# Commutative Rings: New Research by John Lee

By John Lee

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Example text

19, that there exist composable minimal ring extensions A ⊂ B and A ⊂ C such that A ⊂ BC does not have FIP. 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.

11] Dobbs, D. E. Comm. Algebra 35, 773–779 (2007). [12] Dobbs, D. E. Comm. Algebra 37, 604–608 (2009). [13] Dobbs, D. ; and Picavet-L’Hermitte, M. Comm. Algebra 33, 3091–3119 (2005). [14] Dobbs, D. ; Picavet-L’Hermitte, M. Comm. Algebra 36, 2638–2653 (2008). [15] Dobbs, D. ; and Picavet-L’Hermitte, M. J. Algebra and its Appl. 7, 601–622 (2008). [16] Dobbs, D. ; and Picavet-L’Hermitte, M. Int. Electron. J. Algebra 5, 121–134 (2009). [17] Dobbs, D. ; Shapiro, J. J. Algebra 305, 185–193 (2006). [18] Dobbs, D.

Sketch) We treat first the case a = b = 1. Let K be a field with algebraic closure K, and let F, L be distinct minimal field extensions of K inside K. Let V = K + M be a DVR with nonzero maximal ideal M (for instance, V = K[[X ]]). Then the domain R := K + M is Noetherian, as are its integral overrings S := F + M and T := L + M. Note that S, T (as overring extensions of R) give an example of type (1, 1). 15 (c) that S ⊂ ST = V is a minimal ring extension if and only if F ⊂ FL is a minimal field extension.