is a local Noetherian domain R with maximal ideal M such that the completion of R at M is isomorphic to a ring of power series over k. Since k has nilpotent elements/zero divisors, so does a ring of power series over R.
But if M is a maximal ideal of a commutative ring with 1, then it is also a *prime* ideal. So *there* is an example of a Noetherian *domain*, R, and a prime ideal M, such that the completion of R at the prime ideal M is *not* a domain.
So yes: I am saying that it is not always true, even if your initial domain is noetherian and you complete at a prime.
> That's what I am understanding from what you wrote, but I am > very surprised....
> Well, what about the completion of a Noetherian domain at a prime > ideal which is generated by a regular sequence of elements in the > ring...would that be a domain?
I think I remember seeing something about this in Nagata's "Local Rings"; I'll check at the office tomorrow.
> On Mar 14, 6:01 pm, leiko <leikomats...@gmail.com> wrote:
> > On Mar 14, 4:16 pm, Arturo Magidin <magi...@member.ams.org> wrote:
> > > On Mar 12, 2:51 pm, leiko <leikomats...@gmail.com> wrote:
> > > > Thank you a lot! I am looking forward to your post then, whenever it > > > > is... I'll think about it some more also meanwhile.
> > > Unfortunately, I can't quite seem to get the argument to work. We can > > > localize at P so that we are working in a local Noetherian ring and we > > > are localizing at the maximal ideal (here we are implicitly using that > > > P is a prime ideal and not just any old ideal, in order to get R to > > > embed in its localization; this takes care of the Eisenbud > > > counterexample)
> > > And in fact, as I look at Eisenbud's "Commutative Algebra with a View > > > Toward Algebraic Geometry", I see why I can't quite get it to work. > > > It's false!
> > > Specifically, in page 191, it says:
> > > "In fact, a theorem of Larfeldt and Lech [Larfeldt, T. and C. Lech > > > (1981). Analytic ramifications and flat couples of local rings. Acta > > > Math. 146, pp. 201-208, says that if A is any finite-dimensional > > > algebra over a field k (for example, k[x]/(x^2) ), then there is a > > > Noetherian local integral domain R with maximal ideal M such that [the > > > completion of R at M is isomorphic to] A[[x_1,...,x_n]] for some n."
> > > Since A[[x_1,...,x_n]] is not a domain if A has zero divisors (as is > > > the case in the example given), the completion may very well fail to > > > be a domain even when R is a Noetherian local integral domain.
> > > Eisenbud also says: "there have been attempts to define a more special > > > class of rings that would not only be Noetherian, but would also share > > > other good properties of affine rings", such as the completion with > > > respect to a maximal ideal having no nilpotent elements. He mentiones > > > Grothendieck's "excellent rings".
> > > So, that's why I can't get it to work. Sorry for thinking I could. > > Thank you a lot for thinking about it. > > So I did show that if x is not a zero divisor, then its image in the > > completion is not a zero divisor either. > > I know that this is not enough, since we have the counterexample in > > Eisenbud, to conclude that if A is a domain, the completion at any > > ideal is also a domain. > > But I want to show that the completion at a prime ideal is also a > > domain if the initial ring is a domain... Are you saying that this is > > not true?
> I am saying that what Eisenbud says shows that this is not always the > case: pick your favorite finite dimensional algebra over a field that > has nilpotent elements or zero divisors: for example, take k=Q[x]/ > (x^2), with Q the rational numbers. The theorem quoted says that there > is a local Noetherian domain R with maximal ideal M such that the > completion of R at M is isomorphic to a ring of power series over k. > Since k has nilpotent elements/zero divisors, so does a ring of power > series over R.
> But if M is a maximal ideal of a commutative ring with 1, then it is > also a *prime* ideal. So *there* is an example of a Noetherian > *domain*, R, and a prime ideal M, such that the completion of R at the > prime ideal M is *not* a domain.
> So yes: I am saying that it is not always true, even if your initial > domain is noetherian and you complete at a prime.
> > That's what I am understanding from what you wrote, but I am > > very surprised....
> > Well, what about the completion of a Noetherian domain at a prime > > ideal which is generated by a regular sequence of elements in the > > ring...would that be a domain?
> I think I remember seeing something about this in Nagata's "Local > Rings"; I'll check at the office tomorrow.
On Mar 14, 6:01 pm, leiko <leikomats...@gmail.com> wrote:
> Well, what about the completion of a Noetherian domain at a prime > ideal which is generated by a regular sequence of elements in the > ring...would that be a domain?
I think so.
If you localize at the prime ideal P, you get a local Noetherian domain with maximal ideal (the image of) P. The regular sequence of elements (x_1,...x_n) that generates P in R will also generate the localization of P, and will be regular, since the quotient of the localization R_P/(x_1,..,x_i) is the localization of R/(x_1,...,x_i), and localizing a domain at a prime ideal does not introduce any zero divisors.
Now, a local ring R with maximal ideal M of dimension d is regular if and only if the (R/M) vector space M/M^2 is of dimension d; the minimal system of generators will be a regular sequence, and it is a theorem that a regular local ring is a domain; the completion of a regular local ring is a regular local ring, so you will get that the localization of the completion is a domain. Since you are localizing at a prime ideal, you get an embedding from the ring into the localization, so you can embed the completion in the regular local domain.
