Socle degrees of Frobenius powers Lecture 3 February 1, 2006 talk by A. Kustin. I. Demonstrate the Magic machine for turning basis vectors

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1 Today s agenda: Socle degrees of Frobenius powers Lecture 3 February, 6 talk by A. Kustin. Demonstrate the Magic machine for turning basis vectors into socle elements.. Outline the proof of () = ().. Get to work on some of the steps of () = ().. Demonstrate the Magic machine for turning basis vectors into socle elements. Let = k[x,..., x n ] be a polynomial ring, be an ideal of with dim k / finite, F be a resolution of / by free -modules, and G be the (Koszul complex) resolution of k = /(x,..., x n ) by free -modules. The isomorphism (*) H n (F k) = H n (Tot(F G)) = H n (/ G) provides a method for converting the basis elements of F n into socle elements of /. illustrate with an example. Let = (x, xy, y ). n this case, F is ( 3) F f = x y x y ( ) 3 F f =[ y xy x ], F G is ( ) G y g = x ( ) G g =[ x y ], G

2 and F G is F G f F G f F G g g g F G f F G f F G g g g f f F G F G F G. Start with in F G in the lower left hand corner. We see that this element represents an element of the homology of H (F k). One can extend this element to get an element of the homology of H (Tot(F G)): [ ] + x y [ ] y y xy The indicated element of H (Tot(F G)) gives rise to the element y of the socle of /. To answer the question that our freshman ask: Yes, it always works like that. We can use the idea of the snaky game to prove both isomorphisms in (*).. Outline the proof of () = (). Recall that our goal is the following result. Theorem. Let k be a field of positive characteristic p, be the polynomial ring k[x,..., x n ], be the homogeneous complete intersection ideal = (f,..., f c ) in

3 and R be /. Let be a homogeneous ideal in with / a finite dimensional vector space over k. Suppose that the socle degrees of R/R are d d l and that the socle degrees of R/ [p] R are D D L. Then the following statements are equivalent: () L = l and D i = pd i (p )a(r) for all i, and () The ring R/R has finite projective dimension as an R-module. Remark. n the present context a(r) is f i x i. We outline the proof of () = () under the additional hypothesis that / is a Gorenstein ring. Our original proof did have this additional hypothesis. The proof is more direct and the result is better without the additional hypothesis; however, without the additional hypothesis one must make many calculations which involve the canonical module. The canonical module of a Gorenstein ring is itself. f make this additional hypothesis, then can hide the fact that we are making canonical module calculations. Assume (). The following steps will yield (). Step. Tor c (/, /) = : ( f i ). (Actually, you know how to prove this already.) Step. We can connect the generator degrees of : to the socle degrees of /. (This uses Gorenstein duality.) Step 3. Suppose the generators of Tor c (/, /) have degrees {γ i } and the generators of Tor c (/ [p], /) have degrees {Γ i }. Then Γ i = pγ i. (This is a straightforward calculation.) Step 4. Use the generators of Tor c (/, /) to produce the generators of Tor c (/ [p], /). (This is a delicate linear independence argument.) Step 5. Drag the answer to Step 4 through the double complex machine to learn that [p] : = ( : ) [p] (f f c ) p + [p]. ( will leave this step out of these lectures.) Step 6. rove that the conclusion to step 5 implies (This step is very similar to step 4.) [p] = ( ) [p] + [p]. Step 7. rove that the conclusion of Step 6 implies Tor R (R/R, ϕ R) =. (This is a grubby calculation. t is the one piece of the proof in this direction that 3

4 4 included in the notes for last week. copied this calculation into the present notes. might not bother to write them on the board.) Step 8. We are finished by the Theorem of Avramov and laudia Miller (see the last seminar talk given by John Olmo last semester.): f Tor R (R/R, ϕ R) =, then pd R (R/R) <. (There is nothing for us to do here!). Get to work on some of the steps of () = (). roof of Step. Let G be the Koszul complex which resolves /. The end of G is ( c f i ) i= i f. ( c f i ) ( c f c f i ) i= i= i... ( c i= i c f i ).... We may compute Tor c (/, /) by tensoring the above resolution with / (that is setting = ) and then computing homology. So, Tor c (/, /) is the kernel of ( c f i ) i= f.. c ( f c f i ) which is : ( c f i ), as claimed. i= i= i ( c i= i... f i ) ( c f i ), roof of Step. We will use two statements about Gorenstein duality. These statements are not independent; indeed, either one could be used to prove the other. i= i c

