Recognizing simple subextensions of purely transcendental field extensions
Let \(k(X)/k\) be a finitely generating purely transcendental field extension, where \(X\) is algebraically independent over \(k\). It is known that all intermediate fields \(K\) of \(k(X)/k\) with \(\text{transdeg} (K/k)=1\) are simple extensions of \(k\). In this paper, the authors give a criterion that allows one to decide effectively whether the extension \(K/k\) is simple if a finite generating set of \(K\) over \(k\) is known and computations in \(K\) are effective. In the affirmative case, a primitive element of \(K\) over \(k\) is computed. Let \(Z= \{Z_x\mid x\in X\}\), where \(Z_x\) is transcendental over \(K\). Let \(\phi\) be the homomorphism of \(K[Z]\) into \(k(X)\) such that \(\phi(Z_x)= x\) for all \(x\in X\) and \(\phi\) is the identity on \(K\). Let \(B\) denote the kernel of \(\phi\). Let \(G\) be a reduced Gröbner basis of \(B\). Then the authors show that the following statements are equivalent: (i) \(K/k\) is simple. (ii) \(B\) is principal. (iii) \(G= \{n(Z)- fd(Z)\}\) for some \(n(Z),d(Z)\in k[Z]\), \(f\in K\) or \(G= \emptyset\). If (iii) holds and \(G\neq \emptyset\), then \(f\) is a primitive element of \(K/k\). If \(G= \emptyset\), then \(K=k\). The authors show that (i) and (ii) are equivalent if the assumption that \(X\) is finite is dropped.
scientific article; zbMATH DE number 1523456