# A problem in extension fields

Let $E$ be an extension field of a field $F$. (That is, let $F$ be a subfield of the field $E$.) Let $\alpha$, $\beta$ $\in$ $E$ such that $\alpha$ is transcendental over $F$ but algebraic over the simple extension $F(\beta)$ of $F$. Is $\beta$ algebraic or transcendental over $F$? I know that $\beta$ is algebraic over $F(\alpha)$.

Here $F(\beta)$ is effectively the field of rational functions in $\beta$ with co-efficients in $F$.

## 2 Answers

Hint: $[F(\alpha,\beta):F]=[F(\alpha,\beta):F(\beta)][F(\beta):F]$.

• Can you please elaborate on this symbolism? – Saaqib Mahmood Dec 8 '13 at 10:12
• @SaaqibMahmuud If $\alpha$ is transcendental then what is $[F(\alpha,\beta):F]$? If $\alpha$ is algebraic over $F(\beta)$ then what is $[F(\alpha,\beta):F(\beta)]$? What does that mean for $[F(\beta):F]$? You should already be familiar with en.wikipedia.org/wiki/Degree_of_a_field_extension. – anon Dec 8 '13 at 20:16

An algebraic extension of an algebraic extension amounts to an algebraic extension.

• I'm sorry, but I'm afraid I just coouldn't appreciate what your remark says and how it is relevant to my question. – Saaqib Mahmood Dec 8 '13 at 10:14
• @SaaqibMahmuud You said $F(\alpha,\beta)$ is transcendental over $F$ but algebraic over $F(\beta)$ and then asked if $F(\beta)$ is algebraic or transcendental over $F$. If $F(\beta)$ were algebraic over $F$, then since $F(\alpha,\beta)$ is algebraic over $F(\beta)$ this means that $F(\alpha,\beta)$ is algebraic over $F$, a contradiction. Hence $F(\beta)$ cannot be algebraic over $F$. You should already be familiar with en.wikipedia.org/wiki/Algebraic_extension. (A simple extension is algebraic iff it's finite degree and transcendental iff it's infinite degree.) – anon Dec 8 '13 at 20:20