How to find the elements of a finite field?

OK, I am asked to find the elements for the addition and multiplication tables for the finite fields with eight elements. I know I've asked about this question previously, and I've almost gotten a complete understanding of the problem. This is what I've done:

I know that we cannot construct our finite field with a composite, but rather the prime version of it:

$$\mathbb{F_8}=\mathbb{F_{2^3}}$$.

Instead, we reduce the field by finding an isomorphic field to it. We can do this by picking ANY irreducible polynomial:

$$\mathbb{F_{2^3}}\cong \mathbb{F_2}[x]/(x^3+x^2+1)$$

This is where I get stuck. What are the elements of this isomorphic field? If I can get that then I can figure out exactly what the multiplication and addition tables are which is what the problem is asking for.

Someone mentioned that it would be all the polynomials in $\mathbb{Z_2}$ lower than degree 3. Is this true? If so, or not, what would be the best way to find the required polynomials to solve the problem? Thanks so much for the help thus far!

• The statement in your last paragraph is true. It's usually a theorem proved soon after introducing polynomial rings over fields. What reference are you using? Mar 14, 2016 at 19:24
• @Ayman Hourieh It's Terras book on Discrete Fourier Transforms. Mar 14, 2016 at 19:43
• @Dietrich Burde, what would be the correct isomorphism? Mar 14, 2016 at 19:45

The correct isomorphism is $$\mathbb{F}_8\cong \mathbb{F}_2[x]/(x^3+x+1),$$ which contains (by definition of a quotient ring) the equivalence classes of the eight polynomials $$0,1,x,1+x,x^2,1+x^2,1+x+x^2,x+x^2.$$ Classes of higher degree polynomials reduce to these elements, since $x^3=x+1=-(x+1)$ in $\mathbb{F}_2[x]$.

• How do you get those equivalence classes? Also, I had the wrong isomorphism, is it because of the polynomial I've chosen? Mar 14, 2016 at 19:43
• Well, I believe it's read $\mathbb{F_2}$ modded out by $x^3+x+1$. The example in the book is given in the following manner: $\mathbb{F_4}\cong \mathbb{F_2}/(x^2+x+1)=\mathbb{F_\alpha}=\{0,1,\alpha,\alpha+1\}$. I don't understand where those elements in the set is coming from. Mar 14, 2016 at 19:51
• No, this is not correct. You wrote yourself in your other question a different statement: "Terras makes this statement in her book: "If $\alpha$ is a root of $f(x)$, the field obtained by adjoining $\alpha$ to $\mathbb{F}_p$ is $\mathbb{F}_q\cong \mathbb{F}_p[x]/f(x)$...Example: $\mathbb{F}_4\cong \mathbb{F}_2[x]/(x^2+x+1)=\mathbb{F}_2(\alpha)=\{0,1,\alpha,\alpha+1\}$" Mar 14, 2016 at 20:12
• Oh, I'm missing $\mathbb{F_2}[x]$...Sorry, I will edit this. Still, how do you come up with those equivalence classes? Mar 14, 2016 at 21:23
• Take any polynomial, i.e., say, $x^4+1$. Since $x^3=x+1$ we have $x^4=x^2+x$, so that $x^4+1=x^2+x+1$, one of our $8$ elements in $\mathbb{F}_8$. This way we come up with those equivalence classes. Mar 14, 2016 at 23:13