# Notes - Rings and Modules HT24, Free modules

### Flashcards

Suppose:

- $R$ is an integral domain
- $M$ is a finitely generated free $R$-module

What is meant by the rank of $M$?

The size of any basis for $M$.

Suppose:

- $R$ is an integral domain
- $M$ is a finitely generated free $R$-module

Quickly prove that the size of a basis for $M$ is uniquely determined and is known as the rank $\text{rank}(M)$ of $M$.

Overall idea: Fix some basis $X$, and consider a maximal ideal $I$. Then $R/I$ is a field. Can show that $IM = M _ I$ where $M _ I$ is the set of all $I$-linear combinations of basis elements $X$. Then show that $M/IM$ can be viewed as an $R/I$-module, which actually shows that $M/IM$ is a vector space. Then show $q(X)$ (i.e. quotient map from $R$ to $R/I$ applied to $X$, this is where it is relevant that we showed $IM = M _ I$) is a basis for $M/IM$ and is therefore uniquely determined, and finally since $ \vert X \vert = \vert q(X) \vert $ it follows the size of the basis is fixed.

Suppose that $X = \{x _ 1, \cdots, x _ n\}$ is a basis for $M$ and $I$ is a maximal ideal in $R$ (we can always find a maximal ideal of a ring, assuming the axiom of choice). Since $I$ is a maximal ideal, we know that $\mathbb F := R/I$ is a field. We aim to show that $X$ can be turned into the basis for an $\mathbb F$-vector space $\overline X$, and hence has uniquely determined dimension.

Let

\[\begin{aligned} IM &= \langle \\{im \mid i \in I, m \in M\\} \rangle \\\\ M_I &= \\{ \sum^n_{k = 1} i_k x_k \mid i_k \in I \\} \end{aligned}\](the first equality isn’t a definition, just standard notation for multiplication of ideals).

We can show $IM = M _ I$ by double inclusion:

\[\begin{aligned} x \in IM &\iff x = \sum_k i_k m_k \\\\ &\iff x = \sum_k i_k \left(\sum_\ell \alpha_\ell x_\ell\right) &&\alpha_\ell \in I \\\\ &\iff x = \sum_k \beta_k x_k &&\beta_k \in I \\\\ &\iff x \in M_I \end{aligned}\]Now we want to show that the quotient module $M/IM$ is not just an $R$-module, but also an $\mathbb F$-module (and so $\mathbb F$-vector space). This is because if we let $q : M \to M / IM, m \mapsto m + IM$ be the corresponding quotient map, we have the action

\[(r + I) \cdot q(m) := q(rm)\](this is well defined, since if $r + I = r’ + I$, $r - r’ \in I$, so $rm - r’m = (r - r’m) \in IM$ and so $q(r’m) = q(rm)$ as claimed, and this gives $M/IM$ the structure of an $\mathbb F$-module since we have specified a field homomorphism $\phi : \mathbb F \to \text{Hom}(M)$). Hence $M / IM$ can be viewed as a $\mathbb F$-vector space.

Let $\overline X := q(X)$. We want to show two more things:

- $\overline X$ is a basis of $M/IM$; this will imply that $ \vert \overline X \vert $ is independent of $X$ (since the cardinality of bases in vector spaces are uniquely determined), since $ \vert \overline X \vert = \dim _ \mathbb F(M/IM)$.
- $ \vert \overline X \vert = \vert X \vert $, since this will then imply that the cardinality of $X$ is uniquely determined.

To prove (1), note $q(X)$ spans $M/IM$ since $q$ is surjective. Then suppose we have some $c _ k \in \mathbb F = M/IM$ such that

\[\sum^n_{k = 1} c_k q(x_k) = 0 + IM\in M/IM\]Picking representatives $r _ k \in R$ for each $c _ k$, we see

\[\begin{aligned} 0 &= \sum^n_{k = 1} c_k q(x_k) \\\\ &= \sum^n_{k = 1} (r_k + IM) q(x_k) \\\\ &= \sum^n_{k = 1} q(r_k x_k) \\\\ &= q\left(\sum^n_{k = 1} r_k x_k\right) \end{aligned}\]But then

\[\sum^n_{k = 1} r_k x_k \in \ker q \implies \sum^n_{k = 1} r_k x_k \in IM\]Since $IM = M _ I$, this means that $r _ k \in I$ for each $k$, so $c _ k = r _ k + IM = 0$. Hence all coefficients $c _ k$ are $0$, so $\overline X$ is linearly independent and so an $\mathbb F$-basis of $M/IM$ as required.

