We start by mentioning the following result: If a field extension \(K/F\) has degree \(n\), then any \(\alpha \in K\) satisfies a polynomial with degree \(\le n\). Observe that if \(K = F(\alpha)\) then equality always holds, but this need not always be the case. For example, the extension \(\mathbb{Q}(\sqrt[4]{2})/\mathbb{Q}\) is of degree 4, but the element \(\sqrt{2}\) satisfies a degree 2 (< 4) polynomial.
The next major result is that \(F \subseteq K \subseteq L\) then \([L:F] = [L:K][K:F]\). Intuitively, a basis for the vector space of \(L\) over \(F\) is \(\alpha_i \beta_j\), where \(\alpha_i\) and \(\beta_j\) run through bases of \(L\) over \(K\) and \(K\) over \(F\) respectively (one can check this by manual expansion). A special case is where \(K = F(\alpha)\) and \(L = F(\alpha,\beta) = F(\alpha)(\beta)\). The degree \(d\) of \(\beta\) over \(F(\alpha)\) is \(\le\) its degree over \(F\), and an \(F\)-basis for \(L\) is \(\alpha^i \beta^j\), where \(1 \le i \le [K:F]\) and \(1 \le j \le d\).
This leads to the following result: \(K/F\) is finite iff it is generated by finitely many algebraic elements \(\alpha_i\), and its degree is \(\le \prod_{i} \text{deg}\ \alpha_i\). This is useful in proving that the set of algebraic elements of \(K/F\) form a subfield (in particular an algebraic extension of \(F\)). For instance, the set of algebraic elements over \(\mathbb{Q}\) in \(\mathbb{C}\) form an infinite extension of \(\mathbb{Q}\).
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