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Math 711, Fall 2006: Solutions for Polynomial Rings - Prof. Melvin Hochster, Assignments of Algebra

University of Michigan (UM) - Ann ArborAlgebra
Prof. Melvin Hochster

Solutions to problem set #3 for the math 711 course offered in fall 2006. The solutions cover topics such as isomorphisms, ideals, and polynomial rings. How to determine if an element is a zerodivisor, how to expand ideals upon localization, and how to work with tor modules.

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Pre 2010

Uploaded on 09/17/2009

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Math 711, Fall 2006 Problem Set #3 Solutions
1. Work mod Gand so assume G= 0. T[X] has a T-automorphism sending Xto X−F.
Thus, X−Fis simply an indeterminate over T. Change notation and write Xinstead of
X−F. The statement becomes that
0 :TI→XT [X] :T[X](I , X)T[X]/X T [X]
is an isomorphsim. The numerator on the right is the same as XT [X] :T[X]I. But it
is immediate that IF ⊆X T [X] if and only if Ikills the constant term of F, so that
XT [X] :T[X]I= (0 :TI) + X T [X], and the result follows at once.
2. Let G= (g1, . . . , g)T. Then XF −1 is not a zerodivisor in T[X]/GT[X]∼
=(T/G)[X],
since its product with a nonzero element whose lowest nonzero degree term is wwill have
nonzero lowest degree term w.
Then T[X]/(I, X F −1)T[X]∼
=S[X]/(uX −1) ∼
=Su. Evidently, if Gexpands to Iupon
localization at any minimal prime, we have that (G, F X −1)T[X] expands to Jupon
localization at any minimal prime. Now consider the image of
M=(G, F X −1)T[X] :T[X](I, F X −1)T[X]
(G, F X −1)T[X]=(G, F X −1)T[X] :T[X]I
(G, F X −1)T[X]
in Lu, which will be WSu/R. Since the gjdo not involve X, the new Jacobian matrix has
bottom row (0 0 . . . 0F). Therefore, the new Jacobian determinant has image γu, since
Fmaps to u. We claim that
(G, F X −1)T[X] :T[XI= (G:TI)T[X]+(F X −1)T[X].
For the purpose of proving this we may work modulo (F X −1)T[X]. The equality is then
seen to be equivalent to the statement that in the ring TFwe have
GTF:TFITF= (G:TI)TF.
This is a consequence of the fact that TFis flat over Tand Iis finitely generated (Tis
Noetherian here.) It follows that the image of Min Luis the Su-submodule generated by
the images of the gi. Each of these is the same as under the map
G:TI
G→ L,
but multiplied by 1/u, since we are now dividing by γu instead of by γ. Since uis invertible
in Su, the image is (WS/R)u, as required .
3. If r∈I S then with J=I+rR, for some k, (JS)k+1 = (I S)(JS )k, and Jk+1 ⊆
(IJ k)S∩R=I J k. Since IJ k⊆Jk+1 is obvious, we have that Jk+1 =IJ k, and so
u∈I.
pf2

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Math 711, Fall 2006 Problem Set #3 Solutions

  1. Work mod G and so assume G = 0. T [X] has a T -automorphism sending X to X − F. Thus, X − F is simply an indeterminate over T. Change notation and write X instead of X − F. The statement becomes that

0 :T I →

XT [X] :T [X] (I, X)T [X]

/XT [X]

is an isomorphsim. The numerator on the right is the same as XT [X] :T [X] I. But it is immediate that IF ⊆ XT [X] if and only if I kills the constant term of F , so that XT [X] :T [X] I = (0 :T I) + XT [X], and the result follows at once. 

  1. Let G = (g 1 ,... , g)T. Then XF − 1 is not a zerodivisor in T [X]/GT [X] ∼= (T /G)[X], since its product with a nonzero element whose lowest nonzero degree term is w will have nonzero lowest degree term w.

Then T [X]/(I, XF − 1)T [X] ∼= S[X]/(uX − 1) ∼= Su. Evidently, if G expands to I upon localization at any minimal prime, we have that (G, F X − 1)T [X] expands to J upon localization at any minimal prime. Now consider the image of

M =

(G, F X − 1)T [X] :T [X] (I, F X − 1)T [X]

(G, F X − 1)T [X]

(G, F X − 1)T [X] :T [X] I

(G, F X − 1)T [X]

in Lu, which will be WSu/R. Since the gj do not involve X, the new Jacobian matrix has bottom row (0 0... 0 F ). Therefore, the new Jacobian determinant has image γu, since F maps to u. We claim that

(G, F X − 1)T [X] :T [X I = (G :T I)T [X] + (F X − 1)T [X].

