Re: Question about zero divisors in finite rings

Arturo Magidin <magidin@xxxxxxxxxxxxxxxxx> wrote:
G.E. Ivey <george.ivey@xxxxxxxxxxxxx> wrote:

The field with 4 elements? I was under the impression
that a finite field had to have a prime number of elements

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.

and in another post here

In any case, I find it so much easier to think of GF(4) as being
all numbers of the form a+bi with a and b in GF(2)...

Most certainly not. In any domain i^2 = 1 -> i = +-1

In other words, x^2 + 1 = (x+1)^2 is reducible over F2

so F2[i] isn't a field. Correct is F4 = F2[w]/(ww+w+1)

i.e. extend F2 by a primitive cube root of 1, not by i.

--Bill Dubuque
.

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