# Thread: Polynomial rings -- F2[x]

1. ## Polynomial rings -- F2[x]

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

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

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

Peter

2. ## Re: Polynomial rings -- F2[x]

Hi Bernhard.

$\displaystyle F_2 = \{0, 1\}$ is the field $\displaystyle \mathbb Z/2 \mathbb Z$.
Calculations are modulo 2.

3. ## Re: Polynomial rings -- F2[x]

Thanks!

So $\displaystyle {\mathbb{F}}_p$ 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

4. ## 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).

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### what does f2) mean in math

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