# Polynomial rings -- F2[x]

• Dec 27th 2012, 03:35 PM
Bernhard
Polynomial rings -- F2[x]
In Dummit and Foote section 7.4 Exercise 15 reads as follows:

"Let $x^2 + x + 1$ be an element of the polynomial ring $E = {\mathbb{F}_2 [x]$ ... ..."

I am unsure of the meaning of the notation ${\mathbb{F}_2$ - can someone please help?

Peter
• Dec 27th 2012, 04:21 PM
ILikeSerena
Re: Polynomial rings -- F2[x]
Hi Bernhard.

$F_2 = \{0, 1\}$ is the field $\mathbb Z/2 \mathbb Z$.
Calculations are modulo 2.
• Dec 27th 2012, 05:23 PM
Bernhard
Re: Polynomial rings -- F2[x]
Thanks!

So ${\mathbb{F}}_p$ 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, 09:52 PM
Deveno
Re: 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 pn 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 Fk, or sometimes as GF(k) (GF for "galois field", in honor of evariste galois).