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189-570A: Higher Algebra I

Assignment 1: Corrections.



Grading scheme. The first three problems were straightforward (if somewhat tedious) exercises in applying the definitions. Their goal was to get you "warmed up" and familiarised with the language of category theory. Almost everyone did well on those questions, so I did not assign marks for them. I graded each of the remaining 7 questions out of 15 for a maximal total of 105 (but the final grade was out of 100).

Grade distribution. Out of 14 assignments, 5 received a score of 55 or less, and the remaining nine received a score of 80 or more, with several getting a perfect score of 105. This suggests that the class is roughly divided into two groups with unequal backgrounds and prior exposure to the material. If you are part of the first group, you should not get discouraged, but you will need to invest more time and effort in the course. I also encourage you to hand in the questions that you did not have time to complete -- without looking at the solutions below, of course!! -- this will be helpful for you and I will take it into account in your final grade. In the future, it would also be wise to begin working on the asignments as soon as they are posted.

Sketch of solutions to some of the problems.

4. If f is a function (arrow) from A to B, it induces functions
f*: Hom(*,A) --> Hom(*,B), and f*: Hom((B,*) --> Hom(A,*)
by the rules
f*(g) = fg, and f*(g) = gf.
The assignment of hom(*,A) to any object A of C and of f* to any morphism in this category is a covariant functor, and the assignment of hom(A,*) to any object A of C and of f* to any morphism in this category is a contravariant functor. This was straightforward to check; the only subtlety that some missed is that the direction of arrows is reversed with f*, so that
(fg)* = g* f*.
Part c) was straightforward, although admittedly tedious.


5. Showing that the functors Hom(W,-) and Hom(-,W) preserved exactness was a formal argument, applying equally well to modules or vector spaces, except for exactness on the right.

The statement that Hom(W,-) is right exact is simply the statement that every homomorphism f from W to V'' necessarily lifts to a homomorphism from W to V if pi:V-->V'' is a surjective homomorphism. Such a lift could be found for vector spaces by taking a basis for W, sending each basis vector w to a preimage of f(w) under pi, and extending the resulting map by F-linearity.

The statement that Hom(-,W) is right exact is simply the statement that every homomorphism f from V' (contained in a vector space V) to W extends to a homomorphism from V to W. Such an extension is obtained by completing the basis for V' to a basis for V, and sending the additional basis vectors to arbitrarily chosen vectors in W.

In both cases, the existence of bases for vector spaces was used in a crucial way.

In the context of modules, one can find many counterexamples: for example, letting Z/nZ be the Z-module of integers modulo n, the functor Hom(Z/nZ,-) does not preserve exactness on the right, because (for example) the identiy map Z/nZ-->Z/nZ does not lift to a homomorphism Z/nZ --> Z. (There are no such homomorphisms, other than the trivial one!).
Similarly, Hom(-,Z) is not exact, as the homomorphism from nZ to Z sending n to 1 does not extend to a homomorphism from Z to Z, if n>1. (Such an extension would need to send 1 to 1/n.)


6. The representing object for this functor is simply the ring
X = Z[x,y]/(x2-3y5-17),
the quotient of the polynomial ring in two variables with coefficients in Z, by the ideal generated by the relation. Indeed, to give a homomorphism from X to R is "the same thing" as specifying the images of x and y, which must satisfy (in R) the same algebraic relation as x and y satisfy in X. Of course, this phenomenon is completely general: given and set of algebraic equations in a finite number of variables, the functor which to a ring R associates the set of solutions to the system of equation with entries in R is a representable functor.


7. This problem involves a trick: Let
M = { (g,x) such that g x=x } (in G x X.)
Now count the cardinality of M in two different ways. On the one hand it is equal to the sum of FP(g), as G varies ove G; on the other hand it is equal to the sum of the cardinality of Stab(x), as x runs over X. Now evaluate the latter expression by using the counting formula for group actions.... Part b) could be reduced to part a), by consider the action of G on the set Y of ordered pairs of distinct elements of X. (I was surprised than many people, including those who were able to solve the harder part a, did not write a solution to part b.)


8. Use the counting formula arising from the action of a group on itself by conjugation. Since each conjugacy class has size dividing the order of the group (NOT because a conjugacy class is a subgroup, as some people incorrectly wrote, but because a conjugacy class is isomorphic to G/H as a G-set, for some subgroup H of G) this size is a power of p. In particular, each conjugacy class of size >1 has cardinality divisible by p. It follows that p divides the order of the center Z(G) as well.
The proof that every group of order a power of p is solvable now follows by an induction on the size of the group, using the fact that G is solvable if and only if the quotient of G by its center is.


9. To write down an invertible matrix, there are pn-1 possibilities for the first column (the zero vector will not do); then given this choice, there are pn-p choices for the second column; pn-p2 choices for the third column; and so on and so forth. Thus, the total number of invertible matrices is
(pn-1)(pn-p)(pn-p2) ... (pn-pn-1).



10. Here is a solution that some of you wrote down, which is more elegant than what I had in mind: Let X be the collection of ordered sequences of k linearly independent vectors. The cardinality of X (by the same reasoning as in 9) is
(pn-1)(pn-p)(pn-p2) ... (pn-pk-1).
The group GLk(Fp) acts naturally on X (how?) in such a way that the orbits for this action correspond to the distinct k-dimensional spaces. The size of each orbit is equal to the cardinality of GLk(Fp). The result is deduced from this and part 9.