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The details are given next. 20. There exist a field R and elements s,t such that the ring extension R ⊂ R[s,t] does not satisfy FIP even though each of the extensions R ⊆ R[s] and R ⊆ R[t] satisfies FIP. It can be further arranged that both R ⊆ R[s] and R ⊆ R[t] are minimal field extensions (and, hence, minimal ring extensions). Proof. Let R := F(X p ,Y p ), where F is the field F with p elements for some prime number p and X ,Y are algebraically independent indeterminates overF. Let s := X and t := Y .

4]). Let K ⊆ T be a ring extension where K is a field. Then T is a decomposable λ-extension of K if and only if T is isomorphic as a K-algebra to K ×L for some field L such that K ⊆ L is a λ-extension. At the close of Section 3, we noted that nilpotent elements have played an important role in the study of the FIP property. 13. 6]). Let K ⊆ T be a λ-extension where K is a field. If u is a nilpotent element of T and n is the index of nilpotency of u, then n ≤ 3. In [29], Quigley classified the subfields of an algebraically closed fieldL which are maximal among the subfields of L not containing a given element of L.

18] Dobbs, D. ; Shapiro, J. J. Algebra 308, 800–821 (2007). [19] Dobbs, D. ; Shapiro, J. JP J. Algebra, Number Theory and Appl. 9, 241–275 (2007). -P. J.

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