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Topic: Finite Rings
Replies: 28   Last Post: Feb 6, 2013 8:33 AM

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magidin@math.berkeley.edu

Posts: 11,145
Registered: 12/4/04
Re: Finite Rings
Posted: Feb 4, 2013 1:21 AM
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On Sunday, February 3, 2013 10:46:14 PM UTC-6, William Elliot wrote:
> On Sun, 3 Feb 2013, Arturo Magidin wrote:
>

> > On Sunday, February 3, 2013 9:21:19 PM UTC-6, William Elliot wrote:
>
>
>

> > > > If R is a finite commutative ring without multiplicative identity
>
> > > > and if every element is a zero divisor, then does there exist
>
> > > > a nonzero element which annihilates all elements of the ring?
>
> > > Ask-an-Algebraist
>
> > > No - the trivial ring.
>
> >
>
> > Incorrect. The trivial ring *does* have a multiplicative identity. I'll let you figure out what it is and why it *does* satisfy the condition that 1x=x=x1 for all x in the ring.
>
> > In fact, I'll give you three guesses.
>
> > The first two don't count, though.
>
> >
>
> Your trivial ring isn't as trivial as my trivial ring because
>
> your trivial ring is fancied up with a multiplicative identity.



Nonsense (as usual, coming from you). If you have a ring with one element, then that element is a multiplicative identity, simply because it satisfies the definition of being a multiplicative identity. You messed up when trying to act superior. Take your medicine and admit it (you should be used to it by now).


> > > So add the premise that R has a nonzero element.
>
> > Or, perhaps, not.
>
>
>
> Definitely so for OP asked about rings without multiplicative identities.


Which means that the trivial ring DOES NOT satisfy the hypothesis, and therefore is not to be considered, period. The fact that a ring without multiplicative identity must contain a nonzero element need not be a premise, because "does not have a multiplicative identity" IMPLIES, necessarily, the existence of a nonzero identity.

Thus, claiming that such an assumption must be added is incorrect. Adding it would be pleonasmic.

--
Arturo Magidin





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