$F(a) \cong \frac{F[x]}{\langle p \rangle}$ for simple extension with algebraic element

Question

Let $E$ be an extension of the field $F$ by some algebraic element $a$ over $F$ ($a \not \in F$). We want to show that $F(a) \cong \frac{F[x]}{\langle p \rangle}$, where $p \in F[x]$ is the minimal polynomial of $a$ over $F$. The idea is to consider a map $\phi: F[x] \to F(a)$ defined by $f(x) \mapsto f(a)$. Here, $ker \ \phi = \langle p \rangle$, and $\phi$ can be shown to be a ring homomorphism. If I can show that $Im\ \phi = F(a)$, then we can use the first isomorphism theorem (for rings) to conclude that $F(a) \cong \frac{F[x]}{\langle p \rangle}$. My question is - how can I show that $Im \ \phi = F(a)$?

Answer 1

First of all note that the image is a field, because it is isomorphic to the field $F[x]/(p)$. Also, we clearly have $Im(\phi)\subseteq F(\alpha)$. For the converse direction note that $F\subseteq Im(\phi)$, because each $b\in F$ is the image of the constant polynomial $b$. Also, $\alpha\in Im(\phi)$ because it is the image of the polynomial $f(x)=x$. So $Im(\phi)$ is a subfield of $E$ which contains the field $F$ and the element $\alpha$, hence by definition it contains $F(\alpha)$.