Ranks Of Modules In Short Exact Sequences: A Deep Dive

Hey guys! Today, we're diving deep into the fascinating world of commutative algebra, specifically focusing on the ranks of modules within short exact sequences. This is a topic that blends together concepts from module theory, homological algebra, and exact sequences. So, buckle up, and let's get started!

What's the Big Question?

Our main focus today is dissecting the following question:

Suppose we have a short exact sequence of finitely generated free $R$-modules, which looks like this:

0 \to M_1 \to M \to M_2 \to 0

Here, $R$ represents a commutative ring. The core question we're tackling is: Is it always true that the rank of $M$ is equal to the sum of the ranks of $M_1$ and $M_2$? In other words, does rank(M) = rank(M_1) + rank(M_2) hold true? This might seem straightforward, but like many things in abstract algebra, there are nuances and conditions we need to consider.

Let's break this down piece by piece to really understand what's going on and explore why this question is so important.

Delving into the Basics: Short Exact Sequences

First off, what exactly is a short exact sequence? It’s a sequence of modules and homomorphisms (module homomorphisms, to be precise) that satisfies a particular condition regarding their kernels and images. Remember, a module homomorphism is just a structure-preserving map between modules.

A sequence like:

0 \to A \xrightarrow{f} B \xrightarrow{g} C \to 0

is a short exact sequence if it meets these key criteria:

1. The map $f$ is injective. This means that the kernel of $f$ (elements that map to zero) contains only the zero element of $A$. In simpler terms, $f$ is a one-to-one mapping.
2. The map $g$ is surjective. This means that the image of $g$ (the set of elements in $C$ that are mapped to by elements in $B$) is the entire module $C$. So, $g$ covers the whole of $C$.
3. Most importantly, the image of $f$ is equal to the kernel of $g$. This is the crux of exactness. It connects the maps $f$ and $g$ in a very specific way. Mathematically, Im(f) = Ker(g).

In our initial question, the sequence 0 -> M_1 -> M -> M_2 -> 0 being a short exact sequence tells us these things:

• The map from 0 to $M_1$ is trivially injective (it can only map 0 to 0).
• The map from $M_1$ to $M$ is injective.
• The map from $M$ to $M_2$ is surjective.
• And most crucially, the image of the map from $M_1$ to $M$ is precisely the kernel of the map from $M$ to $M_2$.

This interwoven structure is what gives short exact sequences their power and makes them so useful in algebraic arguments.

Free Modules: A Quick Refresher

Now, let's talk about free modules. A free module over a ring $R$ is a module that has a basis. A basis, in this context, is a set of elements that are linearly independent and span the entire module. Think of it like a vector space, but over a ring instead of a field. The number of elements in the basis is the rank of the free module.

For example, consider the ring of integers $\mathbb{Z}$. The module $\mathbb{Z}^n$, which consists of n-tuples of integers, is a free module of rank $n$. The standard basis vectors (like (1,0,0,...,0), (0,1,0,...,0), and so on) form a basis for this module.

In our original question, we're told that $M_1$, $M$, and $M_2$ are finitely generated free $R$-modules. This is a crucial piece of information. It means each of these modules has a finite basis, and we can talk about their ranks in a meaningful way.

The Key Theorem: Rank Additivity in Short Exact Sequences

Okay, so let's circle back to our main question: Does rank(M) = rank(M_1) + rank(M_2)? The answer, spoiler alert, isn't always a straightforward yes. It depends on the ring $R$ we're working with.

Here's the key theorem:

If

0 \to M_1 \xrightarrow{f} M \xrightarrow{g} M_2 \to 0

is a short exact sequence of finitely generated free $R$-modules, then rank(M) = rank(M_1) + rank(M_2) holds true if $R$ is a commutative ring with the invariant basis number (IBN) property.

Whoa, that's a mouthful! Let's break it down. What's this