In short: if the dimension of your domain equals the length of your regular sequence, then "yes": you get a domain when you complete. I don't remember if regular sequences necessarily imply that your ring is regular, though.
> On Mar 14, 6:01 pm, leiko <leikomats...@gmail.com> wrote:
> > Well, what about the completion of a Noetherian domain at a prime > > ideal which is generated by a regular sequence of elements in the > > ring...would that be a domain?
> I think so.
> If you localize at the prime ideal P, you get a local Noetherian > domain with maximal ideal (the image of) P. The regular sequence of > elements (x_1,...x_n) that generates P in R will also generate the > localization of P, and will be regular, since the quotient of the > localization R_P/(x_1,..,x_i) is the localization of R/(x_1,...,x_i), > and localizing a domain at a prime ideal does not introduce any zero > divisors.
> Now, a local ring R with maximal ideal M of dimension d is regular if > and only if the (R/M) vector space M/M^2 is of dimension d; the > minimal system of generators will be a regular sequence, and it is a > theorem that a regular local ring is a domain; the completion of a > regular local ring is a regular local ring, so you will get that the > localization of the completion is a domain. Since you are localizing > at a prime ideal, you get an embedding from the ring into the > localization, so you can embed the completion in the regular local > domain.
> In short: if the dimension of your domain equals the length of your > regular sequence, then "yes": you get a domain when you complete. I > don't remember if regular sequences necessarily imply that your ring > is regular, though.
> -- > Arturo Magidin
Thank you. Thm. A2.18 says that a local ring is regular iff its maximal ideal is generated by a regular sequence. :)
> On Mar 15, 2:39 pm, Arturo Magidin <magi...@member.ams.org> wrote:
> > On Mar 14, 6:01 pm, leiko <leikomats...@gmail.com> wrote:
> > > Well, what about the completion of a Noetherian domain at a prime > > > ideal which is generated by a regular sequence of elements in the > > > ring...would that be a domain?
> > I think so.
> > If you localize at the prime ideal P, you get a local Noetherian > > domain with maximal ideal (the image of) P. The regular sequence of > > elements (x_1,...x_n) that generates P in R will also generate the > > localization of P, and will be regular, since the quotient of the > > localization R_P/(x_1,..,x_i) is the localization of R/(x_1,...,x_i), > > and localizing a domain at a prime ideal does not introduce any zero > > divisors.
> > Now, a local ring R with maximal ideal M of dimension d is regular if > > and only if the (R/M) vector space M/M^2 is of dimension d; the > > minimal system of generators will be a regular sequence, and it is a > > theorem that a regular local ring is a domain; the completion of a > > regular local ring is a regular local ring, so you will get that the > > localization of the completion is a domain. Since you are localizing > > at a prime ideal, you get an embedding from the ring into the > > localization, so you can embed the completion in the regular local > > domain.
> > In short: if the dimension of your domain equals the length of your > > regular sequence, then "yes": you get a domain when you complete. I > > don't remember if regular sequences necessarily imply that your ring > > is regular, though.
> > -- > > Arturo Magidin
> Thank you. > Thm. A2.18 says that a local ring is regular iff its maximal ideal is > generated by a regular sequence. :)
However, I still couldn't quite figure out my other question, in the following particular case: If P is a principal prime ideal generated by a non-zero divisor, is the infinite intersection /\P^n=0? Do you have any ideas? Thanks again so much.
> On Mar 16, 3:40 am, leiko <leikomats...@gmail.com> wrote:
> > On Mar 15, 2:39 pm, Arturo Magidin <magi...@member.ams.org> wrote:
> > > On Mar 14, 6:01 pm, leiko <leikomats...@gmail.com> wrote:
> > > > Well, what about the completion of a Noetherian domain at a prime > > > > ideal which is generated by a regular sequence of elements in the > > > > ring...would that be a domain?
> > > I think so.
> > > If you localize at the prime ideal P, you get a local Noetherian > > > domain with maximal ideal (the image of) P. The regular sequence of > > > elements (x_1,...x_n) that generates P in R will also generate the > > > localization of P, and will be regular, since the quotient of the > > > localization R_P/(x_1,..,x_i) is the localization of R/(x_1,...,x_i), > > > and localizing a domain at a prime ideal does not introduce any zero > > > divisors.
> > > Now, a local ring R with maximal ideal M of dimension d is regular if > > > and only if the (R/M) vector space M/M^2 is of dimension d; the > > > minimal system of generators will be a regular sequence, and it is a > > > theorem that a regular local ring is a domain; the completion of a > > > regular local ring is a regular local ring, so you will get that the > > > localization of the completion is a domain. Since you are localizing > > > at a prime ideal, you get an embedding from the ring into the > > > localization, so you can embed the completion in the regular local > > > domain.
> > > In short: if the dimension of your domain equals the length of your > > > regular sequence, then "yes": you get a domain when you complete. I > > > don't remember if regular sequences necessarily imply that your ring > > > is regular, though.
> > > -- > > > Arturo Magidin
> > Thank you. > > Thm. A2.18 says that a local ring is regular iff its maximal ideal is > > generated by a regular sequence. :)
> However, I still couldn't quite figure out my other question, in the > following particular case: > If P is a principal prime ideal generated by a non-zero divisor, is > the infinite intersection /\P^n=0? Do you have any ideas? > Thanks again so much.
Nope..I realized that's wrong..Nevermind. Thanks a lot for your help.
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