Fri Feb 23 and Mon Feb 26

Friday:
Spec of a ring, as a functor CommRing^op -> Set.

Monday:
Ascending chains of ideals in a PID terminate.
We used that to finally give the first description of a finitely generated module over a PID.
CRT: Let R be a PID. The map R/< p^a m > -> R/< p^a > + R/< m >, where p is a prime not dividing m, is an isomorphism. Showing it was 1:1 used domain, but onto used PID.

HW due tomorrow; lose Q4

I'm ditching question 4 entirely, as it's much too late to properly patch it up now. I do apologize.

HW due Friday Mar 2

1. In class we determined the Spec of C[x], using the fact that complex polynomials of degree > 1 all factor. It had a nice description as C union a weird extra point. (Don't confuse that point with the very different point at infinity in the Riemann sphere!)
Determine the Spec of R[x] (here R is the reals), and describe the map Spec C[x]-> Spec R[x].

2. We showed that a finitely generated torsion free module over an ED is free.
Show that the ring C[z,z^-1] of Laurent polynomials, considered as a module over the subring C[z], is torsion free but not free (i.e. it doesn't have a basis).

3. Let R = F[[z]], power series in one variable, with coefficients in some field F.

a. Show that r in R is invertible iff r not in < z >.

b. What are the elements of Spec R?

c. If F = C, so that we understand Spec C[z], describe the map Spec C[[z]] -> Spec C[z] induced by the inclusion C[z] -> C[[z]].

4. Let f:R->S be a homomorphism and r in R. Show that S @_R (R/< r >) is isomorphic to S/< f(r) > as an S-module.

5. Let V be a finite-dimensional complex vector space, and T:V->V an endomorphism. Consider it as a C[z]-module in the usual way.

a. Let C[z]->C take z |-> lambda, making C a C[z]-algebra.
Show that C @_C[z] V is finite-dimensional over C, and that its dimension is that of the lambda eigenspace of T.

b. Show that C[z,z^-1] @_C[z] V is finite-dimensional over C, and that its dimension is the number of nonzero eigenvalues (with multiplicity) of T.

c. Show that C[[z]] @_C[z] V is finite-dimensional over C, and that its dimension is the lowest degree occurring in T's characteristic polynomial.

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
gr_FR = R/F₁ + F₁/F₂ + ...

Figure out an obvious condition on the {F_k} such that gr_FR is a ring in a "natural" way. Your condition should be general enough to include the case {F_k = I^k} where I is a 2-sided ideal, and it should imply that each F_k is an ideal.
(It was pointed out to me that I didn't define I^k 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" (gr_FR)_k := F_k/F_k+1. Set up the multiplication so that (gr_FR)_k times (gr_FR)_j goes into (gr_FR)_j+k. In particular, each graded piece is a module over (gr_FR)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 F_k = I^k. F is called the I-adic filtration.

a) If I = < x >, show that gr_FR is isomorphic to R.

b) If I = < x² >, show that gr_FR is not isomorphic to R.

3. Let M be a left R-module, and define
gr_FM = M/F₁M + F₁M/F₂M + F₂M/F₃M + ...

a) Show that gr_FM is naturally a gr_FR-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 gr_FM, it doesn't seem to always be 1:1. Show that it is in the case that F_j = < b^j >, i.e. the < b >-adic filtration.

4a) Now let G₀ < G₁ < ... be an increasing filtration, and define gr_GM := G₁M/G₀M + G₂ M/G₁M + ...
The same condition from before makes gr_GR into a ring.

If gr_GM 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 G_i = 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 G_j = {polynomials with highest x power at most j}, H_j = {polynomials with highest y power at most j}

a) Let J = gr_G gr_H I. It is a module over gr_Ggr_HC[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.

Mon Feb 12

Def. Nilpotents, radical ideals, zero divisors, domain, principal ideal, principal ideal domain, Euclidean domain.

Prop. If R has no nilpotents, but ab=0, we can embed R in R/< a > + R/< b >.

Theorem. EDs are PIDs.
Warning: EDs are a stupid concept, not the same level of importance as PIDs.

HW due Fri Feb 16

1a. Let A be an (R,S)-bimodule, and B an abelian subgroup that is both an R-submodule and a (right) S-submodule. Show that A/B is again naturally an (R,S)-bimodule.
(Don't worry about proving naturality.)

b. Let I,J be two-sided ideals of R. Show that the (R,R)-bimodule R/I @_R R/J is isomorphic to R/K for some other ideal K.

