Field theory- Irreducible polynomial- Field extensions

Question

Let $F$ be a field and $p(x)$ be a monic irreducible polynomial over $F$. Suppose $a$ is a root of $p(x)$ in some extension $K$ of $F$. Then $F(a)$ is isomorphic to $F[x]/\langle p(x)\rangle$

Can someone please explain the proof

Answer 1

First, notice that $F[x]/(p)$ is a field, because $(p)$ is a maximal ideal in $F[x]$ (since $p$ is irreducible).

Well, consider the evaluation map $ev: F[x] \to F(\alpha)$ by evaluating a polynomial at $\alpha$. This map is a ring homomorphism.

To show the kernel is $(p)$: You need to prove that if any other polynomial $g$ vanishes at $\alpha$, then $p$ must divide $g$.

Let $I \subset k[x]$ be the collection of polynomials $g$ which satisfy $g(\alpha) = 0$. Given $f,g \in I$, $f+g \in I$ because $(f+g)(\alpha) = f(\alpha) + g(\alpha) = 0 + 0 = 0$. Likewise, if $f \in I$ and $g \in k[x]$, then $f \dot g \in I$ because $f \dot g (\alpha) = f(\alpha) g(\alpha) = 0 \dot g(A) = 0$. Hence, $I \subset k[x]$ is an ideal.

Since $k[x]$ is a PID, there exists a $p \in k[x]$ such that $I = (p)$. Thus, $p$ divides $g$ for every $g \in I$. But, the $p$ which I have found and the $p$ which you have written must be the same upto scaling by units since $p$ is irreducible! monicity guarantees a unique choice (so removes the ``upto scaling part'')

The evaluation map is clearly surjective.

Thus, the evaluation map is an isomorphism.