For the purpose of proving this we may work modulo (F X − 1)T [X]. The equality is then seen to be equivalent to the statement that in the ring TF we have

GTF :TF ITF = (G :T I)TF.

This is a consequence of the fact that TF is flat over T and I is finitely generated (T is Noetherian here.) It follows that the image of M in Lu is the Su-submodule generated by the images of the gi. Each of these is the same as under the map

G :T I G

→ L,

but multiplied by 1/u, since we are now dividing by γu instead of by γ. Since u is invertible in Su, the image is (WS/R)u, as required .

  1. If r ∈ IS then with J = I + rR, for some k, (JS)k+1^ = (IS)(JS)k, and Jk+1^ ⊆ (IJk)S ∩ R = IJk. Since IJk^ ⊆ Jk+1^ is obvious, we have that Jk+1^ = IJk, and so u ∈ I. 

2

  1. Let I = (x 1 ,... , xd− 1 )R and J = (x 1 ,... , xd)R. Consider the long exact sequence · · · → TorRi (R/I, M ) → TorRi (R/I, M ) → TorRi (R/J, M ) → Tori− 1 (R/I, M ) → · · ·. Then for i > 1 the vanishing of Tori(R/J, M ) is a consequence of the fact that, by the induction hypothesis, the two surrounding terms vanish. Moreover, for i = 1, we have · · · → TorR 1 (R/I, M ) → TorR 1 (R/J, M ) → R/I → R/I → R/J → 0. The result follows because TorR 1 (R/I, M ) = 0 by the induction hypothesis and the map given by multiplication by xd is injective on R/I. 
  2. It was shown in class that J ⊆ I is a reduction if and only if the image of J in I/mI generates an ideal primary to the homgeneous maximal ideal M in K ⊗ grI R. Given any reduction J, we may choose a basis for the image of J in I/mI, and then choose elements of J that map to this basis. These elements generate a special reduction contained in J. If J 1 ⊆ J 2 are special reductions and their images in I/mI are the same, we claim that J 1 = J 2. To see this, note that since J 1 + mI = J 2 + mI. we can extend a set S of minimal generators for J 1 to a set of generators of J 2 using elements of J 2 ∩ (mI). This gives a minimal set T of generators of J 2 each of which is in S or is an element of mI. T has the same cardinality as the original set of minimal generators of J 2. T cannot give an image in I/mI of the correct dimension unless we have used all of the elements of S. Since T has the same cardinality as the dimension of the image of J 1 , it follows that T = S, and J 2 = J 1. Thus, if one has a strictly decreasing chain of special reductions, the images in I/mI also decrease strictly, and so the chain cannot be infinite. Given any reduction J with image V in I/mI, choose a subspace W ⊆ V such that no proper subspace of W generates an M-primary ideal. Choose elements of J mapping to a basis for W. These will generate a special reduction of I that does not properly contain any other special reduction. 
  3. The first statement is immediate from 5., since a reduction with an(I) generators must be special, with the image in I/mI spanned by a homogeneous system of linear parameters, and cannot contain a proper reduction, since that would contain a special reduction, and the image would be too small a vector space to span a M-primary ideal. The second statement is clear, because when K is infinite, every reduction has a reduction such that the least number of generators is an(I). 

BONUS Let the homogeneous generators be F 1 ,... , Fn with respective degrees d 1 ,... , dn and let L be the least common multiple of d 1 ,... , dn. Then every mono- mial μ in the Fi of degree D ≥ nL is the product of a monomial of degree L and one of degree D − L: the fact that μ = F a^1 · · · F (^) ka khas degree D implies that

∑n i=1 diai^ ≥^ nL, and so at least one diai ≥ L. Then we can choose b ≤ ai such that dib = L, and F (^) ib has degree L and is a factor of μ. If μ has degree nLh for h > 1 we can iterate this n(h − 1) times to write μ as a product of the n(h − 1) forms of degree L and one of degree nL. The former term can be written as a product of h − 1 forms of degree nL by grouping. Thus, every monomial of degree nLh is a product of h monomials of degree nL, and we may take d = nL.