NOTES, EXPOSITIONS, AND SLIDES

I tend to fall write something down when I’m learning something new, but also when I’m trying to solidify my understanding of a concept, which I might consider to have good grasp of. Here I’ll keep some of those writings—I hope they may be of use to those that seek to read about whatever the paper is about. (Warning: many of them are incomplete and and might not be giving full justice to the depth of the paper’s main topic.)

• Derived Functors and Derived Categories: Written to summarize the main construction of $\text{Ext}$ and $\text{Tor}$, and give an introduction to derived categories from the perspective of abstracting derived functors. The first time I came to learning about $\operatorname{Ext}^i _A (-, M)$, I didn't really understand what we really meant once we consider some $A$-module $N$ and take a projective resolution, trucate the resolution, take (co)homology, and yadda yadda yadda; we get $\operatorname{Ext}^i _A(N,M)$. To understand $\operatorname{Ext}^i _A(N, M)$, we should actually look at why we call it "Ext": this comes from looking at so-called extensions. To keep this simple, we consider $\operatorname{Ext}^1 _A(N,M)$. By an extension of $B$ by $A$ for some $R$-modules $A$ and $B$, we mean a short exact sequence:

  $$ 0 \to A \to D \to B \to 0$$

  So, for example, $0 \to \mathbf Z \to \mathbf Q \to \mathbf Q/\mathbf Z \to 0$ is an extension (with the obvious maps) of $\mathbf Q / \mathbf Z$ by $\mathbf Z$. Now, the "Ext" functor essentially measures how many inequivalent extensions there are (whereby we define equivalence by isomorphisms of short exact sequence that are the identity). (For the details of this equivalence, see the linked document above.) We define $\mathfrak{ext} (B,A)$ as the set of extensions of $B$ by $A$ up to the equivalence described. To keep this short, there is an isomorphism between $\mathfrak{ext} (B,A) \cong \operatorname{Ext}_R ^1(B,A) $ (considered as groups, where the binary operation of $\mathfrak{ext} (B,A)$ is something called Baer multiplication), where the $\operatorname{Ext} _R^1 (B,A)$ corresponds to the derived functor perspective, which preserves the trivial element. Although these derived functors are defined quite abstractly, they represent an answer to a quite simple question! For example, as $\mathfrak{ext} (B,A) \cong \operatorname{Ext} _R ^1(B,A)$, then we can consider $\mathfrak{ext} (\mathbf Q / \mathbf Z,\mathbf Z) \cong \operatorname{Ext} _{\mathbf{Z}}^1(\mathbf Q / \mathbf Z,\mathbf Z)$. We have that $\mathbf Q / \mathbf Z$ is a divisble abelian group and thereby an injective object in $\mathsf{Ab}$, so $\mathbf Q / \mathbf Z$ is acyclic, so $\mathfrak{ext} (\mathbf Q / \mathbf Z,\mathbf Z) \cong \operatorname{Ext} _{\mathbf{Z}} ^1(\mathbf Q / \mathbf Z,\mathbf Z) = (0)$. As this isomorphism carries the trivial extension, i.e. $\mathbf Z \oplus \mathbf Q / \mathbf Z$, then there essentially only one way to fill in the sequence:

  $$ 0 \to \mathbf Z \to \ast \to \mathbf Q / \mathbf Z \to 0$$

  to get an exact sequence (up to the equivalence described)!