Re: Question about zero divisors in finite rings

In article <17654468.1189595212126.JavaMail.jakarta@xxxxxxxxxxxxxxxxxxxxxx>,
G.E. Ivey <george.ivey@xxxxxxxxxxxxx> wrote:
In article
<kuiee39s3pni8ah478mk7fcqv0tt9rejdb@xxxxxxx>,
Brian VanPelt <brvanpelt@xxxxxxxxxxxxx> wrote:

[...]

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.

Z/nZ is a field if and only if p is prime for those reasons, but the
quotients of Z are not the only finite rings.

In particular, (Z/2Z)[i] = { 0, 1, i, 1+i} with i^2 = 1 is a field.

--
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"It's not denial. I'm just very selective about
what I accept as reality."
--- Calvin ("Calvin and Hobbes" by Bill Watterson)
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Arturo Magidin
magidin-at-member-ams-org

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