With a lot of free time on my hands, I've been looking at bits of abstract algebra I never studied as a student (25+ years ago). I think I'm getting to grips with the ideas of rings, ideals and splitting fields, but in modifying a problem I've seen, I've come across a phenomenon I don't understand. I'd appreciate if someone could explain the structures involved, and how they fit together.
Consider the polynomial ring $R=\mathbf F_2[X]$ where $\mathbf F_2 = \{0,1\}$ with the usual field operations.
Now consider the polynomial $p(X) = X^4 + X^2 + 1$.
$P$ is irreducible, so we can construct the quotient ring $Q = R / (p(X))$
Now define $\alpha = \pi(X) \in Q$, which is a root of $p(x)$. We can show (I think) that $$(\alpha^3 + 1)^2 = \alpha^6 + 1 = (\alpha^2+1) (\alpha^4 + \alpha^2 + 1) = 0$$ so therefore $\alpha^3 = 1$.
Similarly, we can check that $$ p(x) = \left[(x+\alpha)(x+\alpha^2)\right]^2$$ so all $\mathbf F_2[\alpha,\alpha^2]$ is the splitting field.
So, my question is, where does $\pi(X^3)$ fit into all this?
If
$\alpha^3 = 1$, but $X^3$ is not congruent to $1$, does this mean that $\pi(X)^3 \ne \pi(X^3)$, and so the splitting field and the quotient ring are not isomorphic? Because $2^4-1$ is not prime?
If $p$ was indeed irreducible, then $p$ would be the minimal polynomial of $\alpha$ which you contradicted.
It follows that $p$ is reducible, and you can check that indeed $$p(x)=(x^2+x+1)^2\,. $$