Exercise 4.1.7

Proof. As in Proposition 4.1.14, we assume that $F\subset L$ is a field extension, and that $\alpha \in L$. Suppose that $\alpha \in L$ is algebraic over $F$, where $p\in F[x]$ is the minimal polynomial of $\alpha$ over $F$, and $\beta \in F[\alpha ],\beta \ne 0$.

There exists $g\in F[x]$ such that $\beta =g(\alpha )$.

(a)

The minimal polynomial $p$ of $\alpha$ is irreducible over $F$ (Prop. 4.1.5).

Let $u\in F[x]$ such that $u\mid p,u\mid g$. Then $p=\mathit{uq},q\in F[x]$, and since $p$ is irreducible over $F$, $u$ or $q$ is a constant of $F^{\text{∗}}$.

If $q=\lambda \in F^{\text{∗}}$, then $u={\lambda }^{-1}p$ and $p$ divides $u$, which divides $g$, thus $p$ divides $g$. In this case, since $p(\alpha )=0$, $\beta =g(\alpha )=0$, in contradiction with the hypothesis $\beta \ne 0$.

So $u=\mu \in F^{\text{∗}}$ , $u\mid 1$. Consequently, for all $u\in F[x]$, $(u\mid p,u\mid g)\Rightarrow u\mid 1$: $p,g$ are relatively prime.

(b)

Then there exists a Bézout’s relation between these two polynomials: $$\mathit{Ap}+\mathit{Bg}=1,A,B\in F[x].$$

The evaluation of these polynomials in $\alpha$, since $p(\alpha )=0$, gives

$$B(\alpha )g(\alpha )=1,B(\alpha )\in F[\alpha ]$$

So $B(\alpha )$ is the multiplicative inverse of $\beta =g(\alpha )\ne 0$ in $F[\alpha ]$: $F[\alpha ]$ is a field.

Note: We have proved in Exercise 3.5.1 that $F[x]∕⟨f⟩$, where $f$ is irreducible over $F$, is a field with the same argumentation. Here $f=p$ is the minimal polynomial of $\alpha$ over $F$, so it is irreducible over $f$.