Vakil

Here are my exercise solutions and notes for the beginning of Vakil’s Foundations of Algebraic Geometry.

Chapter 1: Category Theory

1.2.A.

The set $\text{Aut}(A)$ 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 $A$ is isomorphic to $B$ then $\text{Aut}(A)$ is isomorphic to $\text{Aut}(B).$

Suppose $\phi$ is an isomorphism $A\to B.$ Let $f\in\text{Aut}(A).$ Observe that $\phi\circ f\circ\phi^{-1}$ is a morphism $B\to B$ with inverse $\phi\circ f^{-1}\circ\phi^{-1}.$ So $\phi\circ f\circ\phi^{-1}\in\text{Aut} (B,B).$

Let $\phi&39;:\text{Aut}(A)\to\text{Aut}(B)$ be defined by $f\mapsto\phi\circ f\circ\phi^{-1}.$ Note that this map is surjective by the argument above. Moreover, for $f,g\in\text{Aut}(A)$ , observe that $\phi&39;(fg)=\phi&39;(f)\phi&39;(g).$ Thus $\phi&39;$ is a homomorphism. Finally, suppose that $f\mapsto\text{id}_B.$ Then $\begin{aligned}[t] f &= (\phi^{-1}\circ\phi)\circ f\circ (\phi^{-1}\circ\phi)\\ &= \phi^{-1}\circ\text{id}_B\circ\phi\\ &= \phi^{-1}\circ\phi\\ &= \text{id}_A. \end{aligned}$ Thus $\phi&39;$ 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 $A, B$ be two initial objects. Let $\varphi:A\to B$ , $\psi:B\to A$ be the only morphisms. Observe that both $\text{id}_A$ and $\psi\circ\varphi$ are morphisms $A\to A$ , thus they must be equal since $A$ is initial. Similarly for $B.$ Thus $\varphi,\psi$ are isomorphisms.

A similar argument follows for $A, B$ final.

1.3.B. In Sets and Top the initial object is $\emptyset$ and the final object is a singleton $\{a\}.$ In Rings the initial object is $\mathbb{Z}$ and the final object is the zero ring.

1.3.C. Let $\varphi:A\to S^{-1}A$ be the natural homomorphism. Observe that $\varphi(a)=\varphi(b)$ if and only if $s(a-b)=0$ for some $s\in S.$ Thus $\ker\varphi=\{a\in A \ : \ as=0 \ \text{for some} \ s\in S\}$ , and $\varphi$ injective if and only if there exist no zero divisors in $S.$

1.3.D. Suppose $f:A\to B$ is an $A$ -algebra homomorphism sending each element of $s$ to a unit in $B.$ Observe that if $as=0$ for some $s\in S$ , then $f(a)=f(a)f(s)f(s)^{-1}=f(as)f(s)^{-1}=0.$ Thus $\ker f\subseteq\ker\varphi.$ Hence there exists a unique map $\widetilde{f}:S^{-1}A\to B$ such that $f=\widetilde{f}\circ\varphi$ (see Lemma 1.2 in my [Fulton notes][BroFulton]).

1.3.E. Clearly $S^{-1}M$ as defined in the hint is an abelian group under the defined addition, with identity $0/1\equiv 0/s$ for all $s\in S.$ Moreover, the defined ring action defined satisfies the axioms for $S^{-1}M$ to be an $S^{-1}A$ -module.

1.3.F.

(a) Define $\varphi:S^{-1}(M_1\times\ldots\times M_n)\to S^{-1}M_1\times\ldots S^{-1}M_n$ by $(m_1,\ldots,m_n)/s\mapsto (m_1/s,\ldots,m_n/s).$ Observe that $\begin{aligned}[t] \varphi\left(\frac{(m_1,\ldots,m_n)}{s}+\frac{(k_1,\ldots,k_n)}{t}\right) &= \varphi\left(\frac{(tm_1+sk_1,\ldots,tm_n+sk_n)}{st}\right)\\ &=\left(\frac{tm_1+sk_1}{st},\ldots,\frac{tm_n+sk_1}{st}\right)\\ &=\left(\frac{m_1}{s},\ldots,\frac{m_n}{s}\right) + \left(\frac{k_1}{t},\ldots,\frac{k_n}{t}\right)\\ &=\varphi\left(\frac{(m_1,\ldots,m_n)}{s}\right)+ \varphi\left(\frac{(k_1,\ldots,k_n)}{t}\right). \end{aligned}$ Therefore $\varphi$ is a group homomorphism, and thus clearly an $S^{-1}R$ -module homomorphism.

Suppose that $(m_1,\ldots,m_n)/s\mapsto 0.$ Then $s_1m_1=s_2m_2=\ldots=s_nm_n=0$ for some $s_1,\ldots,s_n\in S.$ Therefore $s_1\ldots s_n m_i=0$ for all $i$ and thus $s_1\ldots s_n(m_1,\ldots,m_n)=(0,\ldots,0).$ Hence $(m_1,\ldots,m_n)/s=0\in S^{-1}(M_1\times\ldots\times M_n).$ Thus $\ker\varphi=0.$

Let $(m_1/s_1,\ldots,m_n/s_n)\in S^{-1}M_1\times\ldots\times S^{-1}M_n.$ Let $s=s_1s_2\ldots s_n$ and $t_i=s_1\ldots s_{i-1}\ldots s_{i+1}\ldots s_n.$ Then observe that $(t_1m_1,\ldots,t_nm_n)/s\mapsto (m_1/s_1,\ldots,m_n/s_n).$ Thus $\varphi$ is surjective.

(b) The same argument above works equivalently for arbitrary direct sums.

(c) Let $S=\{q^n \ : \ n\in\mathbb{N}\}$ for some prime $q\in\mathbb{Z}.$ Then $\prod_{i=1}^k M_i.$

See the corresponding exercise in my solutions to Fulton’s algebraic curve.