If $0\neq \alpha \in E$ is algebraic over F, then $\frac{1}{a}$ is also algebraic over F.

Question

Exercise: Show that if $0\neq \alpha \in E$ is algebraic over F, then $\frac{1}{a}$ is also algebraic over F.

My first idea is to somehow use the fact that the field extension $F(\alpha)$ is isomorphic to the field $F[x]/<p(x)>$ where $p(x)$ is the minimal polynomial of $\alpha$ over $F$. However, I continuously run into $\frac{1}{0}$ which is problematic for obvious reasons. Any hints to start this proof would be very helpful.

Answer 1

Let $P(x)$ be a polynomial with non-zero constant term and coefficients in $F$ such that $P(\alpha)=0$.

Suppose that $P$ has shape $b_nx^n+b_{n-1}x^{n-1}+\cdots+b_0$. Then $\dfrac{1}{\alpha}$ is a root of the polynomial $b_0x^n+b_1x^{n-1}+\cdots+b_n$. To show this, divide $P(x)$ by $x^n$.

Answer 2

Hint: $\frac{1}{\alpha}\in F(\alpha)$

Answer 3

Start with your algebraic number $a$. There is $b\in F$ so that $ab=1$. Then for any $k$, we have $a^kb^k=1$. Using the fact that $a$ is algebraic, there is some polynomial with coefficients $c_i$ that $a$ satisfies. Explicitly, we have

$$\sum_{k=1}^n c_ia^k=0.$$

Substituting in for $b$, we have

$$\sum_{k=1}^n c_i(1/b)^k=0.$$

Can you now show that $b$ is algebraic?