# Notes - Rings and Modules HT24, Structure theorems

### Flashcards

Can you state the structure theorem for finitely generated modules over a Euclidean domain?

Suppose:

- $R$ is a Euclidean domain
- $M$ is a finitely ganerated $R$-module

Then

- $\exists r, s \in \mathbb N$
- $\exists c _ 1, c _ 2, \cdots, c _ r \in R$ nonzero nonunits such that $c _ 1 \vert c _ 2 \vert \cdots \vert c _ r$ and

Quickly prove the “structure theorem”, i.e. if:

- $R$ is a Euclidean domain
- $M$ is a finitely generated $R$-module

Then

- $\exists r, s \in \mathbb N$
- $\exists c _ 1, c _ 2, \cdots, c _ r \in R$ nonzero nonunits such that $c _ 1 \vert c _ 2 \vert \cdots \vert c _ r$ and

\[M \cong \left( \bigoplus^r_{i = 1} \frac{R}{c_iR} \right) \oplus R^s\]

Since $R$ is a Euclidean domain, it is in particular a PID.

We can then find a resolution (a presentation where $\psi$ is also injective) for $M$ (e.g. via picking a generating set $\{m _ 1, \cdots, m _ n\}$ then using $\phi : R^n \to M$ via $e _ i \mapsto m _ 1$, and then taking a basis for $\ker \phi$ like $\{x _ 1, \cdots, x _ m \}$ which will exist because this is a submodule of a free module, then letting $\psi : R^m \to R^n$ given by $e _ i \mapsto x _ i$ and writing $\psi$ as a matrix since it’s a homomorphism between free modules), i.e. a surjective homomorphism $\phi : R^n \to M$ and an injective homomorphism $\psi : R^m \to R^n$ such that $\ker \phi = \text{Im } \psi$ and therefore by the first isomorphism theorem $R \cong \frac{R^n}{\text{Im } \psi}$.

Consider the matrix $A$ representing the homomorphism $\psi$. By the Smith Normal Form theorem, we can transform $A$ into a diagonal matrix $D$ with diagonal entries $d _ 1 \mid d _ 2 \mid \cdots \mid d _ m$ using EROs and ECOs. These correspond to pre- and post-multiplying $A$ by invertible matrices, and hence changes of bases for $R^n$ and $R^m$. So there are bases of $R^n$ and $R^m$ such that $\psi$ has matrix $D$.

Let $\{f _ 1, \cdots, f _ n\}$ denote the basis of $R^n$. Then the $\text{Im } \psi$ is spanned by $\{d _ 1 f _ 1, \cdots, d _ m f _ m\}$ (which encodes the fact we have the relations $d _ i f _ i = 0$ for each $i$).

Define the following map, which is the one we actually use to show the isomorphism we want:

\[\theta : R^n \to \left(\bigoplus^m_{i = 1} \frac{R}{d_i R}\right) \oplus R^{n - m}\]where for any element of $M$, say $p = \sum^n _ {i = 1} a _ i f _ i \in M$ we define

\[\theta(p) = \theta\left(\sum^n_{i = 1} a_i f_i\right) := (a_1 + d_1 R, \cdots, a_m + d_m R, a_{m + 1}, \cdots, a_n)\]$\theta$ is surjective (since $\alpha _ i$ are just arbitrary members of $R$), and we see that $\ker \theta$ is the submodule generated by $\{d _ i f _ i \mid 1 \le i \le m\}$ which is equal to $\text{Im } \psi$ (since if we’re in the span of this set, then $\alpha _ i + d _ i R = 0$, and $\alpha _ {m+1}, \ldots, \alpha _ n$ are $0$).

Then:

\[\begin{aligned} M &\cong \frac{R^n}{\text{Im } \psi} \\\\ &= \frac{R^n}{\ker \theta} \\\\ &\cong \left(\bigoplus^m_{i = 1} \frac{R}{d_i R}\right) \oplus R^{n - m} \end{aligned}\]Since $\psi$ is injective, each $d _ i$ is nonzero, and if $d _ i$ is a unit, $R / d _ i R$ is $0$ and hence can be ommitted. Thus we get a reduced set of nonzero nonunits $c _ 1, \cdots, c _ r$ for some $r$ such that

\[M \cong \left(\bigoplus^r_{i = 1} \frac{R}{c_iR} \right) \oplus R^s\]The structure theorem for finitely generated modules over a Euclidean domain states that if

- $R$ is a Euclidean domain
- $M$ is a finitely generated $R$-module

Then

- $\exists r, s \in \mathbb N$
- $\exists c _ 1, c _ 2, \cdots, c _ r \in R$ nonzero nonunits such that $c _ 1 \vert c _ 2 \vert \cdots \vert c _ r$ and

\[M \cong \left( \bigoplus^r_{i = 1} \frac{R}{c_iR} \right) \oplus R^s\]
Can you state another theorem which relates this decomposition to the torsion properties of the module, and in particular what this implies about finitely generated torsion-free modules?

