Re: Question about zero divisors in finite rings
- From: Brian VanPelt <brvanpelt@xxxxxxxxxxxxx>
- Date: Wed, 12 Sep 2007 19:39:49 -0400
On Wed, 12 Sep 2007 04:55:22 GMT, Gerry Myerson
<gerry@xxxxxxxxxxxxxxxxxxxxxxxxx> 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.
Sorry, I don't know what it means either and I wrote it!
What I meant was to suppose that R was a ring with unity, and n was
composite. Say the unity is called 1. Then k*1 represents 1 added to
itself k times. If n = km, and * is the ring multiplication, then
k*1 * m*1 = (km)*1
which would be zero in this ring.
I didn't see how to do this without a unity element and I honestly
don't know it is true.
It looks like the poster wants to know that is i) and ii) are both
true, with no other restrictions on the ring, then the ring will
contain zero divisors. Now, I haven't read all the responses yet to
know if a counter-example was exhibited, but I did want to clarify
what I said.
Sorry,
Brian
.
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