To prove (2), assume that $\exists i, j$ where $i \ne j$ and $q(x _ i) = q(x _ j)$. Then $q(x _ i - x _ j)$ = 0, so $x _ i - x _ j \in IM$ since $\ker q = IM$. Then this implies

\[x_i - x_j = \sum_k i_k x_k\]where not all $i _ k$ are $0$ (since by assumption $x _ i, x _ j$ distinct) and none of them are $1$, since if $i _ k = 1$, $1 \in I$, which would imply that $I$ were not a proper ideal (which is, since it is the maximal ideal). But then rearranging we get

\[\sum_k \hat i_k x_k\]for some $\hat i _ k$ not all $0$, so we have a contradiction on $X$ being a basis. Hence $ \vert X \vert = \vert \overline X \vert $. Then we are done since

\[|X| = |\overline X| = \dim_\mathbb F(M/IM)\]and we have shown (by arguing $M _ I = IM$) that $IM$ is independent of $X$.

Suppose:

- $R$ is a PID (
**this condition is important, it doesn’t hold for general rings**)
- $M$ is a finitely generated free $R$-module of rank $n$

Then what can you deduce about the submodules of $M$?

**this condition is important, it doesn’t hold for general rings**)They are also finitely generated, free and have a rank of at most $n$.

Suppose:

- $R$ is a PID
- $M$ is a finitely generated free $R$-module of rank $n$
- $N$ is a submodule of $M$

Then quickly prove that:

- $N$ is also free, and has a rank of at most $n$.

Let $X = \{e _ 1, \cdots, e _ n\}$ be a basis of size $n$ for $M$. We proceed by induction on $n = \vert X \vert $.

If $n = 1$, then $\phi : R \to M$ given by $r \mapsto re _ 1$ shows that $M \cong R$. Hence it suffices to consider the ideals of $R$, since the submodules of a ring over itself are the ideals.

Since $R$ is a PID by assumption, any ideal $N \trianglelefteq R$ is of the form $N = aR$ for some $a \in R$. Then $\{a\}$ is a basis of $N$ unless $a = 0$, so $N$ is free of rank $0$ or rank $1$ as required.

Now suppose $n > 1$. Consider

\[\begin{aligned} W &:= Re_1 + \cdots + Re_{n-1} \\\\ &= \langle e_1, \cdots, e_{n-1} \rangle \end{aligned}\]and let

\[N_1 := N \cap W\]Then $N _ 1$ is a submodule of the free module $W$. Since $\text{rank}(W) = n - 1$, by the inductive hypothesis $N _ 1$ is free of rank $k \le n - 1$. Hence we can assume $N _ 1$ has a basis $\{v _ 1, \cdots, v _ k\}$.

The second isomorphism theorem shows that

\[\frac{N}{N_1} = \frac{N}{N\cap W} \cong \frac{N + W}{W} \subseteq \frac{M}{W}\]But $M / W$ has basis $\{e _ n + W\}$, so we see that $N / N _ 1$ is either $0$ or free of rank $1$.

In the former case, $N = N _ 1$ and we are done. In the latter we may pick some $v _ {k + 1}$ so that $v _ {k + 1} + N _ 1$ is a basis of $N / N _ 1$.

We claim $\{v _ 1, \cdots, v _ {k+1}\}$ is a basis of $N$.

To see that it spans, suppose $m \in N$, then since $\{v _ {k+1} + N _ 1\}$ is a basis of $N / N _ 1$, there exists $r _ {k + 1} \in R$ such that

\[m + N_1 = r_{k+1}v_{k+1} + N_1\]But then $m - r _ {k+1} v _ {k + 1} \in N _ 1$, hence $\exists r _ 1, \cdots, r _ k \in R$ such that

\[m - r_{k+1} v_{k+1} = \sum^k_{i = 1} r_k v_k\]Hence

\[m = \sum^{k+1}_{i = 1}r_i v_i\]For linear independence, suppose that

\[\sum^{k+1} _ {i = 1} s _ i v _ i = 0\]for some $s _ i \in R$. Then

\[\begin{aligned} 0 + N _ 1 &= \left(\sum^{k+1} _ {i = 1} s _ i v _ i\right) + N_1 \\\\ &= s _ {k+1}v _ {k+1} + N_1 \end{aligned}\]Then

\[s_{k+1} = 0\]since $\{v _ {k + 1} + N _ 1\}$ is a basis of $N / N _ 1$. But then

\[\sum^{k+1} _ {i = 1} s _ i v _ i = 0 \implies \sum^k _ {i = 1} s _ k v _ k = 0\]and hence $s _ i = 0$ for all $i$, since $\{v _ 1, \cdots, v _ k\}$ is a basis of $N _ 1$.

Then $\{v _ 1, \cdots, v _ {k+1}\}$ is a basis of $N$, and we are done, since

\[k + 1 \le (n - 1) + 1 = n\]