Describe the ring obtained from the field $F2$ by adjoining an element α satisfying the given relation (listing out all elements of each ring)

Question

a) $α^2 + α + 1 = 0$

b) $α^2+ 1=0$

c) $α^2 +α = 0$

I want to explicitly list out all the elements obtained by adjoining alpha satisfying the relations.

From the answers I have, it seems that a) has elements $\{0,1, α, 1+α\}$. I know that $0$ and $1$ are already in $F_2$. But how do we conclude that $α$ and $1+α$ are everything in $F_2[x]/(x^2 + x + 1)$?

Also can someone give me brief pointers to the other two?

Thanks so much.

Answer 1

for $a)$ We have $x^2+x+1$ is an irreducible polynomial in $\mathbb Z_2$, so $\mathbb Z_2[x]/(x^2+x+1)\cong \mathbb Z_2(\alpha)$,then $ \{1,\alpha \} $ is a basis of $\mathbb Z_2(\alpha)$ over $\mathbb Z_2$, i.e. $\mathbb Z_2(\alpha)=\{ a+b\alpha; a,b\in\mathbb Z_2 \} =\{0,1,\alpha,1+\alpha \}$

for $b)$ we have $x^2+1 $ is a reducible polynomial in $\mathbb Z_2$ where $x^2+1=(x+1)^2$ , then $\mathbb Z_2[\alpha]/(\alpha^2+1)=\mathbb Z_2[\alpha]/(\alpha+1)^2.$ for $(c)$: $Z_2[\alpha]/{(\alpha^2+\alpha}) \cong Z_2^2$

Answer 2

For the first part prove that the polynomial is irreducible over $\mathbb{F}_2$. This means that $\mathbb{F}_2/\langle x^2 + x + 1 \rangle \cong \mathbb{F}_2(\alpha)$. Now what's the basis of this field?

For the other cases prove that each polynomial is reducible in the field and hence $\mathbb{F}_2(\alpha) = \mathbb{F}_2$