Homework 7

Problem 1

Let $R$ be an integral domain and $M$ be a $R$ -module.

Suppose that $M$ has rank $n$ and $S = {m_{1}, \dots, m_{n}}$ is a maximal linearly independent set of elements in $M$ . Let $N$ be the submodule generated by $S$ . Prove that $N ≃ R^{n}$ and $M / N$ is a torsion $R$ -module.
Conversely, prove that if $M$ contains a submodule $N$ that is free of rank $n$ such that the quotient $M / N$ is a torsion $R$ -module, then $M$ has rank $n$ .

Hints

Show that for any $m \in M$ there is a nonzero $r \in R$ such that $r m = r_{1} m_{1} + \dots + r_{n} m_{n}$ for some $r_{i} \in R$ .
Let ${m_{1}, \dots, m_{n + 1}}$ be a set of $n + 1$ elements of $M$ . Find some nonzero $r_{i} \in R$ so that $r_{i} m_{i}$ can be written using a basis for $N$ . Then show the $r_{i} m_{i}$ (and hence $m_{i}$ ) are linearly dependent.

Solution

Let $A = {x_{1}, \dots, x_{k}}$ be a fixed set with $n$ elements and define an $R$ -module morphism $f : F (A) \to M$ by $x_{i} \mapsto m_{i}$ . The image of $f$ is exactly the submodule $N$ and the kernel of $f$ is the set of all formal sums $r_{1} x_{1} + \dots + r_{k} x_{k}$ such that $r_{1} m_{1} + \dots r_{k} m_{k} = 0_{M}$ . Since the set ${m_{1}, \dots, m_{k}}$ is $R$ -linearly independent, it follows that $\ker (f) = (0)$ and so by the First Isomorphism Theorem for modules $F (A) ≃ F (A) / \ker (f) ≃ im (f) = N$ . This proves $N ≃ F (A) ≃ R^{k}$ .

We next prove $M / N$ is torsion. Take any coset $m + N$ in $M / N$ . The set $T = {m_{1}, \dots, m_{k}, m}$ strictly contains the set $S$ , the latter of which is a maximal $R$ -linearly independent subset of $M$ , so the set $T$ must be $R$ -linearly dependent. This implies there are some $r_{i} \in R$ (not all zero) such that

$r_{1} m_{1} + \dots + r_{k} m_{k} + r_{k + 1} m = 0_{M} .$

We cannot have $r_{k + 1} = 0_{R}$ , since the above relation would then yield a nontrivial relation on the set $S$ . So we must have $r_{k + 1} \neq 0_{R}$ . But now notice that $- r_{k + 1} m = r_{1} m_{1} + \dots + r_{k} m_{k} \in N$ , so

$- r_{k + 1} (m + N) = - r_{k + 1} m + N = N = 0_{M / N} .$

Thus, the element $m + N$ is torsion in $M / N$ .
First notice that since $M$ contains a submodule $N$ of rank $k$ , the rank of $M$ must be at least $k$ . (Any independent set of elements in $N$ is also an independent set of elements in $M$ .) So we only need to prove $M$ does not have an independent subsets consisting of more than $k$ elements. To that end, suppose ${m_{1}, \dots, m_{k + 1}}$ is any set of $k + 1$ elements of $M$ . Since $M / N$ is torsion, for each $i$ the coset $m_{i} + N$ is torsion in $M / N$ . This implies there is some nonzero $r_{i} \in R$ with $r_{i} (m_{i} + N) = 0_{M / N}$ ; i.e., $r_{i} m_{i} \in N$ . But then we have a set ${r_{1} m_{1}, \dots, r_{k + 1} m_{k + 1}}$ of $k + 1$ elements in the module $N$ (which has rank $k$ ), so this set must be $R$ -linearly dependent. This implies there are $r_{1}^{'}, \dots, r_{k + 1}^{'} \in R$ (not all zero) such that

$r_{1}^{'} (r_{1} m_{1}) + \dots + r_{k + 1}^{'} (r_{k + 1} m_{k + 1}) = 0_{N} .$

Since $0_{N} = 0_{M}$ , the above equality gives the relation

$(r_{1}^{'} r_{1}) m_{1} + \dots + (r_{k + 1}^{'} r_{k + 1}) m_{k + 1} = 0_{M} .$

Every $r_{i}$ is nonzero and at least one $r_{j}^{'}$ is nonzero, so (since $R$ is a domain) the elements $r_{i}^{'} r_{i}$ are not all zero. Thus, the above relation implies the set ${m_{1}, \dots, m_{k + 1}}$ is $R$ -linearly dependent. We've proven that $M$ does not contain any linearly independent sets with more than $k$ elements, and hence the rank of $M$ is no more than $k$ . By our earlier remarks, this proves the rank of $M$ is exactly $k$ .

Problem 2

Let $R$ be an integral domain. Prove that if $A$ and $B$ are $R$ -modules of ranks $m$ and $n$ , respectively, then $A \oplus B$ is an $R$ -module of rank $m + n$ .

Hints

Use an alternate characterization of rank.