5 Furthermore, maybe the real key statement is that if / is Gorenstein and a finite dimensional vector space, then / is an injective /-module (which means that the functor Hom / (, /) is an exact functor.) At Bard ollege, Lars hristenson tossed off that most of us think of this as the definition of Gorenstein. Assume that / is a finite dimensional vector space and is a Gorenstein ring. Let N be the socle degree of /. Let M be a finitely generated /-module. Then A. Hom / (Hom / (M, /), /) = M, and B. dim k Hom / (M, /) d = dim k M N d for all d. Of course, the point is that Hom / (, /) exactly turns / modules upside down! Anyhow, claim that if {δ i } are the generator degrees of :, then δ i = N d i. roof. n this argument, Hom means Hom / and means /. Use : Nakayama s Lemma to see that the generator degrees of are equal to the : degrees of +m( : ). Recall that R/R is the same as /( + ); and therefore, the socle of R/R is equal to and by A, this is equal to (+) : m + = Hom( m, + ) Hom( m, Hom(Hom( +, ), )) = Hom( : (+) m, Hom(, )) = Hom( m, Hom( :, )). Now use the Adjoint somorphism Theorem, which says to see that the socle of R/R is equal to Hom(A B, ) = Hom(A, Hom(B, )), Hom( m :, ) = Hom( : +m( : ), ). 5 Finally, we use B to complete the proof. roof of Step 3. We have the socle degrees of R/R are {d i }, the socle degrees of R/ [p] R are {D i },

6 6 D i = pd i (p )( f j x j ), the generator degrees of Tor c (/, /) are {γ i }, the generator degrees of Tor c (/ [p], /) are {Γ i }, Tor c (/, /) = : ( f i ), Tor c (/ [p], /) = [p] : ( f i ), the generator degrees of : are {N d i}, and the generator degrees of [p] : are { pn (p )( x j ) D i }. (Recall that the () () part of the proof tells us that if the socle degree of / is N, then the socle degree of / [p] is pn (p )a( ). This explains the formula inside the box.) Our job is to Do the Math. find it convenient to let {δ i } be the generator degrees of : and { i} be the generator degrees of [p] :. We have Γ i = i + f j = pn (p )( x j ) D i + f j = pn (p )( x j ) as claimed. ( pd i (p )( f j ) x j ) + f j = pn pd i + (p ) f j + f j = pn pd i + p f j = p((n d i ) + f j ) = p(δ i + f j ) = pγ i, roof of Step 7. We want to prove that (***) [p] = ( ) [p] + [p] implies Tor R (R/R, ϕ R) =. We show that Tor R (R/R, ϕ R) = by showing that (**) R b d R b d R R/ = R b R b is exact R R/ [p] is exact. We show that (***) implies (**). make my calculation at the -level. Let a,..., a b d = [ a... a b ] generate in ; so,

7 7 and We think of d as having two pieces: where = [ ap... a p b ]. d = [ d d ] b d b d is exact (and d is all of the extra columns that describe elements of which are also in.) Recall that Kunz s Theorem (ingredient (B) of the other direction) tells us that is exact. Suppose v is in b b (d )[p] b with (v). n other words, (v) [p] = ( ) [p] + [p]. So, there exist s,..., s t ; α,..., α t in ; and c,... c b Of course, there exists v i b So, So, t b (v) = α i s p i + a p i c i. (v) = d[p] v i= i= in so that with d (v i ) = s i (and therefore also v[p] = s p i ). t i= t i= α i v [p] i α i v [p] i + c.. c b c. c b. is killed by ; hence is in the image of (d ) [p]. Finally, d (v i ) = s i, so v i = d (w i) for some w i ; hence, v [p] i = (d )[p] (w [p] i ). Thus, as desired. v im + b,