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Math Forum / Mathematics / General Topics / October 2008A local ring problem
Hi! I have a question: Let R be a commutative ring with identity. Let M be a maximal ideal in R, and n a positive integer. Consider the product ideal M^n=M . ... . M. How can I prove that R/(M^n) is a local ring (i.e. has a has a unique maximal ideal)? Evidently, if n=1, we are done! Well, I'm trying to prove that the quotient ring has a unique prime ideal. (Maybe this is too much). I was trying to use the fact: every prime ideal in R/I is of the form P/I, where P is a prime ideal in R that contains I. Thanks for helping. Max. > Hi! > [quoted text clipped - 5 lines] > Well, I'm trying to prove that the quotient ring has a unique prime > ideal. (Maybe this is too much). No, this is ok. > I was trying to use the fact: every prime ideal in R/I is of the form > P/I, where P is a prime ideal in R that contains I. This is a good starting point. What does the inclusion M^n subset P tell you? ;) > Thanks for helping. > Max.  SignatureBest wishes, J. > On 05.10.2008 23:39, maxblac...@gmail.com wrote: > [quoted text clipped - 15 lines] > This is a good starting point. What does the inclusion M^n subset P tell > you? ;) Thanks Jannick, If x.y belongs to M^n then x or y belongs to P, and R/P is integral. Also, x.y = (m_11 . ... . m_1n) + ... + (m_k1. .. .m_kn), for some positive k and m_ij in M. How can I use the maximality of M? I'm still very lost. Best, Max. >> On 05.10.2008 23:39, maxblac...@gmail.com wrote: >> [quoted text clipped - 20 lines] > How can I use the maximality of M? > I'm still very lost. If P is a prime ideal containing M^n, try to show that M is contained in P. Then use the maximality of M to get M = P. > Best, > Max. HTH. SignatureBest wishes, J. > On 06.10.2008 00:43, maxblac...@gmail.com wrote: > [quoted text clipped - 25 lines] > If P is a prime ideal containing M^n, try to show that M is contained in > P. Then use the maximality of M to get M = P. Ok, m in M, then m^n in M^n; so, m in P (P is prime) and M subset P subset R. Using THE FACT for maximal ideals, M=P. Therefore any prime ideal in R/I must come from M. Thanks a lot! Best, Max. I'd like to ask something else. If I + J = R (for I,J ideals in R), why then I^n + J^n = R? Kind regards, Max. : I'd like to ask something else. : If I + J = R (for I,J ideals in R), why then I^n + J^n = R? Use the fact that an ideal = R iff it contains 1. So if I + J = R, then for some x in I, y in J, we have x + y = 1. Continue from there. Ted > maxblac...@gmail.com <maxblac...@gmail.com> wrote: > [quoted text clipped - 5 lines] > > Ted Ok, I proved that! Now, mutiplying x+y=1 by x or y (left or right), I got x^2 + 2xy + y^2 = 1 (and all other 3 possibilities). I don't know how to move on. Thanks for your help Ted. Best wishes, Max. > maxblac...@gmail.com <maxblac...@gmail.com> wrote: > [quoted text clipped - 5 lines] > > Ted Ok, I proved that! Now, mutiplying x+y=1 by x or y (left or right), I got x^2 + 2xy + y^2 = 1 (and all other 3 possibilities). I don't know how to move on. Thanks for your help Ted. Best wishes, Max. > maxblac...@gmail.com <maxblac...@gmail.com> wrote: > [quoted text clipped - 5 lines] > > Ted Ok, I proved that! Now, mutiplying x+y=1 by x or y (left or right), I got x^2 + 2xy + y^2 = 1 (and all other 3 possibilities). I don't know how to move on. Do I have to suppose that R is commutative? Thanks for your help Ted. Best wishes, Max. : > maxblac...@gmail.com <maxblac...