Exercise 5.1.16 (Polynomial of transcendental)

Solution: We have that $K(\alpha )=\left\{\frac{f(\alpha )}{g(\alpha )}:f,g\in F[t]\right\}$, which is just $K(t)$, the field rational expressions. Therefore it is not polynomial is $\alpha$ over $K$.

Let $\alpha \in M$ be a transcendental element over $K$. Reasoning by contradiction, assume that $1∕\alpha$ is represented as a polynomial in $\alpha$, i.e.

Then $1-\mathit{\alpha p}(\alpha )=0$. Define $q(t)=1-\mathit{tp}(t)$. Then $q(t)\in K(t),q(\alpha )=0$ and $q\ne 0$ (because the constant coefficient is $1\ne 0$). This shows that $\alpha$ is algebraic over $K$, and this is a contradiction with the assumption.

There are some elements in $K(\alpha )$, such as $1∕\alpha$, which cannot be represented as a polynomial in $\alpha$ over $K$.