In Dummit and Foote section 7.4 Exercise 15 reads as follows:

"Let be an element of the polynomial ring ... ..."

I am unsure of the meaning of the notation - can someone please help?

Peter

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- Dec 27th 2012, 04:35 PMBernhardPolynomial rings -- F2[x]
In Dummit and Foote section 7.4 Exercise 15 reads as follows:

"Let be an element of the polynomial ring ... ..."

I am unsure of the meaning of the notation - can someone please help?

Peter - Dec 27th 2012, 05:21 PMILikeSerenaRe: Polynomial rings -- F2[x]
Hi Bernhard.

is the field .

Calculations are modulo 2. - Dec 27th 2012, 06:23 PMBernhardRe: Polynomial rings -- F2[x]
Thanks!

So would mean Z/<p> where p is prime and the notation would not be used when p is composite since then Z/<p> would not be a field

Is that correct?

Peter - Dec 27th 2012, 10:52 PMDevenoRe: Polynomial rings -- F2[x]
yes, it turns out that for finite fields, the possibilities are more severely restricted than for other structures: there are NO fields of order 6, for example.

here is why: every field has a minimal subfield, called its prime field, which is the subfield generated by 1. if a field is finite, then the (additive) order of 1 is likewise finite.

this number (called the characteristic of F....if 1 is of infinite order, we use 0 as the characteristic) must be a prime number for a finite field, because if mn*1 = 0, then (m*1)(n*1) = 0, so we have zero divisors.

furthermore, any finite field K which contains a prime field isomorphic to Z/(p), is a vector space (of necessarily finite dimension n) over Z/(p), and hence has exactly p^{n}elements, meaning finite fields must be of prime power order.

moreover, every two such fields are isomorphic as fields, meaning a finite field is essentially (up to isomorphism) determined by its size. it is customary to refer to such a field as F_{k}, or sometimes as GF(k) (GF for "galois field", in honor of evariste galois).