Q:

Kronecker's Theorem - what can be deduced if $f$ is reducible?

I am happy with the statement of Kroncker's Theorem as follows. Let $\:$ $ f \in K[X]$. Then there exists a simple field extension $L = K(\alpha)$ of $K$ with $f(\alpha) = 0$. If $f$ is irreducible over $K$, then the extension $L$ is unique in sense that the natural map
$$K[X]/fK[X] \rightarrow L, \: \: \: \: \: \: g + fK[X] \rightarrow g(\alpha)$$ constitutes an  isomorphism.
However, I am trying to prove something at present and need to understand what can be said if $f$ is actually reducible.  So I am wondering to what extent we can reason the other way, i.e. if we have some polynomial $f$ of degree $n$ that is not irreducible, then is it impossible that it generates a field of size $p^n$ via the quotient group $F_p[X]/fF_p[X]$ and hence can we conclude that $f$ must be irreducible?

A:

If $f$ is reducible, then it does not generate a field at all. For example, if $f = X^2$ then there is no polynomial $g$ such that $Xg = 1$ modulo $f$. This generalises to other $f$ and does not depend on the field $K$.
However, the irreducible factors of $f$ do generate a field.

