- Suppose that for an extension field \(F\) over \(K\) and for \(a \in F\), we have that \(b \in F\) is algebraic over \(K(a)\) but transcendental over \(K\). Prove that \(a\) is algebraic over \(K(b)\).
Solution
1. Because \(b\) is algebraic over \(K(a)\), there is a nonzero polynomial\[ q(Y)=c_0+c_1Y+\cdots+c_mY^m\in K(a)[Y] \]such that \(q(b)=0\).
This is the definition of \(b\) being algebraic over \(K(a)\).
2. Clearing denominators in the coefficients \(c_i\in K(a)\) yields a nonzero polynomial \(P(X,Y)\in K[X,Y]\) such that\[ P(a,b)=0. \]
Write each \(c_i=r_i(a)/s_i(a)\) with \(r_i,s_i\in K[X]\) and \(s_i(a)\neq0\). Multiplying \(q(b)=0\) by the product of the denominators gives a relation \[ \sum_i R_i(a)b^i=0 \] with \(R_i\in K[X]\), not all zero. Set \(P(X,Y)=\sum_iR_i(X)Y^i\).
3. The polynomial \(P(X,b)\in K(b)[X]\) is nonzero.
Write \[ P(X,Y)=\sum_{j=0}^d p_j(Y)X^j \] with \(p_j\in K[Y]\), not all zero. Since \(b\) is transcendental over \(K\), the evaluation map \(K[Y]\to K(b)\), \(Y\mapsto b\), is injective. Hence not all \(p_j(b)\) vanish, so \(P(X,b)\neq0\) in \(K(b)[X]\).
4. Therefore \(a\) is algebraic over \(K(b)\).
By 2, \[ P(a,b)=0, \] and by 3, \(P(X,b)\) is a nonzero polynomial with coefficients in \(K(b)\). Thus \(a\) satisfies a nonzero polynomial over \(K(b)\), which is exactly algebraicity over \(K(b)\).