Splitting fields of distinct polynomials that agree on all inputs

Question

I'm trying to construct the splitting field $f(x) = x^3 - 1\in \mathbb{Z}_2[x]$. But, $1$ is the only zero of $f(x)$, and if we define $g(x) = x-1\in\mathbb{Z}_2[x]$, then $f(0) = g(0)$ and $f(1) = g(1)$. So, why wouldn't $f(x)$ and $g(x)$ have the same splitting field, i.e., just $\mathbb{Z}_2$?

Answer 1

What you are missing is that it is not sufficient for $f$ and $g$ to agree on the elements of $\mathbb{F}_2$: this condition is actually pointless, since there are infinitely many polynomials in $\mathbb{F}_2[x]$. What matters for $f$ to have splitting field $\mathbb{F}_2$ is that all its roots are in $\mathbb{F}_2$, and this is not the case.

First of all, computing $f'$ shows that $f$ has distinct roots in an algebraic closure of $\mathbb{F}_2$: since $\mathbb{F}_2$ is the field with $2$ elements, the pigeonhole principle tells you that its splitting field is strictly larger. More concretely, just write the decomposition of $f$ in irreducible factors and notice that $f$ and $x^2+x+1$ have the same splitting field. From here, you should be able to come to the conclusion yourself.