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.
Best Answer
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 $\alpha=1 \in \mathbb Z_2$, so $\mathbb Z_2[\alpha]=\mathbb Z_2 $. similar way for $(c)$