Connection between splitting field and quotient ring of polynomials

Question

Suppose that $f(x) \in \mathbb{F}_p[x]$ is an irreducible polynomial. We will call a field $L$ a splitting field, if $f(x) = (x-a_1) \ldots (x-a_n)$ in $L$. Also i know how to construct a field from $f(x)$ and ring of polynomials $\mathbb{F}_p[x]$. Just by taking a quotient ring $\mathbb{F}_p[x] / (f)$, and doing this we add a root of $f(x)$. Is it true that $\mathbb{F}_p[x] / (f)$ is a splitting field of $f(x)$? If no, what is a counterexample?And what is if we look at an arbitrary field $F$ instead of $\mathbb{F}_p$?

Answer 1

$$K=\mathbb F_p[x]/(f(x))$$ is a field of $p^d$ elements, with $d=\deg f.$ The field has, as automorphisms:

$$\phi_k:\alpha\mapsto \alpha^{p^k}, k=0,1,\dots,d-1$$ which fix $\mathbb F_p.$

This means $\phi_k(\langle x\rangle)$ are all roots of $f.$ If $f$ doesn’t split, then you must have $$\phi_i(\langle x\rangle)=\phi_j(\langle x\rangle),$$ for some $0\leq i<j<d.$

Then $$\langle x\rangle^{p^i}=\langle x\rangle ^{p^j}$$ or $x^{p^j}-x^{p^i}$ is divisible by $f(x)$ in $\mathbb F_p[x].$ But $$x^{p^j}-x^{p^i}=(x^{p^{j-i}}-x)^{p^i}$$ in $\mathbb F_p[x].$

So this means $f$ must divide $x^{p^k}-x$ for $1\leq k=j-i <d-1.$

But $\phi_k(\langle x\rangle)=\langle x\rangle$ means, for any polynomial $p(x)\in\mathbb F_p[x],$ $$\phi_k(\langle p( x)\rangle)=\langle p(x)\rangle.$$ But that means $x^{p^k}-x$ has every element of $K$ as a root, and $|K|=p^d>p^k.$

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The finite fields are special. This is true for irreducible $f\in \mathbb F_q[x]$ over any finite field $\mathbb F_q,$ but is not in general true over other fields, even fields of finite characteristic.

For example, in the field $F=\mathbb F_p(x),$ the field of fractions of $\mathbb F_p[x],$ the polynomial $y^p-x$ does not split in $F[y]/(y^p-x).$