Let $\mathbb{F}=\mathbb{F}_3$ Find an irreducible polynomial of degree 2 and construct a field of 9 elements as a quotient.

Question

Find an irreducible polynomial p of degree 2, and use it construct a field of 9 elements as a quotient. Describe the cosets in the quotient explicitly, and use them to construct the addition and multiplication tables for the field obtained by this quotient.

I'm not going to ask anyone to construct the tables. I'm more just unsure of how to go about solving this. I have an irreducible polynomial $[2]x^2+x+[1]$,

I know the elements should be remainders when dividing by other polynomials of degree less then 2. So I should have to use long division by this polynomial I believe I get remainders $\{0,1,x,x+1,x+2,2x,2x+1,2x+2,2\}$

but I'm not sure exactly how I go about this from here. I've done this for 4 elements but it seemed having so few elements made it obvious how to make it work.

Answer 1

The polynomial $p(x)=2x^2+x+1$ is irreducible because it has no roots: indeed $p(0)=1$, $p(1)=1$ and $p(2)=2$. It would be more complicated to find, say, an irreducible polynomial of degree four or more, because the “no root” criterion doesn't apply.

The remainders are $\{0,1,2,x,x+1,x+2,2x,2x+1,2x+2\}$. How do you multiply two of them? Actually, it's better to think of them as the elements $$ 0,1,2,\xi,\xi+1,\xi+2,2\xi,2\xi+1,2\xi+2 $$ with $2\xi^2+\xi+1=0$, that is, $\xi^2=\xi+1$. Then $$ (\xi+2)(2\xi+1)=2\xi^2+\xi+\xi+2=2(\xi+1)+2\xi+2=\xi+1 $$ (recall that $2+2=1$ in this field). Similarly for the other pairs.

With your polynomial, the remainders are the same, and the same is the relation.

Answer 2

You have several remainders with degree 2. How do you get a remainder of degree 2 when what you're dividing by has degree 2?

The short answer is that you don't. You get every single possible first degree (or lower) polynomial as remainders: any such polynomial appears as a remainder, for instance when trying to divide that polynomial itself by $2x+x+1$, and any remainder has at most degree 1 because second degree or higher terms may be divided (possibly with some remainder, but that's the point). There are nine of those polynomials, so those are the elements of your field.

Answer 3

$F_3[x]/(2x^2+x+1)\cong F_3(\alpha)$ is the vector space over $F_3$ with basis $\{1,\alpha\}$, where $2\alpha^2+\alpha +1=0$.

Two polynomials in $F_3[x]$ are equivalent if they differ by an element of $(2x^2+x+1)$ (which is to say by a multiple of $2x^2+x+1$, since $F_3[x]$ is commutative).

Now to complete the exercise you have only to write out the multiplication and addition tables for the $9$ elements.