HW due Fri Mon Feb 26
The "+" in what follows are direct sums.
1. Let R be a ring, and F0 > F1 > F2 > ... be a succesion of abelian subgroups of the additive group R. Assume F0 = R. Write
grFR = R/F1 + F1/F2 + ...
Figure out an obvious condition on the {Fk} such that grFR is a ring in a "natural" way. Your condition should be general enough to include the case {Fk = Ik} where I is a 2-sided ideal, and it should imply that each Fk is an ideal.
(It was pointed out to me that I didn't define Ik in class. It is generated by all k-fold products of elements from I.)
HINT. The "gr" stands for "graded", where the "kth graded piece" (grFR)k := Fk/Fk+1. Set up the multiplication so that (grFR)k times (grFR)j goes into (grFR)j+k. In particular, each graded piece is a module over (grFR)0 = R/I, and hence a module over R.
This condition makes F into a decreasing filtration.
2. Let R = C[x], I an ideal, and Fk = Ik. F is called the I-adic filtration.
a) If I = < x >, show that grFR is isomorphic to R.
b) If I = < x2 >, show that grFR is not isomorphic to R.
3. Let M be a left R-module, and define
grFM = M/F1M + F1M/F2M + F2M/F3M + ...
a) Show that grFM is naturally a grFR-module.
b) Indeed, figure out how to use "extension of scalars" to make this happen automatically. (Lots has to be invented for this -- e.g. where's the ring homomorphism?)
CORRECTION. While there is a map from the tensor product suggested by extension of scalars to grFM, it doesn't seem to always be 1:1. Show that it is in the case that Fj = < bj >, i.e. the < b >-adic filtration.
4a) Now let G0 < G1 < ... be an increasing filtration, and define grGM := G1M/G0M + G2 M/G1M + ...
The same condition from before makes grGR into a ring.
If grGM is finitely generated, it doesn't follow that M is finitely generated. Find a nice condition on G under which one can draw the conclusion of finite generation.
HINT. It can fail when R = C[x] and Gi = C for all i.
HINT. This should remind you of the "N and M/N f.g. => M f.g." question from last week.
4. Let I be an ideal in C[x,y]. Let Gj = {polynomials with highest x power at most j}, Hj = {polynomials with highest y power at most j}
a) Let J = grG grH I. It is a module over grGgrHC[x,y] ~ C[x,y]. Show that J is generated by monomials.
b) [Combinatorics.] Show that any ideal in C[x,y] generated by monomials is generated by finitely many such.
c) Infer that all ideals are finitely generated as C[x,y]-modules.
1. Let R be a ring, and F0 > F1 > F2 > ... be a succesion of abelian subgroups of the additive group R. Assume F0 = R. Write
grFR = R/F1 + F1/F2 + ...
Figure out an obvious condition on the {Fk} such that grFR is a ring in a "natural" way. Your condition should be general enough to include the case {Fk = Ik} where I is a 2-sided ideal, and it should imply that each Fk is an ideal.
(It was pointed out to me that I didn't define Ik in class. It is generated by all k-fold products of elements from I.)
HINT. The "gr" stands for "graded", where the "kth graded piece" (grFR)k := Fk/Fk+1. Set up the multiplication so that (grFR)k times (grFR)j goes into (grFR)j+k. In particular, each graded piece is a module over (grFR)0 = R/I, and hence a module over R.
This condition makes F into a decreasing filtration.
2. Let R = C[x], I an ideal, and Fk = Ik. F is called the I-adic filtration.
a) If I = < x >, show that grFR is isomorphic to R.
b) If I = < x2 >, show that grFR is not isomorphic to R.
3. Let M be a left R-module, and define
grFM = M/F1M + F1M/F2M + F2M/F3M + ...
a) Show that grFM is naturally a grFR-module.
b) Indeed, figure out how to use "extension of scalars" to make this happen automatically. (Lots has to be invented for this -- e.g. where's the ring homomorphism?)
CORRECTION. While there is a map from the tensor product suggested by extension of scalars to grFM, it doesn't seem to always be 1:1. Show that it is in the case that Fj = < bj >, i.e. the < b >-adic filtration.
4a) Now let G0 < G1 < ... be an increasing filtration, and define grGM := G1M/G0M + G2 M/G1M + ...
The same condition from before makes grGR into a ring.
If grGM is finitely generated, it doesn't follow that M is finitely generated. Find a nice condition on G under which one can draw the conclusion of finite generation.
HINT. It can fail when R = C[x] and Gi = C for all i.
HINT. This should remind you of the "N and M/N f.g. => M f.g." question from last week.
4. Let I be an ideal in C[x,y]. Let Gj = {polynomials with highest x power at most j}, Hj = {polynomials with highest y power at most j}
a) Let J = grG grH I. It is a module over grGgrHC[x,y] ~ C[x,y]. Show that J is generated by monomials.
b) [Combinatorics.] Show that any ideal in C[x,y] generated by monomials is generated by finitely many such.
c) Infer that all ideals are finitely generated as C[x,y]-modules.
posted by Allen Knutson @ 11:23 AM
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