Spec o' Math

Derived Functors

October 27, 2009, 8:43 am
Filed under: Category Theory, Homological Algebra | Tags: Abelian Categories, complexes, Derived Functors, Dummit & Foote, Functors, Weibel

While studying commutative algebra and module theory, you’ll eventually be introduced to projective and injective modules. Then, suddenly, people will start throwing around words like derived functors, chain complexes, Ext and Tor. It can all be pretty daunting, especially if homological algebra and category theory is only taught as a ‘side-dish’ on a need to know basis, supplementary to whatever material is typically meant to be covered.

On the other hand, you could also be studying Algebraic Topology, where chain complexes first came about. Either way, we’ll be approaching the topic in a more or less abstract manner (hopefully) appropriate for both situations.

So, where do you go when many homological algebra books don’t talk about derived functors until the second half of the book? I’d recommend either An Introduction to Homological Algebra by C.A. Weibel or, my personal preference, Section 17.1 of Abstract Algebra by Dummit & Foote, which most students already own. Both start out from (co)chain complexes and go straight to derived functors.

The only downside to using Dummit & Foote is that they do proofs using specific functors ( $\textnormal{Hom}(\cdot,\cdot), \cdot \otimes \cdot$ ) as opposed to arbitrary left/right exact functors as Weibel does it.

Here I’ll give a basic introduction on how to get from an understanding of commutative algebra with a little category theory (including: functors, objects, (mono/epi)morphisms, exact sequences) to understanding what a derived functor is and how it is derived from left/right exact functor. Giving proof outlines and/or citations as needed.

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“The Map”

Let $X$ be a compact Hausdorff space and let $C(X)$ denote the ring of all real-valued continuous functions on $X$ (add and multiply functions by adding and multiplying their values). For each $x \in X$ , let $\mathfrak{m}_x$ be the set of all $f \in C(X)$ satisfied by $x$ , i.e. that $f(x) = 0$ . The ideal $\mathfrak{m}_x$ is maximal because it’s the kernel of the homomorphism $C(X) \to \mathbb{R}$ which takes $f$ to $f(x)$ .

If $\tilde{X}$ denotes $\textnormal{Max}(C(X))$ , we can define a map $\mu : X \to \tilde{X}$ that takes $x \mapsto \mathfrak{m}_x$ . We’ll be showing neat stuff about this map.

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Boolean rings and Spec

October 15, 2009, 12:03 am
Filed under: Commutative Algebra | Tags: Boolean rings, Commutative Algebra, Commutative Rings, Max Spec, Spec

Boolean Ring

A Boolean ring is a ring $A$ with every element satisfying $x^2 = x .$ From this we can deduce a couple other quick facts about Boolean rings.

$2x = 0$ for all $x \in A$
every Boolean ring is commutative
every prime ideal $\mathfrak{p}$ is maximal, and $A/\mathfrak{p}$ is a field with two elements
every finitely generated ideal in $A$ is principal

For the first one just expand out $(x+x)^2$ . And for the second (though I always assume commutativity) expand out $(x+y)^2$ remembering that $-x = x$ because of (1).

Now, assuming $\mathfrak{p}$ is a prime ideal, $A/\mathfrak{p}$ must be an integral domain. But every element in $A$ , and hence in $A/\mathfrak{p}$ , satisfies $x(x-1) = 0$ . So it must be that $A/\mathfrak{p} \cong \mathbb{F}_2$ , a finite field of two elements.

Let $(x,y)$ be a finitely generated ideal in $A$ , then we can check that $x(x+y-xy) = x$ and $y(x+y-xy) = y$ , hence $(x,y) = (x+y-xy)$ . So we can reduce any finitely generated ideal to a principal ideal.

Spec

Let $\phi : A \rightarrow B$ be a ring homomorphism, $X = \textnormal{Spec}(A)$ and $Y = \textnormal{Spec}(B)$ . If $\mathfrak{q} \in Y$ , then $\phi^{-1}(\mathfrak{q}) \in X.$ Hence $\phi$ induces a mapping $\phi^*: Y \rightarrow X$ .

The map $\phi^*: Y \rightarrow X$ is continuous.

proof: If $\mathfrak{q} \in \phi^{*-1}(X_f)$ , then $f \not\in \phi^{-1}(\mathfrak{q})$ , which implies that $\phi(f) \not\in \mathfrak{q}$ , hence $\mathfrak{q} \in Y_{\phi(f)}$ .

On the other hand, if $\mathfrak{q} \in Y_{\phi(f)}$ , then $\phi(f) \not\in \mathfrak{q}$ and so $f \not\in \phi^{-1}(\mathfrak{q})$ . Hence $\phi^{*}(\mathfrak{q}) \in X_f$ and $\mathfrak{q} \in \phi^{*-1}(X_f)$ . Finally showing that $\phi^{*-1}(X_f) = Y_{\phi(f)}$ and proving that $\phi^*$ is continuous.

Some other tedious, but good-to-know facts are that $\phi^{*-1}(V(\mathfrak{a})) = V(\phi(\mathfrak{a}))$ and $\overline{\phi^*(V(\mathfrak{b}))} = V(\phi^{-1}(\mathfrak{b})$ .

A much more interesting piece of information is that if $\phi$ is surjective, then $\textnormal{Spec}(B)$ is homeomorphic to a subset $V(\ker(\phi))$ of $\textnormal{Spec}(A)$ . No lengthy proof required, simply observe that $A/\ker(\phi) \cong B$ , hence $V(\ker(\phi)) \cong \textnormal{Spec}(A/\ker(\phi)) \cong \textnormal{Spec}(B)$ .

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