For modules one has the isomorphism theorem $(A/C)/(B/C) \cong A/B$ for $C\leq B\leq A$. One way to remember it is through analogy with canceling of fractions. Another way to remember and prove it is to put all the modules in a 3×3 commutative diagram

$$

\begin{matrix}

C & \to & B & \to & B/C\\

\downarrow & ~ & \downarrow & ~ & \downarrow\\

C & \to & A & \to & A/C\\

\downarrow & ~ & \downarrow & ~ & \downarrow\\

0 & \to & A/B & \to & (A/C)/(B/C)

\end{matrix}

$$

where there are also zero arrows on the edges of the diagram, but I have omitted them for ease of typesetting. All the columns are exact, and the first two rows are exact, so the remaining row is exact giving the required isomorphism via the 3×3-lemma.

Here is a fact from linear algebra: if $ V$ is a finite dimensional vector space over a field $ k$ then $ V$ is naturally isomorphic to its double dual. Of course, it is also isomorphic to its dual as well, but it is the natural isomorphism that is more interesting.

Write $ V^* = \mathrm{Hom}_k(V,k)$ for the dual space of $ V$.

This is not true in general for infinite dimensional vector spaces. Before I actually talk about the subject matter, let us describe a counterexample:if we use the field $ k = \mathbb{Q}$, and $ I$ is a countable set, then $ \mathbb{Q}^{(I)}$ is also countable (here, the notation $ \mathbb{Q}^{(I)}$ refers to the direct sum of $ I$ copies of $ \mathbb{Q}$). The dual space of any vector space has cardinality at least as great as the original space, but $ \mathrm{Hom}_k(\mathbb{Q}^{(I)},\mathbb{Q})\cong\prod_{i\in I}\mathbb{Q}$, which is uncountable.

Anyways, the point is $ V\cong V^{**}$. Why is this? Of course, one can provide a short and direct proof: define $ \Phi:V\to V^{**}$ by $ \Phi(v)(f) = f(v)$. This map is injective and since $ V$ is finite dimensional, it is surjective.

Is this really satisfying though? Sometimes in mathematics, it is not completely satisfying just to prove something. Instead, one must also find the proper context for results. The proper context should be enlightening in a variety of situations. For instance, that we know $ V\cong V^{**}$ where $ V$ is a finite-dimensional vector space isn’t all that enlightening. We have a natural isomorphism, so we ought to ask: is there an equivalence of categories hiding somewhere? Indeed there is, and it’s called Morita theory!

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I’m going to talk a bit about flatness. I created the following exercise for myself while studying flatness: describe some good examples of flat modules that are not projective.

We call a left $ R$-module $ F$ flat if the functor $ -\otimes F$ is exact, and of course the analogue can be made for right R-modules. As the tensor functor $ -\otimes F$ is always right exact, this is a rather natural definition to make. Recall that there is a similar definition for the $ \mathrm{Hom}(P,-)$ functor: we call an $ R$-module $ P$ projective if $ \mathrm{Hom}(P,-)$ is exact.

Projective modules have a rather nice characterisation. That is, a module $ P$ is projective if and only if it is the direct summand of a free module. In particular free modules are projective. Mac Lane in his book “Homology” gives an example of a projective module that is not free: take the ring $ \mathbb{Z}\oplus\mathbb{Z}$ and consider the submodule $ \mathbb{Z}\oplus 0$ of it.

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