## Decide whether a given element is algebraic or transcendental over a field

Let $F$ be a field. Suppose $\alpha$ is algebraic over $F$, $\gamma$ is transcendental over $F$, that $\beta$ is algebraic over $F(\alpha)$, and $\delta$ is transcendental over $F(\alpha)$. Decide whether the following elements are algebraic or transcendental over the given field.

1. $\alpha + \beta$ over $F$
2. $1/\alpha$ over $F$
3. $\alpha+\delta$ transcendental $F$
4. $\gamma$ over $F(\alpha)$
5. Is $\gamma+\delta$ transcendental over $F$?

Note that $F(\alpha,\beta) = F(\theta)$ is a finite extension of $F$. By Theorem 5.5 in TAN, $\alpha+\beta$ is algebraic over $F$.

Since $1/\alpha \in F(\alpha)$ and $F(\alpha)$ is an algebraic (hence finite) extension of $F$, by Theorem 5.5, $1/\alpha$ is algebraic over $F$.

Suppose $\alpha+\delta$ is algebraic over $F$. Now $F(\alpha + \delta)$ is an algebraic, hence finite, extension of $F$. Then $F(\alpha+\delta,\alpha)$ is also a finite extension. But $(\alpha+\delta) - \alpha = \delta \in F(\alpha+\delta,\alpha)$, so that (by Theorem 5.5) $\delta$ is algebraic over $F$. So $\delta$ is algebraic over $F(\alpha)$, a contradiction. Thus $\alpha+\delta$ is transcendental over $F$.

Suppose $\gamma$ is algebraic over $F(\alpha)$. Now $F(\alpha,\gamma)$ is a finite extension of $F$, so that $\gamma$ is algebraic over $F$– a contradiction. Thus $\gamma$ is transcendental over $F(\alpha)$.

Suppose $\delta = -\gamma$; then $\gamma + \delta = 0$ is (trivially) algebraic over $F$. So $\gamma + \delta$ need not be transcendental over $F$. More generally, consider $\alpha+\gamma$ and $-\gamma$.