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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: 11,697 Registered: 12/4/04
Re: Finite Rings
Posted: Feb 4, 2013 1:21 AM

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?
>
>
> > > 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.
>
>
>

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