Notes - Rings and Modules HT24, Presentations
Flashcards
Suppose $M$ is a finitely-generated $R$-module. What is a presentation of $M$?
Two homomorphisms $\phi : R^n \to M$ and $\psi : R^m \to R^n$ such that
\[\text{Im } \phi = M\](i.e. it’s surjective) and
\[\text{Im } \psi = \ker \phi\]Suppose:
- $M$ is a nonzero finitely generated $R$-module
- $R$ is a PID
Quickly prove that there is then a presentation (actually a resolution) for $M$, i.e.
- $\exists n \in \mathbb N$ such that there is a surjective homomorphism $\phi : R^n \to M$
- There exists a free module $R^m$ with $m \le n$ and an injective homomorphism $\psi : R^m \to R^n$ such that $\text{Im } \psi = \ker \phi$. (The fact that $\psi$ is injective means it is it a resolution)
And in particular, these imply that
\[M \cong R^n/(\text{Im } \psi)\]
- For the first part, consider the map $\phi : R^n \to M, e _ i \mapsto m _ i$ where $e _ i$ is a standard basis for $R^n$ and $m _ i$ is a collection of generators for $M$. Then this is a surjective homomorphism.
- For the second part, since $R$ is a PID, the submoudle $\ker \phi$ is a free submodule of rank $n \le m$ (this is the general result that the submodules of a finitely generated free module over a PID have rank at most the rank of the big module). Hence we can pick a basis $\{x _ 1, \cdots, x _ m\}$ and define $\psi : R^m \to R^n$ by sending $x _ i \mapsto e _ i$. This is an injective homomorphism.
Suppose $A$ is an $n \times m$ presentation matrix for a module $M$, corresponding to the pair of maps
\[\begin{aligned}
&\phi : R^n \to M \\\\
&\psi : R^m \to R^n
\end{aligned}\]
where $\ker \phi = \text{Im } \psi$. Describe four ways that you can modify the presentation matrix so that it still presents the same module.
- Can pre- or post- multiply by an invertible matrix, since this represents a change of basis in either $R^n$ or $R^m$ (in particular, this implies you can multiply rows by a unit).
- Can remove a row of zeroes, since this corresponds to the trivial relation $0 = 0$, which is not neccessary.
- If a row looks like some basis vector like $e _ i$, we can remove both the column and row $i$. This represents the relation $m _ i = 0$ for some member of the generating set $m _ i$, and so any relation in which is appears can be shortened.
Suppose $A$ is an $n \times m$ presentation matrix for a module $M$, corresponding to the pair of maps
\[\begin{aligned}
&\phi : R^n \to M && \text{surjective hom.}\\\\
&\psi : R^m \to R^n && \text{injective hom.}
\end{aligned}\]
where $\ker \phi = \text{Im } \psi$. If $\{e _ 1, \ldots, e _ n\}$ is a basis for $R^n$ and $\phi$ is defined so that $e _ i \mapsto m _ i$ for some generating set $\{m _ 1, \cdots, m _ n\}$, what does $\ker \phi$ concretely represent?
The “module of relations”, all the ways that the generators $m _ i$ can be combined to get $0$.
Suppose $A$ is an $n \times m$ presentation matrix for a module $M$, corresponding to the pair of maps
\[\begin{aligned}
&\phi : R^n \to M && \text{surjective hom.}\\\\
&\psi : R^m \to R^n && \text{injective hom.}
\end{aligned}\]
where $\ker \phi = \text{Im } \psi$. Let $\{e _ 1, \ldots, e _ n\}$ is a basis for $R^n$ and $\phi$ is defined so that $e _ i \mapsto m _ i$ for some generating set $\{m _ 1, \cdots, m _ n\}$. If the presentation matrix has a column $(\alpha _ 1, \alpha _ 2, \alpha _ 3)^\top$, what does this concretely represent?
The fact that
\[\alpha_1 m_1 + \alpha_2 m_2 + \alpha_3 m_3 = 0\]Suppose we have a $\mathbb F[t]$-module $M$ where the action of $f(x) \in \mathbb F[t]$ corresponds to $f(A) v$. In order to do things like find the JNF or CNF, we need to find a presentation of $M$ in the form of a presentation matrix. How can this be done?
The module is generated by the basis vectors $e _ 1, \cdots, e _ n$. For each vector we have the relation that
\[xe_i - Ae_i = 0\]which can be collected into the presentation matrix
\[xI - A\](For example…)
\[A = \begin{pmatrix}1 & -1 & 1 \\\\ 0 & 0 & 1 \\\\ 0 & 1 & 0\end{pmatrix}\]has presentation matrix
\[xI - A = \begin{pmatrix}x-1 & 1 & -1 \\\\ 0 & x & -1 \\\\ 0 & -1 & x\end{pmatrix}\]since, e.g. for the second column,
\[e_1 + x e_2 - e_3 = e_1 + Ae_2 - e_3 = e_1 - e_1 + e_3 - e_3 = 0\]