# What is a free module?

Among modules over a commutative ring $R$, the module $R^n$ has a nice feature: homomorphisms from $R^n$ to another module $M$ are entirely determined by where they send the $n$ elements $(1,0,\dots,0)$, $(0,1,0,\dots, 0)$, up to $(0,\dots,0,1)$. Moreover, any choice of where we want those basis elements to be sent in $M$ extends uniquely to a homomorphism $R^n\to M$. Thus homomorphisms $R^n\to M$ correspond bijectively to $n$-tuples of elements of $M$. This is an example of a universal property: $R^n$ is the universal module equipped with an $n$-tuple of elements, also called the free module on $n$ elements.

What about the free module on an infinite collection of elements? Continue reading

# Interlude: principal ideal rings

Recently I mentioned that Noetherian rings can be characterized either by the fact that their finitely generated modules don’t have non-finitely generated submodules, or by the a priori weaker condition that their ideals are all finitely generated. In other words, if you want a counterexample to the statement that a submodule of a finitely generated module $M$ is again finitely generated, it’s sufficient to consider the case $M = R$.

I also claimed that the function $\mu_R$ sending an $R$-module to its minimum number of generators is not generally increasing: for $R$-modules $N\subseteq M$ we don’t necessarily have $\mu_R(N)\leq \mu_R(M)$, even when $R$ is Noetherian. For example, over $R = \mathbb{C}[x,y]$, the ideal $I = (x,y)\subseteq R$ has $\mu_R(I)=2$, but $\mu_R(R) = 1$.

These two facts led me to ask the following question:

If we want a commutative ring $R$ to have the property that $\mu_R$ is increasing, is it again sufficient to look for counterexamples among the ideals of $R$? In other words, are the rings whose ideals can all be generated by single elements (the principal ideal rings) exactly the rings with this property?

# What is an exact sequence?

I’ve heard it said that the concept of modules was invented to put ideals of rings on the same footing as quotient rings. Both the inclusion of an ideal $I$ into a ring $R$ and the quotient map $R \to R/I$ are homomorphisms of $R$-modules. In fact, these two homomorphisms have a special relationship: the elements of $R$ that are sent to zero in $R/I$ are exactly those in the image of $I\to R$.

# What is a Noetherian ring?

Noetherianness is about making sure small modules don’t have big parts. But how do you measure how big a module is? Remember that while every vector space has a dimension—the size of a basis—most modules don’t have a basis. We can still talk about a generating set for an $R$-module $M$, though, like $\{1, 1/2, 1/3, \dots\}$ for the $\mathbb{Z}$-module $\mathbb{Q}$: a subset of $M$ for which every element can be written as an $R$-linear combination of those generating elements.

Question: For a vector space, its dimension is its minimum number of generators. Can we use the minimum number of generators as a measure of the size of a module in general?

# What is a module?

The most familiar example of a module is probably a vector space like $\mathbb{R}^n$, in which you can add and subtract vectors, or multiply a vector by a real number (a scalar). More generally, the abstract axioms for a vector space $V$ over a field $K$ look like this: Continue reading