Suppose:

- $R$ is a Euclidean domain
- $M$ is a finitely generated $R$-module

Then:

- $\bigoplus^r _ {i = 1} \frac{R}{c _ iR} \cong M^\text{tor}$
- $R^s \cong M/M^\text{tor}$
- $s = \text{rank}(M / M^\text{tor})$

Hence if $M$ is torsion-free (and finitely generated), the decomposition is just

\[M \cong R^s\]so $M$ is in fact free (in summary, if $M$ is a finitely generated torsion-free module over a Euclidean domain, $M$ is free).

The structure theorem for finitely generated modules over a Euclidean domain states that if

- $R$ is a Euclidean domain
- $M$ is a finitely generated $R$-module

Then

- $\exists r, s \in \mathbb N$
- $\exists c _ 1, c _ 2, \cdots, c _ r \in R$ nonzero nonunits such that $c _ 1 \vert c _ 2 \vert \cdots \vert c _ r$ and

\[M \cong \left( \bigoplus^r_{i = 1} \frac{R}{c_iR} \right) \oplus R^s\]
Quickly prove the corollary that:

- $\bigoplus^r _ {i = 1} \frac{R}{c _ iR} \cong M^\text{tor}$
- $R^s \cong M/M^\text{tor}$
- $s = \text{rank}(M / M^\text{tor})$

(in other words, the torsion part of the module is isomorphic to the non-free part of the above decomposition, and the torsion-free part of the module is isomorphic to the free part of the above decomposition).

Being a bit sloppy with the notation (using equality symbols rather than isomorphism) write the following:

\[\begin{aligned} F &= R^s \quad \text{(the Free part})\\\\ N &= \bigoplus^r_{i = 1} \frac{R}{c_iR} \quad \text{(the Non-free part)} \end{aligned}\]so that

\[M = N \oplus F\]we want to show that

\[N = M^\text{tor}\]Recall

\[M^\text{tor} = \\{m \in M \mid \exists r \in R \text{ s.t. } rm = 0\\}\]First we show $N \subseteq M^\text{tor}$. If $a \in R / c _ iR$, then since $c _ i \mid c _ r$, $c _ r a = 0$. Extending to all of $N$, if

\[(a_1, \cdots, a_r) \in N, \quad a_i \in R/c_iR\]it follows

\[c_r (a_1, \cdots, a_r) = (c_r a_1, \cdots, c_ra_r) = (0, \cdots, 0)\]So $N \subseteq M^\text{tor}$. For the other direction, if we have some $(n, f) \in M^\text{tor}$ where $n \in N$ and $f \in F$ so that $\exists r \in R$ such that

\[r(f, n) = (rf, rn) = (0, 0)\]it would have to be the case that $f = 0$, since $F$ is free. Hence we see that $M^\text{tor} \subseteq N$. Then $M^\text{tor} = N$.

By the second isomorphism theorem,

\[\frac{M}{M^\text{tor} } = \frac{N \oplus F}{N} \cong \frac{F}{N \cap F} \cong \frac{F}{\\{0\\} } \cong F\](here we replace $M^\text{tor}$ with $N$ since they are equal). Hence

\[s = \text{rank}(F) = \text{rank}(M / M^\text{tor})\]The structure theorem for finitely generated modules over a Euclidean domain states that if

- $R$ is a Euclidean domain
- $M$ is a finitely generated $R$-module

Then

- $\exists r, s \in \mathbb N$
- $\exists c _ 1, c _ 2, \cdots, c _ r \in R$ nonzero nonunits such that $c _ 1 \vert c _ 2 \vert \cdots \vert c _ r$ and

\[M \cong \left( \bigoplus^r_{i = 1} \frac{R}{c_iR} \right) \oplus R^s\]
By considering Abelian groups as appropriate modules, can you state a special case of the above theorem known as the “structure theorem for finitely generated abelian groups”?

Suppose:

- $A$ is a finitely generated Abelian group

Then

- $\exists r, s \in \mathbb Z _ {\ge 0}$
- $\exists c _ 1, c _ 2, \cdots, c _ s \in \mathbb Z _ {\ge 1}$ such that $c _ 1 \mid c _ 2 \mid \cdots \mid c _ s$ and

(this follows from the fact every Abelian group can be considered as a $\mathbb Z$-module).

