$\Bbb{Z}_{2}(\alpha)$ as splitting field

Question

i have problems with an exercise:

let $\alpha$ be a root of the polynomial $X^{3}+X^{2}+1$ in $\Bbb{Z}_{2}$.
Prove that $\Bbb{Z}_{2}(\alpha)$ is the splitting field of this polynomial over $\Bbb{Z}_{2}$.

It is the only exercise i have about splitting fields in $\Bbb{Z}_{p}$, and i don't know how to approach them. If someone could give me a hint or help me start that would be awesome.

Answer 1

Notice that $\mathbf{Z}_{2}(\alpha)$ is also a finite field, a finite extension of $\mathbf{Z}_{2}$. Hence it is Galois over $\mathbf{Z}_{2}$. Hence the minimal polynomial of $\alpha$ splits completely in this extension. But the polynomial $x^{3}+x^{2}+1$ is irreducible as it has no roots (Degree 3), and is satisfied by $\alpha$ and hence is its minimal polynomial. You can also show that $\alpha^{2}$ and $\alpha^{4}$ are roots of this polynomial, as an alternative solution, if i m not mistaken. More generally, but not obvious at all, and needs work to show, say you have a finite field, for simplicity I will pick $\mathbf{F}_{p}=\mathbf{Z}_{p}$ p prime, and an irreducible polynomial f(x) in $\mathbf{F}_{p}[x]$ of degree n. Let $\alpha$ be a root of this polynomial. $\mathbf{F}_{p}(\alpha)$ is a splitting field of $f(x)$ over $\mathbf{F}_{p}$, and the roots are $\alpha$, $\alpha^{p}$,$\alpha^{p^{2}} \cdots \alpha^{p^{n-1}}$. \ $\alpha$ is a root of $x^{3}+x^{2}+1$, so $\alpha^{3}+\alpha^{2}+1=0$. Let's square both sides, keeping in mind that we are in characteristic 2, so $(\alpha^{3}+\alpha^{2}+1)^{2} =0$. So $\alpha^{6}+\alpha^{4}+1=0$, hence $(\alpha^{2})^{3}+(\alpha^{2})^{2}+1=0$. So $\alpha^{2}$ is also a root!. By squarring the last relation you can also get that $\alpha^{4}$ is also a root!

Answer 2

You can easily see that $X^3 + X^2 + 1$ is irreducible over $\mathbb F_2 = \mathbb Z/2\mathbb Z$ since it has no roots (namely $0$ and $1$ are not). Since $[\mathbb F_2(\alpha):\mathbb F_2] = 3$ (the degree of the irreducible polynomial), this means $\mathbb F_2(\alpha)$ has $8$ elements. There is a unique finite field $\mathbb F_{2^3}$ of size $8$, so $\mathbb F_8 = \mathbb F_2(\alpha)$. We know by the theory of finite fields that $\mathbb F_8$ is the set of roots (and the splitting field) of $X^8-X$, and we have a factorization for this polynomial : $$ X^8-X = X(X-1)(X^3 + X + 1)(X^3 + X^2 + 1). $$ (In case you wonder, this is a general result ; the polynomial $X^{p^n}-X$ is the product of all irreducible polynomials in $\mathbb F_p[X]$ of degree $d$ for $d$ dividing $n$. This follows from the fact that $\mathbb F_{p^n}$ is the set of roots of $X^{p^n}-X$ by grouping the factors $(X-\alpha)$ according to their minimal polynomials of degree $d$, which has to divide $n$ since $\mathbb F_{p^d} \subseteq \mathbb F_{p^n}$ if and only if $d$ divides $n$.)

In particular, $\mathbb F_2(\alpha)$ has all the roots of $X^3 + X^2 + 1$, so it is a splitting field.

Hope that helps,