Re: Question about zero divisors in finite rings


Gerry Myerson wrote:
In article <kuiee39s3pni8ah478mk7fcqv0tt9rejdb@xxxxxxx>,
Brian VanPelt <brvanpelt@xxxxxxxxxxxxx> wrote:

On Tue, 11 Sep 2007 18:07:54 -0700, Rotwang <sg552@xxxxxxxxxxxxx>
wrote:

While reading an old thread the other day I saw an article in which a
poster asserted that any ring with 6 elements necessarily contains
zero divisors. Having thought a bit about why this is true I think I
can show that a ring with n elements must have zero divisors provided

i) n is composite, and
ii) no prime appears in the prime factorisation of n with exponent
greater than one.

My question is: is this correct, and is it a special case of a more
general fact?

If n is composite with unity, then i) is all you need. If n = km,
then the k*1 times m*1 is the ring zero.

I don't know what it means for a number to be "composite with unity,"
but the field of 4 elements is a ring with a composite number of
elements and no zero divisors.

Hi Gerry,

I think the O.P. may have meant 'ring that isn't also a field' when
he said
"ring with n elements". At least this is what is implied by his
conditions i) & ii)
which forbids n to be a prime power. Your example, (GF(2^2)), is
forbidden by condition
ii)

But perhaps this is not correct, since it follows immediately that
such a ring
must have zero divisors from the definition. (i.e. it isn't an
integral domain)

Perhaps he is trying to prove that every finite field must have a
prime or prime
power number of elements????

I too do not know what he means by "composite with unity".

.

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