Re: algebra again

In article <1113336793.697650.61010@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
nonton <scrilla_12_1999@xxxxxxxxx> wrote:
>But how would i show that M=S?

You don't have to. You were asked to show that EITHER M=S or ELSE that
M is contained in Z. Under the assumptions you have provided, "M
contained in Z" is ->always<- true.

Which suggests to me that, yet again, you have failed to accurately
write down the problem you were asked to solve.

>> >m= sms^(-1)for every element m of M and every nonzero element s of
>S.
>>
>> Then this is trivial! Multiply both sides by s on the right, and we
>> have that
>>
>> ms = sm for every m in M and every nonzero elemetn s of S.
>>
>> That means that M is contained in the center of S, since
>>
>> Z = {y in S: yx = xy for every x in S}
>>
>> and clearly m0 = 0m for every m in M.


--
======================================================================
"It's not denial. I'm just very selective about
what I accept as reality."
--- Calvin ("Calvin and Hobbes")
======================================================================

Arturo Magidin
magidin@xxxxxxxxxxxxxxxxx

.

Relevant Pages

  • Re: "It is easy to see...."
    ... all this stuff cold, but it's been too many years... ... what I accept as reality." ... Arturo Magidin ... Prev by Date: ...
    (sci.math)
  • Re: Set of irrationals closed under sum
    ... and let A be the Z-span of S. ... what I accept as reality." ... Arturo Magidin ... Prev by Date: ...
    (sci.math)
  • Re: Dedekind Domain
    ... Sorry I can't be more detailed; I'm away from the office, and plan to ... what I accept as reality." ... Arturo Magidin ... Prev by Date: ...
    (sci.math)
  • Re: Split or Irreducible
    ... If u is a root for f, ... what I accept as reality." ... Arturo Magidin ... Prev by Date: ...
    (sci.math)
  • Re: A few more GRE questions
    ... Hmmm... ... what I accept as reality." ... Arturo Magidin ... Prev by Date: ...
    (sci.math)