
Quotient ring structure
Problem: Describe the ring structure of F_2[x]/(x^4 + x^2 + 1).
Quotient rings are very difficult for me. What are the elements of this quotient ring? Is this polynomial irreducible? How do I determine this? Do I try to divide out by the degree 1 polynomials in this quotient ring? If the poly is irreducible, then this is a field, right? How do the elements in this ring differ from F_2[x]/(x^4 + x + 1)?
Sorry for being so stupid! :<
Sam

Hi
$\displaystyle X^4+X^2+1$ has no root in $\displaystyle \mathbb{F}_2,$ so it can't be written as a product of irreducible polynomials of degree 1 and 3. The other and last possibility for it to be reducible is to be the product of two irreducible polynomials of degree 2. The only irreducible polynomial of degree 2 in $\displaystyle \mathbb{F}_2$ is $\displaystyle X^2+X+1,$ (exercice), and we have $\displaystyle X^4+X^2+1=(X^2+X+1)^2.$
Therefore $\displaystyle X^4+X^2+1$ is reducible, and:
($\displaystyle \mathbb{F}_2[X]/(X^4+X^2+1)\cong \mathbb{F}_2[X]/(X^2+X+1)\times \mathbb{F}_2[X]/(X^2+X+1)$)(wrong) which is not a field, since it's not an integral domain.
With the same idea of proof, you can show that $\displaystyle X^4+X+1$ is irreducible in $\displaystyle \mathbb{F}_2[X],$ and then, since this ring is a principal ideal domain, $\displaystyle \mathbb{F}_2[X]/(X^4+X^2+1)$ is a field.
$\displaystyle \mathbb{F}_2[X]/(X^2+X+1)$ and $\displaystyle \mathbb{F}_2[X]/(X^4+X^2+1)$ are fields, and $\displaystyle [\mathbb{F}_2[X]/(X^2+X+1):\mathbb{F}_2]=2,\ [\mathbb{F}_2[X]/(X^4+X^2+1):\mathbb{F}_2]=4$
By unicity up to isomorphism for finite fields, $\displaystyle \mathbb{F}_2[X]/(X^2+X+1)\cong F_4$ and $\displaystyle \mathbb{F}_2[X]/(X^4+X^2+1)\cong F_{16}$
In conclusion, ($\displaystyle \mathbb{F}_2[X]/(X^4+X^2+1)\cong \mathbb{F}_4\times \mathbb{F}_4$)(wrong) while $\displaystyle \mathbb{F}_2[X]/(X^4+X^2+1)\cong F_{16}$

Thank you so much again clicclac! This is by far the most helpful forum I've found; I'm donating to it right away.

I'm sorry I can't write that like that. When $\displaystyle K$ is a field and $\displaystyle P,Q$ are two polynomials in $\displaystyle K[X],$ if $\displaystyle P$ and $\displaystyle Q$ are relatively prime, then $\displaystyle K[X]/(PQ)\cong K[X]/(P)\times K[X]/(Q).$
But of course, $\displaystyle X^2+X+1$ and itself are not relatively prime polynomials.
What remains true is that $\displaystyle \mathbb{F}_2[X]/(X^4+X^2+1)$ is not an integral domain: $\displaystyle a=X^2+X+1$ is such that $\displaystyle a^2=0$ but $\displaystyle a\neq 0.$ (indeed $\displaystyle X^2+X+1\notin (X^4+X^2+1)$ ).
One more thing: to see that the "isomorphism" quoted is wrong, one may note that the first ring has a nonzero nilpotent element, $\displaystyle a$, while the second, as a product of fields, has no nonzero nilpotent element.