VakilPosted on September 15, 2018
Here are my exercise solutions and notes for the beginning of Vakil’s Foundations of Algebraic Geometry.
Chapter 1: Category Theory
The set in a groupoid with one object is a group, and vice versa.
Pick any groupoid with two or more objects?
1.2.B. Identity and associativity under the operation of composition follows from the definition of morphisms in a category. Thus the set of invertible automorphisms forms a group.
The automorphism groups in the category of sets are the symmetric groups. Assuming vector spaces are finite dimensional, their automorphism groups are general linear groups. For infinite dimensional vector spaces more complicated groups would arise.
If is isomorphic to then is isomorphic to
Suppose is an isomorphism Let Observe that is a morphism with inverse So
Let be defined by Note that this map is surjective by the argument above. Moreover, for , observe that Thus is a homomorphism. Finally, suppose that Then Thus has trivial kernel and is hence injective.
The point of the “isomorphic, but not canonically isomorphic” comment is that although the automorphism groups are pairwise isomorphic, there is not necessarily a “canonical” way of identifying the groups; that is, there might be distinct isomorphisms between the same two automorphism groups. This is classically the case with pointed fundamental groups in the situation that Vakill describes.
1.3.A. Let be two initial objects. Let , be the only morphisms. Observe that both and are morphisms , thus they must be equal since is initial. Similarly for Thus are isomorphisms.
A similar argument follows for final.
1.3.B. In Sets and Top the initial object is and the final object is a singleton In Rings the initial object is and the final object is the zero ring.
1.3.C. Let be the natural homomorphism. Observe that if and only if for some Thus , and injective if and only if there exist no zero divisors in
1.3.D. Suppose is an -algebra homomorphism sending each element of to a unit in Observe that if for some , then Thus Hence there exists a unique map such that (see Lemma 1.2 in my [Fulton notes][BroFulton]).
1.3.E. Clearly as defined in the hint is an abelian group under the defined addition, with identity for all Moreover, the defined ring action defined satisfies the axioms for to be an -module.
(a) Define by Observe that Therefore is a group homomorphism, and thus clearly an -module homomorphism.
Suppose that Then for some Therefore for all and thus Hence Thus
Let Let and Then observe that Thus is surjective.
(b) The same argument above works equivalently for arbitrary direct sums.
(c) Let for some prime Then
See the corresponding exercise in my solutions to Fulton’s algebraic curve.