By Samuel Moy

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**Extra resources for An introduction to the theory of field extensions**

**Example text**

VR ) ≤ 1; (ii) Ext1R (V, V (A) ) = 0 for all sets A; (iii) Ker(Hom R (V, )) ∩ V ⊥ = 0. Proof. If VR is generalized tilting and Hom R (V, M) = 0 = Ext1R (V, M), then M ∈ V ⊥ = Gen(VR ); therefore M = 0, proving (iii). 4. Conversely, assume that (i), (ii), and (iii) hold. 4, (i) and (ii) imply that Gen(VR ) ⊆ V ⊥ . So assume M ∈ V ⊥ . Since Gen(VR ) ⊆ V ⊥ , TrV (M) ∈ V ⊥ . Hence, the exact sequence 0 → β Tr(M) → M → M/ Tr(M) → 0, where β is the inclusion, induces an exact sequence Hom(V,β) 0 → Hom R (V, TrV (M)) −→ Hom R (V, M) → Hom R (V, M/TrV (M)) → 0 where Hom(V, β) is epic by definition of trace, and so Hom R (V, M/ Tr(M)) = 0.

An inductive argument shows that, if 0 → M1 −→ M2 −→ · · · −→ Mn−1 −→ Mn → 0 is exact, then n (−1)i [Mi ] = 0 i=1 in K 0 (mod-R). The Tilting Theorem, together with this observation and the following lemma, allows us to prove that the Grothendieck groups of a pair of right noetherian rings R and S are isomorphic if there is a tilting bimodule S VR . 2. If R and S are right noetherian rings and S VR is a tilting bimodule, then (1) H = Hom R (V, M) and H = Ext1R (V, M) belong to mod-S whenever M ∈ mod-R; (2) T = N ⊗ S V and T =Tor1S (N , V ) belong to mod-R whenever N ∈ mod-S.

H M) ≤ m. dim. dim. S ≤ n + 1. The most interesting case occurs when R is right hereditary, for then the torsion theory (S, E) on Mod-S induced by a tilting bimodule S VR splits in the sense that each N ∈ Mod-S is a direct sum of a module in S and a module in E. 2. If S VR is a tilting bimodule that induces the torsion theory (T , F ) in Mod-R, then, for all L , M ∈ T and all integers i ≥ 0 ExtiS (H L , H M) ∼ = ExtiR (L , M). Proof. 8), there is an exact sequence d3 d2 d1 d0 · · · −→ V2 −→ V1 −→ V0 −→ L → 0 with each Vi ∈ Add(VR ).

### An introduction to the theory of field extensions by Samuel Moy

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