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I have this question which I don't know how to approach:

Let ${F}_{2} = {Z}/2Z$, find representatives for the residue classes of ${F}_{2}[X]$ modulo the polynomial $f(x)$ and compute the multipication table for the ring ${F}_{2}[X]/(f(x))$ where

  1. $f(x) = x+1$
  2. $f(x) = x^2+x+1$
  3. $f(x) = x^2+1$

which are those rings are fields?

I understand $F_2$ and ${F}_{2}[X]$ but I don't know what the residues are and how to compute the multiplication table or how multiplication is defined in this ring. I suppose that checking if something is a field is easy once you have the multiplications table.

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up vote 3 down vote accepted

Do you know how you can define multiplication $ab$ in the ring ${\bf Z}/7{\bf Z}$ by multiplying $a$ times $b$ as integers, dividing by 7, and reporting the remainder as the answer? Well, it's the same thing: if $a$ and $b$ are in ${\bf F}_2[X]$ then you multiply them as polynomials, divide by $f(x)$, and report the remainder as the answer. For example, if $f(x)=x^2+x+1$, and you want to multiply $x$ by $x$, you get $x^2$, then divide by $x^2+x+1$, and the remainder is $x+1$ (well, $-x-1$, but, over ${\bf F}_2$, it's the same thing), so that's the answer.

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let $f=ax^n+\ldots+a_0$ where $a_i\in F$ and $a_n\neq 0$ then we have $F[x]/(f(x))=\{b_{n-1}x^{n-1}+\ldots +b_0\}$

Now you are in $F_2$ so how many choices we have for $b_{n-1}$,...,$b_0$ ?

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I guess $2^n$. How would the ${b}_{i}$s and the ${a}_{i}$ relate to one another? – Sonia Feb 13 '12 at 5:42
what is n for the first case? second case? third case ? – steven Feb 13 '12 at 6:23

Just as Z/pZ is a field when p is prime, the situation above is a field when the polynomial, f(x) can not factor (irreducible) over the base field. Of the polynomials you mentioned, only the second one has not roots in Z_2, hence it is irreducible, and will give rise to a field.

Since we are modulo a second degree polynomial now, because of the division algorithm, the remainders wil be of degree 1 or less, so of the form ax+b. There are only two choices for a and two choices for b; hence this filed has 4 elements.

0, 1, x, and x+1. Since x^2+x+1 =0 mod x^2+x+1 then x^2=-x-1 but over Z_2 it is equal to x+1. This is helpful when computing the multiplication table for the new field, and another reason the field has at most degree one elements, since one could continuously knock the degree down of elements, using this substitution.

hope this helps,

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