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.

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