Let $f(x)\in\Bbb F_p[x]$ be irreducible such that $\alpha$ is a root of $f$. And consider $ev_{\alpha}:\Bbb F_p[x]\mapsto\Bbb F_p(\alpha)$ that sends a polynomial to its image by $\alpha$.

Clearly $f\in$Ker$(ev_{\alpha})$ and since the kernel is a ring $(f(x))\subset$Ker.

But my question is why should the Kernel be a subset of $(f(x))$?

That will be enough to prove the isomorphism $\Bbb F_p(\alpha)\cong\Bbb F_p[x]/(f(x))$ by the first isomorphism theorem...

So far I have: if Ker is not a subset of $(f(x))$ then $\exists g(x)\in$Ker such that $g(x)\notin (f(x))$ and $g(\alpha$)=0

  • $\begingroup$ I don't remember exactly, but I think that $f$ being irreducible and $\alpha$ a root, the $f$ is the minimal polynomial that $\alpha$ is a root $\endgroup$
    – ett
    May 23, 2018 at 9:34
  • $\begingroup$ so $g$ must be a multiple of $f$? $\endgroup$
    – H. Walter
    May 23, 2018 at 9:37

1 Answer 1


Indeed, $\ker(\text{ev}_{\alpha}) = \langle f \rangle$. This is because $\mathbb{F}_p$ is a field, and the ring of polynomials over any field is a Euclidean domain. All Euclidean domains are principal ideal domains, which means we'll have $\ker(\text{ev}_{\alpha}) = \langle g \rangle$ for some $g \in \mathbb{F}_p[x]$ (kernels of homomorphisms are ideals). As you've pointed out, $f \in \ker(\text{ev}_{\alpha})$, so if $g \neq f$, then we'd need to have $gh = f$ for some other polynomial $h$; this contradicts the irreducibility of $f$.

Note that this result extends to any choice of field.

Footnote: To be entirely accurate, this generator is only unique up to a constant multiple: we could have $g = cf$ for any nonzero $c \in \mathbb{F}_p$ due to the multiplicative closure property for ideals. No matter, $\langle cf \rangle = \langle f \rangle$ for all $c \neq 0$.


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