@gmail.com> wrote: : > [quoted text clipped - 3 lines] : > Use the fact that an ideal = R iff it contains 1. o if I + J = R, then : > for some x in I, y in J, we have x + y = 1. ontinue from there. : Ok, I proved that! : Now, mutiplying x+y=1 by x or y (left or right), I got : x^2 + 2xy + y^2 = 1 (and all other 3 possibilities). : I don't know how to move on. What can you do with x+y=1 to get elements of I^n or J^n to appear? : Do I have to suppose that R is commutative? No, but it might be easier to think of that case first. Ted > If I + J = R (for I,J ideals in R), why then I^n + J^n = R? Max M > (i^m,j^n) => M > (i,j) = 1 =><= --Bill Dubuque >> If I + J = R (for I,J ideals in R), why then I^n + J^n = R? > >Max M > (i^m,j^n) => M > (i,j) = 1 =><= Of course, this is implicitly assuming at least a weak version of the Axiom of Choice, to warrant the existence of a maximal ideal containing the given ideal. But this can be done without assuming that every proper ideal is contained in a maximum ideal: pick a in I and b in J with a+b = 1, and consider (a+b)^k for sufficiently high k. Signature====================================================================== "It's not denial. I'm just very selective about what I accept as reality." --- Calvin ("Calvin and Hobbes" by Bill Watterson) ====================================================================== Arturo Magidin magidin-at-member-ams-org > pick a in I and b in J with a+b = 1, and consider (a+b)^k for > sufficiently high k. Doesn't this method need that the ring is commutative? I am just wondering if the assertion holds at all if R is not commutative. SignatureBest wishes, J. >> pick a in I and b in J with a+b = 1, and consider (a+b)^k for >> sufficiently high k. > >Doesn't this method need that the ring is commutative? In its most obvious form, yes; and that was the context in which the question was asked. >I am just >wondering if the assertion holds at all if R is not commutative. I think you can make it work for arbitrary rings; the monomials will all be words in a and b; since I and J are ideals, you can group the terms to get elements of I^i or of J^j. For example, if you had abaaababbab you can think of it as (ab)*a*a*(ab)*(abb)*(ab) so this element is in I^5; though writing out (a+b)^k in the noncommutative case is probably tedious, it should still work. If you raise it to the kth power, each monomial will contain k letters; if k>n, then either you have at least n b's, or you have at least n a's. Etc. Signature====================================================================== "It's not denial. I'm just very selective about what I accept as reality." --- Calvin ("Calvin and Hobbes" by Bill Watterson) ====================================================================== Arturo Magidin magidin-at-member-ams-org >>> pick a in I and b in J with a+b = 1, and consider (a+b)^k for >>> sufficiently high k. [quoted text clipped - 21 lines] > k>n, then either you have at least n b's, or you have at least n > a's. Etc. Thanks. That means we can cope with one-sided ideals as well. SignatureBest wishes, J. >>>> pick a in I and b in J with a+b = 1, and consider (a+b)^k for >>>> sufficiently high k. [quoted text clipped - 23 lines] > >Thanks. That means we can cope with one-sided ideals as well. Well, you'll have to be careful there, since you'll need to have not only the "correct" number of, say, a's, but also have them on the "correct" side. As I said, probably very tedious... Signature====================================================================== "It's not denial. I'm just very selective about what I accept as reality." --- Calvin ("Calvin and Hobbes" by Bill Watterson) ====================================================================== Arturo Magidin magidin-at-member-ams-org [ R a ring with 1, A,B left ideals with R=A+B. Is R=A^2 + B^2? ] > Thanks. That means we can cope with one-sided ideals > as well. I don't think the argument handles one sided ideals. (a+b)^n = a^(n-1)*b + b^(n-1)*a mod (Ra)^2 + (Rb)^2 > [ R a ring with 1, A,B left ideals with R=A+B. > Is R=A^2 + B^2? ] [quoted text clipped - 5 lines] > I don't think the argument handles one sided ideals. > (a+b)^n = a^(n-1)*b + b^(n-1)*a mod (Ra)^2 + (Rb)^2 Actually this is foolish. If 1=a+b, then a and b commute. In general though (RaR + RbR)^n = (RaR)^n + (RbR)^n while (Ra+Rb)^n need not be equal to (Ra)^n + (Rb)^n. >> [ R a ring with 1, A,B left ideals with