Prove that $x^3+x+1$ splits in $Z_2[x]/(x^3+x+1)$

Question

Let $f(x)=x^3+x+1$ and set $E=z_2[x]/(x^3+x+1)$ prove that $f(x)$ splits in E that is find 3 distinct roots of $f(x)$ in $E$

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Questions found $[x^2+1]$ $[x^2+x+1]$ on my sloppy work. There are more than 3 roots if my notes are correct. If that is the case do we just use the irreducibles??

Another question I have is that what is [x]?? Is it a variable, some sort of constant. I know the book uses idertimenatn and has special properties.

Answer 1

Here's an answer that uses a bit more theory. First some generalities. Suppose $F$ is a field, $f \in F[x]$ is irreducible and $K = \frac{F[x]}{(f(x))} = F(\theta)$ is the extension of $F$ obtained by adjoining a root $\theta$ of $f$ (which in your notation is $[x]$, the image of $x$ in the quotient). Given $\sigma \in \text{Aut}(K/F)$, then it follows by the properties of ring homomorphisms and the fact that $\sigma$ is the identity on $F$ that $\sigma(\theta)$ is also a root of $f$, where $$ \text{Aut}(K/F) = \{\sigma : K \to K \mid \sigma \text{ is a field automorphism}, \sigma|_F = \text{id}_F\} \, . $$

Alright, now back to your question. Let $\mathbb{F}_2 = \mathbb{Z}/2\mathbb{Z}$. Then $\frac{\mathbb{F}_2[x]}{(x^3 + x + 1)} = \mathbb{F}_2(\theta)$ is a field extension where $\theta$ is the image of $x$ in the quotient. Now, over $\mathbb{F}_2$ (more generally we can replace $2$ by any prime $p$) we have one very special automorphism called the Frobenius map: \begin{align*} \sigma: \mathbb{F}_2(\theta) &\to \mathbb{F}_2(\theta)\\ \alpha &\mapsto \alpha^2 \, . \end{align*} (This is a homomorphism by the freshman's dream.) Then, by the first paragraph, $\sigma(\theta) = \theta^2$ and $$ \sigma^2(\theta) = \theta^4 = \theta \theta^3 = \theta (\theta + 1) = \theta^2 + \theta $$ are also roots of $f$. (Note that $\sigma^3 = \text{id}_K$ and higher powers of $\sigma$ will just repeat the identity, $\sigma$, and $\sigma^2$.) In fact one can show that any finite extension of finite fields has cyclic Galois group generated by the Frobenius map; cf. this article by Keith Conrad.

Answer 2

In the field $\Bbb F_2[x]/(x^3+x+1)$, one root of $X^3+X+1$ is $\alpha=x$, or more precisely the coset of $x$ with respect to the ideal $(x^3+x+1)$. You should be able to check quickly that the other two roots are $\alpha^2$ and $\alpha^4$.

The reason for this is that $z\mapsto z^2$ is an automorphism of every finite field of characteristic two, and is identity on the prime field $\Bbb F_2$. So it carries any root $\rho$ of a polynomial $g(X)\in\Bbb F_2[X]$ to a root $\rho^2$ of $g$.