Showing that $\alpha^{-1} \in F[\alpha]$:

Question

Let $E$ be a field extension of $F$. If $\alpha \in E\setminus F$ is algebraic over $F$ show that $\alpha^{-1}$ is in $F[\alpha]$

Answer 1

Since $\alpha$ is algebraic over $F$. there is a polynomial $p(x)$ with coefficients in $F$ such that $p(\alpha)=0$. Without loss of generality, we can assume that $p(x)$ has non-zero constant term. By multiplying the polynomial by an appropriate constant, we can assume the constant term is $-1$.

Thus we can assume that $p(x)$ has shape $$p(x)=c_0x^n+c_1x^{n_1}+\cdots+c_{n-1}x-1.$$ We have $p(\alpha)=0$. It follows that $$\alpha(c_0\alpha^{n-1}+c_1\alpha^{n-2}+\cdots +c_{n-1})=1.$$ We have found an explicit expression for $\alpha^{n-1}$ as a linear combination of powers of $\alpha$, with coefficients in $F$.

Answer 2

Let $I$ be the ideal of all polynomials in $E[x]$ with $\alpha$ as a root.

Then $I=(f)$, because $E[x]$ is a principal ideal domain. I is a prime ideal, because $(gg')(\alpha)=0\Longrightarrow g(\alpha)=0\lor g'(\alpha)=0$, so $f$ is irreducible.

$(f)$ is a maximal ideal, because $f\ne0$ and for any $g\notin(f)$ we have $(f,g)=(h)$, so $h\mid f$ and $h\ne f$, because $f\nmid g$, therefore $(h)=(1)$.

So $E[x]/(f)\simeq F[\alpha]$ is a field and there exists $\alpha^{-1}\in F[\alpha]$.