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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.

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    The same question was asked in the last couple days. – 2017-02-02

2 Answers 2

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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 $(c)$

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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 $\mathbb{F}_2(\alpha) = \mathbb{F}_2$

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    Thank you, but I specifically want to list out the elements in each of those rings, like the first example I showed. I am wondering how I should go about that. – 2017-02-02
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    @Lana Finding the base of the field in the first part should be easy, as it should be included in the most introductory books on this topic. Then every element is a linear combination of the element in the basis, where the scalars are elements of $\mathbb{F}_2$. – 2017-02-02
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    @Lana The other two parts are easier, as we know that $\mathbb{F}_2(\alpha) = \mathbb{F}$ – 2017-02-02
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    Hello, this is exactly what I'm asking about though. Finding the basis. Sorry if it's too elementary. – 2017-02-02
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    @Lana The basis is given by $\{1,a\}$. But if you're not familiar with it and the reasoning behind it, then you need to read something more on it, as I doubt it can be explained in an MSE answer or comment. – 2017-02-02
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    Ok, so the answer for the first problem is that it has those elements: {0,1,α,1+α}. I am not seeing how you obtain 1+α from any of the elements in this basis. – 2017-02-02
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    @Lana The answer is right. As I told you all elements of the field are linear combinaton of the elements in the basis, i.e. of the form $c_1\cdot 1 + c_2 \cdot \alpha$, where $c_1, c_2 \in \mathbb{F}_2$ – 2017-02-02