The structure theorem for finitely generated modules over a Euclidean domain states that if

- $R$ is a Euclidean domain
- $M$ is a finitely generated $R$-module

Then

- $\exists r, s \in \mathbb N$
- $\exists c _ 1, c _ 2, \cdots, c _ r \in R$ nonzero nonunits such that $c _ 1 \vert c _ 2 \vert \cdots \vert c _ r$ and

\[M \cong \left( \bigoplus^r_{i = 1} \frac{R}{c_iR} \right) \oplus R^s\]
By applying the Chinese Remainder Theorem, can you state the “primary decomposition” form of the above theorem?

Suppose:

- $R$ is a Euclidean domain
- $M$ is a finitely generated $R$-module

Then

- $\exists r, s \in \mathbb N$
- $\exists p _ 1, \cdots, p _ r \in R$ irreducibles and $\exists q _ i \in \mathbb N, 1 \le i \le r$ such that

**Quick proof**: We use the following generalisation of the Chinese Remainder Theorem

If $d _ 1, \cdots, d _ k \in R$ are pairwise coprime elements of a PID, then

\[\frac{R}{(d_1 d_2 \cdots d_k )R} \cong \bigoplus^k_{i=1} \frac{R}{d_i R}\](the proof of this is uses induction on $k$, and the fact that in a PID, if $d _ i$ and $d _ j$ are coprime, then Bezout’s lemma shows that $\langle d _ i \rangle + \langle d _ j \rangle = R$ and $\langle d _ i \rangle \cap \langle d _ j \rangle = \langle d _ i d _ j \rangle$). Then since each $c _ i$ in the original structure theorem is in a Euclidean domain, it can be written as a product of primes $p^{k _ 1} _ 1 \cdots p^{k _ n} _ n$. Then $R/\langle p^{k _ 1} _ 1 \cdots p^{k _ n} _ n\rangle \cong \bigoplus^k _ {i = 1} R/\langle p^{k _ i} _ i \rangle$. Applying this to each term gives the primary decomposition form.

What’s the difference between the invariant factors and elementary divisors in a finitely generated module over a PID?

- Invariant factors come from the decomposition given by the structure theorem
- Elementary divisors come from the decomposition given by the primary decomposition form of the structure theorem

How can you work out how many Abelian groups there are of order $n$ using the structure theorem?

- Any finite Abelian group $G$ of order $n$ must have a decomposition using the primary decomposition version of the structure theorem, where $G \cong \bigoplus^n _ {i = 1} \frac{\mathbb Z}{p _ i^{k _ i} \mathbb Z}$
- Find the prime factorisation of $n$
- Consider all the ways you could come up with a group of order $n$ that matches the decomposition $G \cong \bigoplus^n _ {i = 1} \frac{\mathbb Z}{p _ i^{k _ i} \mathbb Z}$
- This just corresponds to grouping primes together in different ways. E.g. for $n = 240$, the prime factors are $2 \cdot 2 \cdot 2 \cdot 2 \cdot 3 \cdot 5$. The possible groupings are:
- $(2) \cdot (2) \cdot (2) \cdot (2) \cdot (3) \cdot (5)$
- $(2 \cdot 2) \cdot (2) \cdot (2) \cdot (3) \cdot (5)$
- $(2 \cdot 2) \cdot (2 \cdot 2) \cdot (3) \cdot (5)$
- $(2) \cdot (2 \cdot 2 \cdot 2) \cdot (3) \cdot (5)$
- $(2 \cdot 2 \cdot 2 \cdot 2) \cdot (3) \cdot (5)$

The structure theorem states that if:

- $R$ is a Euclidean domain
- $M$ is a finitely ganerated $R$-module

Then:

- $\exists r, s \in \mathbb N$
- $\exists c _ 1, c _ 2, \cdots, c _ r \in R$ nonzero nonunits such that $c _ 1 \vert c _ 2 \vert \cdots \vert c _ r$ and

\[M \cong \left( \bigoplus^r_{i = 1} \frac{R}{c_iR} \right) \oplus R^s\]
In the proof, we consider a presentation of $M$, which is given by a pair of maps

\[\begin{aligned}
&\phi : R^n \to M \\\\
&\psi : R^m \to R^n
\end{aligned}\]
where $\ker \phi = \text{Im } \psi$. Using Smith Normal Form, we can then find a particularly simple form for $\psi$ which then induces the isomorphism above. However, it’s necessary that $\psi$ is actually injective and hence the presentation is actually a “resolution”. Why is this an important assumption?

It ensures that the diagonal entries in the Smith Normal Form of $\psi$ are nonzero.