R=A+B. >> Is R=A^2 + B^2? ] [quoted text clipped - 7 lines] > >Actually this is foolish. If 1=a+b, then a and b commute. D'oh! So, of course, if I+J = (1), then I^n + J^n = 1, since you can expand (a+b)^{2n+1} using the binomial theorem and conclude that there exist r in I^n and s in J^n such that r+s = 1. For one sided ideals, you still get the same thing: If I and J are, say, left ideals, and I+J = (1), then there exist a in I and b in J with a+b=1; since they commute, (a+b)^{2n+1} can be expanded using the binomial theorem, and each monomial is either of the form k*a^n or j*b^n, with k,j in R, hence in I^n and in J^n, respectively, so I^n+J^n=(1). (I^n would be the ideal of all sums of n-fold products of elements of I, which is still a left ideal). Signature====================================================================== "It's not denial. I'm just very selective about what I accept as reality." --- Calvin ("Calvin and Hobbes" by Bill Watterson) ====================================================================== Arturo Magidin magidin-at-member-ams-org >>> [ R a ring with 1, A,B left ideals with R=A+B. >>> Is R=A^2 + B^2? ] [quoted text clipped - 7 lines] > > D'oh! Autsh!!! I haven't seen this either! These non-commutative thingies are really not my domain. > So, of course, if I+J = (1), then I^n + J^n = 1, since you can expand > (a+b)^{2n+1} using the binomial theorem and conclude that there exist [quoted text clipped - 9 lines] > (I^n would be the ideal of all sums of n-fold products of elements of > I, which is still a left ideal). ... just out of curiosity: Do we have something similar for three and more ideals - or does the result above tremendously depend on "descent" (to the commutative subring Z[a,b]) for two summands only? I must confess that I cannot see anything useful while staring at the commutators [a,b+c]=[b,a+c]=[c,a+b]=0 given 1 = a + b + c. SignatureBest wishes, J. [If R is a commutative ring with 1 and R = A + B for ideals A,B of R, then does R = A^n + B^n for each positive integer n?] >> pick a in I and b in J with a+b = 1, and consider >> (a+b)^k for sufficiently high k. > > Doesn't this method need that the ring is commutative? > I am just wondering if the assertion holds at all if > R is not commutative. Let R be a ring with 1, and A,B ideals such that R=A+B. Choose a in A, b in B with 1 = a+b. Notice that (RaR)^n is contained in A^n and (RbR)^n is contained in B^n. But (RaR)^n contains all elements of R that can be expressed as a product containing at least n copies of a. In particular, every term in the expansion of (a+b)^(2n) is contained in either (RaR)^n or (RbR)^n, so 1 = 1^(2n) in (RaR)^n + (RbR)^n so R = A^n + B^n. If one drops the assumption of unity, then the proposition is false. Choosing R to be the Klein four group with zero multiplication and A,B to be two maximal ideals gives R=A+B but A^2 + B^2 = 0. >>> If I + J = R (for I,J ideals in R), why then I^n + J^n = R? >> [quoted text clipped - 5 lines] > every proper ideal is contained in a maximum ideal: pick a in I and b > in J with a+b = 1, and consider (a+b)^k for sufficiently high k. If your prefer the constructive route then it'd be simpler to be a bit more radical, i.e. rad(I^m + J^n + ...) > I + J + ... = 1 => I^m + J^n + ... = 1 --Bill Dubuque >>>> If I + J = R (for I,J ideals in R), why then I^n + J^n = R? >>> [quoted text clipped - 12 lines] > > => I^m + J^n + ... = 1 Personally, I have no objection to the Axiom of Choice and will use it willingly; in deference to those that do have misgivings, though, I try to be explicit about its use, that's all. Signature====================================================================== "It's not denial. I'm just very selective about what I accept as reality." --- Calvin ("Calvin and Hobbes" by Bill Watterson) ====================================================================== Arturo Magidin magidin-at-member-ams-org |
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