By Christian Peskine
Peskine does not supply loads of causes (he manages to hide on 30 pages what frequently takes up part a ebook) and the workouts are difficult, however the booklet is however good written, which makes it lovely effortless to learn and comprehend. urged for everybody keen to paintings their manner via his one-line proofs ("Obvious.")!
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Extra resources for An Algebraic Introduction to Complex Projective Geometry: Commutative Algebra
That Ae n keru = (0) is obvious. Furthermore, if x E L, we have x = u(x)e (x - u(x)e)E Ae ker U . Clearly, Ae’ n (ker u n M ) = (0) is also obvious. ) = ba. This shows z = be’ + ( z - be’) E Ae’ (keru n M ) . From the decomposition M = Ae’@(keru n M ) ,we deduce that the maximal number of linearly independent elements in (ker u n M ) is strictly smaller than the corresponding number for M . By the induction hypothesis (ker U n M ) is free, hence so is M and we have proved (i). We now prove (ii) by induction on n = rk(L).
To begin with, note that if the evaluation homomorphism eD,M = 0, then f ( x ) = 0 for all f E HomA(M,D) and all x E M . This shows HomA(M, D ) = (0). Let M be a maximal ideal. Since HomA(A/M, D ) f 0, then eD,A/M : A / M HomA(HomA(A/M, D ) ,D ) (ii) + (iii). Using (*) twice, we find + is different from zero. Note next that (iii) + (iv). By (*), lA(HomA(M,D ) ) 5 ~ A ( Mfor ) all finitely generated A-modules M. If M is finitely generated, there exists an integer n and an exact sequence 0 + K 4 nA 4 M -+0.
Ii) We denote by M‘ the submodule of JJME~UPPm(M)MM formed by all ( z M / s M ) ~ ~ s such ~ ~ that ~ ~ X( M~S M) ‘ = X M ~ S M for all M, M ’ ~ S u p p ~ ( A 4 ) . It is clear that f ( M ) c M’. Note that (0) : M = (0) : M’. 29, hence Supp(M’) = Supp(M). Consider an element x = ( Z M / S M ) M E M’. We have S M X M , / S M ! = Z M / ~E MM, for all M‘ E Suppm(M). kfh This shows S M X E f(M), hence (f(M)M = for all M E Suppm(M). When M is finitely generated, we have Supp(M’) = Supp(M) and this proves f ( M ) = M’ When M is not finitely generated, we assume that A is Noetherian.