Re: Question about zero divisors in finite rings

In article
<17654468.1189595212126.JavaMail.jakarta@xxxxxxxxxxxxxx
orum.org>,
"G.E. Ivey" <george.ivey@xxxxxxxxxxxxx> 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.
The field with 4 elements? I was under the impression that a finite field had to have a prime number of elements for exactly the reasons expressed before.

Here is the field with four elements.

+| 0 1 x| 0 1
------- ------
0| 0 1 0| 0 1
1| 1 0 1| 1 1

A finite field must have a prime character.

(I think he meant the number was composite and the ring had "unity", a multiplicative identity.)

--
Michael Press
.

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