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Gilmer , Two cmistructions of Prüfer domams
Lemma 3. 5, there is a nonzero finitely generated ideal of / such that А1^= Л^ + М^ for each i. Hence [л{А) = oc and ^ is onto. If P is the subgroup of F consisting of zero principal fractional ideals of /, then it is clear that P g ker //. However, \i F ^ ker /^, then choose a nonzero element d oi J such that dF — 5 is an integral ideal of /. Then BJ^ = d • FJ^ is principal in J^ for each i. By Theorem 3. 5 b), it follows that Б, and hence F, is principal. Thus ker /л = P and F/P, the class group of /, is isomorphic to G as asserted.
We remark that in case D^ is Prüfer with quotient field ÜC, D^ and J^ have the same class group ([7], p. 562).
Acknowledgement . The author's investigation of the construction in § 2 was motivated by a remark of W. Heinzer, who reported to the author that A. Dress, during a visit to Louisiana State University, had stated that if Z is a field in which — 1 is not a square, then the subring of К generated by {1/(1 + t^)}t^K i^ ^ Prüfer domain whose set of valuation overrings is the set of valuation rings on К for which — 1 is not a square in the residue field.
References
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Florida State University, Department of Mathematics, Tallahassee, FL 32306, USA
Eingegangen 11. Februar 1969