Why is \([K(\alpha): K]\) equal to the degree of the minimal polynomial of \(\alpha\) when it is algebraic?
Solution
Let \(m\in K[x]\) be the minimal polynomial of \(\alpha\), of degree \(d\).
1. \(K(\alpha)=K[\alpha]\cong K[x]/(m)\).
Evaluation \(K[x]\to K[\alpha]\), \(f\mapsto f(\alpha)\), is surjective with kernel \((m)\), since \(m\) generates the ideal of polynomials vanishing at \(\alpha\). Since \(m\) is irreducible, \((m)\) is maximal, so \(K[\alpha]\) is a field and equals \(K(\alpha)\).
2. \(1,\alpha,\dots,\alpha^{d-1}\) is a \(K\)-basis of \(K(\alpha)\).
They span: for \(f\in K[x]\), division gives \(f=qm+r\) with \(\deg r<d\), so \(f(\alpha)=r(\alpha)\). They are independent: a relation \(\sum_{i<d}c_i\alpha^i=0\) is a polynomial of degree less than \(d\) vanishing at \(\alpha\), hence zero by minimality of \(m\).
3. Q.E.D.
By step 2, \([K(\alpha):K]=d=\deg m\).