Past exam of the mathematics course of the University of Cambridge / 2005 / ii / Paper 1 / 18G / Solution

An element $x$ of an extension $L/K$ is an algebraic element over $K$ when some nonzero polynomial in $K[t]$ vanishes at $x$. Let its monic minimal polynomial have degree $d$. Polynomial division shows $K[x]$ is spanned by $1,x,\dots ,x^{d-1}$, and minimality makes these vectors independent. The minimal polynomial is irreducible; for any nonzero $q(x)$, Bezout's identity between $q$ and that polynomial gives a polynomial representative for $q(x{)}^{-1}$. Thus $K[x]$ is a field, equals $K(x)$, and has degree $d$.

Conversely, if $K(x)$ has finite vector-space dimension $d$, then $1,x,\dots ,x^d$ are linearly dependent over $K$, giving a nonzero annihilating polynomial. HenceAn algebraic extension is one in which every element is algebraic. Suppose $M/L$ and $L/K$ are algebraic, and take $\alpha \in M$. A polynomial relation for $\alpha$ over $L$ uses only finitely many coefficients $a_1,\dots ,a_r$. Each is algebraic over $K$, so adjoining them successively gives a finite extension $E=K(a_1,\dots ,a_r)$, by the proved criterion and the tower law. The same relation makes $\alpha$ algebraic over $E$, so $E(\alpha )/E$ is finite. Therefore $E(\alpha )/K$ is finite, and its subspace $K(\alpha )$ is finite-dimensional. The criterion makes $\alpha$ algebraic over $K$. Since $\alpha$ was arbitrary, $M/K$ is algebraic.