First Tor of a cyclic quotient 2026-10-14
Start a free resolution of with and a free module surjecting onto . The next differential presents the relations in . Right exactness of the tensor product of modules identifies with . Taking degree-one homology then gives the displayed kernel formula. Hence vanishing for every ideal is equivalent to the flatness criterion using ideals.
Past exam of the mathematics course of the University of Cambridge 2010 iii Paper 3 5 3 Solution Created 2026-10-03 Updated 2026-10-14
For an ideal , the canonical multiplication maphas image . Consequently the asserted isomorphism is the natural surjection being injective. This is the flatness criterion using ideals, whose proof follows from the extension criterion above.
Under the character Tensor-hom adjunction, the dual map identifies withIndeed, for , its restriction corresponds to . If is injective, this restriction is surjective, so the cogenerator criterion makes injective. Conversely, if every is injective, every restriction map in this display is surjective. The proved Baer criterion then makes injective. ThusIn particular, (1) also implies (3) directly by tensoring . No finite-generation condition on , and no Noetherian condition on , has been inserted.