2. Let M > N be a pair of R-modules. If N and M/N are finitely generated, show M is also finitely generated.

3. Let I be a 2-sided ideal of R, and for any module M, let IM := { sum_j i_j m_j }.

a. Show that IM is an R-submodule.

b. Show that M/IM is naturally an R/I-module (not just an R-module).
(Don't worry about proving naturality.)

c. Figure out how to describe M/IM as a degenerate sort of "extension of scalars" (!).

d. If M a free module over R, show M/IM is a free module over R/I.

4. Let R be the (commuting) polynomial ring C[x,y] and I the ideal < x, y > generated by the variables. Show I is not a free module.

Wed Feb 7 and Fri Feb 9

Friday:

"Bilinearity" is only a good notion for commutative rings, and is then closely related to tensor products.

How to show a tensor product is big or small.

If R,S are k-algebras, and A,B are modules over them, then A @_k B is a module over R @_k S.
In the R=S case, sometimes there are interesting "comultiplications" R -> R @_k R.

Def. Symmetric algebra, group algebra, monoid algebra.

The co-commutative comultiplication on a monoid algebra.

Modified question 4

Sorry for this being so late, but only now do I see why question 4 was so nonsensical. It's been modified.

Mon Feb 5

Part of the proof that tensor is adjoint to Hom.
Extension of scalars.
The tensor algebra of an abelian group -- the left adjoint of Ring -> Ab.
The tensor algebra of a (k,k)-bimodule -- the left adjoint of k-Alg -> (k,k)-BiMod.

HW due Fri Feb 9Mon Feb 12 (to avoid the complex analysis midterm)

1a. Let R be a subring of S, and A_S, _SB be two modules. Define a surjection A@_RB ->> A@_SB.

1b. Can you extend this to a natural transformation between two functors between something-or-other? This question is intentionally ill-defined, and I imagine different people will experience different creative urges.

2. Let A,B be right R-modules and C,D left R-modules. Let + denote direct sum. Define isomorphisms (A + B) @_R C ~ A@_RC + B@_RC and B @_R (C + D) ~ B@_RC + B@_RD.

...a very dull question, but it leads to...

3. Let M_n(C) denote the ring of nxn complex matrices, and C^{k x n} denote the (M_k(C),M_n(C))-bimodule of kxn complex matrices. Show that matrix multiplication
C^{k x n} x C^{n x m} -> C^{k x m}
induces a map
C^{k x n} @_Mn(C) C^{n x m} -> C^{k x m}
and that this map is an isomorphism.

4. An isomorphism D: R -> R^op is called an anti-automorphism of R. For example, the identity function is an anti-automorphism iff R is commutative.

a. Let M_R be a right R-module. Use D to define a left module structure on it.


b. Let _RN be a left R-module. Using D, we can make N^* := Hom_R(N,R) a left R-module. Using D again, we can make (N^*)^* a left R-module. Find a condition on D that guarantees that this new left R-module structure on N^** agrees with the old one.

c. Show the converse: if this new left R-module structure on N^** agrees with the old one for all N, then D satisfies that condition.


b. Even if D is just an anti-homomorphism, not necessarily invertible, say how to use it to turn any right R-module into a left R-module.

c. Let M be a left R-module, which we turn into a right module, then into a left module again. Determine the condition on D under which this new module structure agrees with the old one (for arbitrary M).

5. Forgetting module theory for a moment, an inner product < , > on a real vector space V induces an isomorphism V -> V^*, taking v |-> (w |-> < v,w >). This looks weird to us now, in that it's an isomorphism of a left module with a right module.

a. Use question 4 to state the definition of an inner product on a real vector space, being careful about left vs. right modules. (This being an algebra class, you can skip nonalgebraic conditions like "positive definite".)

b. Use question 4 to define a Hermitian inner product on a complex vector space, being careful about left vs. right modules.

6. Let A,B be two finite abelian groups (aka (Z,Z)-bimodules!). Show that A @_Z B is again a finite abelian group, and say how to calculate it.

Fri Feb 2

Representable functors.
Yoneda lemma.
Given modules _SA, _RB_S, we get a functor
Hom_S(A,Hom_R(B,*)) -> Set,
which we claim is representable by "B @_S A".
Definition of @.
Example: Reals @_Z Z ~ Reals.

Wed Jan 31

Generating a module. Finite generation. Cyclic modules.
A -> Hom_R^op(Hom_R(A,R)).
Right modules, bimodules.
Hom_R(_RA_S,_RB_T) is an (S,T)-bimodule.