A local ring problem
maxbl...@gmail.com
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.
Jannick Asmus
> 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).
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.
--
Best wishes,
J.
maxbl...@gmail.com
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.
Jannick Asmus
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.
--
Best wishes,
J.
maxbl...@gmail.com
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.
Ted Hwa
: 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
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 =><=
--Bill Dubuque
Ted Hwa
: On 5 out, 23:09, Ted Hwa <hwath...@xenon.Stanford.EDU> wrote:
: > maxblac...@gmail.com <maxblac...@gmail.com> wrote:
: >
: > : I'd like to ask something else.
: > : If I + J = R (for I,J ideals in R), why then I^n + J^n = R?
: >
: > 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
Arturo Magidin
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.
--
======================================================================
"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
Jannick Asmus
> 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.
--
Best wishes,
J.
Arturo Magidin
Jannick Asmus <jannick.ne...@web.de> wrote:
>On 06.10.2008 15:43, Arturo Magidin wrote:
>
>> 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.
Jannick Asmus
> In article <gcd6bs$gak$1...@online.de>,
> Jannick Asmus <jannick.ne...@web.de> wrote:
>> On 06.10.2008 15:43, Arturo Magidin wrote:
>>
>>> 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.
Thanks. That means we can cope with one-sided ideals as well.
--
Best wishes,
J.
Arturo Magidin
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...
Jack Schmidt
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.
Jack Schmidt
> Is R=A^2 + B^2? ]
>
> > Thanks. That means we can cope with one-sided
> ideals
> > as well.
>
> (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.
Jack Schmidt
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.
Arturo Magidin
Jack Schmidt <Jack.Schmi...@gmail.com> wrote:
>> [ 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
>
>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).
Jannick Asmus
> In article <23197006.1223305598...@nitrogen.mathforum.org>,
> Jack Schmidt <Jack.Schmi...@gmail.com> wrote:
>>> [ 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
>> Actually this is foolish. If 1=a+b, then a and b commute.
>
> 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
> 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).
... 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.
--
Best wishes,
J.
Bill Dubuque
>Bill Dubuque <w...@nestle.csail.mit.edu> wrote:
>>"maxbl...@gmail.com" <maxbl...@gmail.com> wrote:
>>>
>>> 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.
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
Arturo Magidin
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.