- Show that \(\alpha/F\) is algebraic \(\iff F(\alpha)/F\) is a finite extension.
Solution
1. If \(\alpha\) is algebraic over \(F\), then \(F(\alpha)/F\) is finite.
Let \(m_\alpha(x)\in F[x]\) be the minimal polynomial of \(\alpha\), of degree \(n\). Evaluation induces an isomorphism \[ F[x]/(m_\alpha)\xrightarrow{\sim}F(\alpha). \] Hence \[ 1,\alpha,\dots,\alpha^{n-1} \] is an \(F\)-basis of \(F(\alpha)\), so \([F(\alpha):F]=n<\infty\).
2. If \(F(\alpha)/F\) is finite, then \(\alpha\) is algebraic over \(F\).
If \([F(\alpha):F]=n\), then the \(n+1\) vectors \[ 1,\alpha,\dots,\alpha^n \] are linearly dependent over \(F\). Thus there exist coefficients, not all zero, such that \[ c_0+c_1\alpha+\cdots+c_n\alpha^n=0. \] Therefore \(\alpha\) is a root of a nonzero polynomial in \(F[x]\), so it is algebraic.
Hence \[ \alpha/F\text{ is algebraic} \iff [F(\alpha):F]<\infty. \]