Solution

Let $i_{A} : A \to A \oplus B$ and $i_{B} : B \to A \oplus B$ be the natural injective $R$ -module morphisms (that are part of the construction of the direct sum $A \oplus B$ ). At the level of elements, these maps are $a \mapsto a \oplus 0_{B}$ and $b \mapsto 0_{A} \oplus b$ , respectively.

Let ${a_{1}, \dots, a_{m}}$ and ${b_{1}, \dots, b_{n}}$ be $R$ -linearly independent subsets of $A$ and $B$ , respectively. We first prove that the set $S = {i_{A} (a_{1}), \dots, i_{A} (a_{m}), i_{B} (b_{1}), \dots, i_{B} (b_{n})}$ is an $R$ -linearly independent subset in $A \oplus B$ . To see this, suppose we have $r_{i}, r_{j}^{'} \in R$ such that

$r_{1} i_{A} (a_{1}) + \dots + r_{m} i_{A} (a_{m}) + r_{1}^{'} i_{B} (b_{1}) + \dots r_{n}^{'} i_{B} (b_{n}) = 0_{A \oplus B} .$

The above equation is equivalent to the equality

$i_{A} (r_{1} a_{1} + \dots + r_{m} a_{m}) + i_{B} (r_{1}^{'} b_{1} + \dots + r_{n}^{'} b_{n}) = 0_{A \oplus B} .$

By the definition of $i_{A}$ and $i_{B}$ , the above is equivalent to the equality of formal sums

$(r_{1} a_{1} + \dots + r_{m} a_{m}) \oplus (r_{1}^{'} b_{1} + \dots + r_{n}^{'} b_{n}) = 0_{A} \oplus 0_{B} .$

This is last equality is equivalent to the pair of equalities

$r_{1} a_{1} + \dots + r_{m} a_{m} = 0_{A} and r_{1}^{'} b_{1} + \dots + r_{n}^{'} b_{n} = 0_{B} .$

Since the sets ${a_{1}, \dots, a_{m}}$ and ${b_{1}, \dots, b_{n}}$ are $R$ -linearly independent, the above equalities imply all $r_{i} = 0_{R}$ and $r_{j}^{'} = 0_{R}$ . Thus, the set $S$ is indeed $R$ -linearly independent.

Let $N$ be the submodule of $A \oplus B$ generated by the set $S$ . Then $N$ is a free module of rank $m + n$ . We will now show that the quotient $(A \oplus B) / N$ is torsion, at which point it will follow that the rank of $A \oplus B$ is exactly $m + n$ . Take any element in $(A \oplus B) / N$ , say a coset $(a \oplus b) + N$ . The sets ${a_{1}, \dots, a_{m}, a}$ and ${b_{1}, \dots, b_{n}, b}$ must both be $R$ -linearly dependent (since $A$ and $B$ have ranks $m$ and $n$ , respectively). This implies there are $r_{i}, r_{j}^{'} \in R$ such that

$r_{1} a_{1} + \dots + r_{m} a_{m} + r_{m + 1} a = 0_{A} and r_{1}^{'} b_{1} + \dots + r_{n}^{'} b_{n} + r_{n + 1}^{'} b = 0_{B}$

Since the sets ${a_{1}, \dots, a_{m}}$ and ${b_{1}, \dots, b_{n}}$ are $R$ -linearly independent, the elements $r_{m + 1}$ and $r_{n + 1}^{'}$ must both be nonzero. Let $r = - r_{m + 1} r_{n + 1}^{'}$ , which is nonzero since $R$ is an integral domain. The above equalities then imply

$r a = (r_{1} r_{n + 1}^{'}) a_{1} + \dots + (r_{m} r_{n + 1}^{'}) a_{m} and r b = (r_{1}^{'} r_{m + 1}) b_{1} + \dots + (r_{m}^{'} r_{m + 1}) b_{m},$

and so $r (a \oplus b) = r a \oplus r b \in N$ . This, in turn, implies that the $r ((a \oplus b) + N) = N = 0_{(A \oplus B) / N}$ , and hence the coset $(a \oplus b) + N$ is torsion. We've therefore proven the module $(A \oplus B) / N$ is torsion.

By our work above, we can now conclude that the rank of $A \oplus B$ is exactly $m + n$ .

Problem 3

Let $R$ be an integral domain, $M$ an $R$ -module, and $N$ a submodule of $M$ . Prove that the rank of $M$ is the sum of the ranks of $N$ and $M / N$ .

(You may assume $M$ has finite rank.)

Hints

For part (2), let ${m_{1} + N, \dots, m_{s} + N}$ be a maximal $R$ -linearly independent set in $M / N$ and ${n_{1}, \dots, n_{l}}$ be a maximal $R$ -linearly independent set in $N$ . Show that the set ${m_{1}, \dots, m_{s}, n_{1}, \dots, n_{l}}$ is an $R$ -linearly independent set in $M$ . Then use an alternate characterization of rank.

Solution

Let $k$ be the (finite) rank of the module $M$ . Let $N$ be any submodule of $M$ , and let $l$ be the rank of $N$ . We immediately have that $l \leq k$ . We claim that $M / N$ also has finite rank $s$ and that $s \leq k$ .

To see this, note that if ${m_{1} + N, \dots, m_{s} + N}$ is any $R$ -linearly independent set in $M / N$ , then the set ${m_{1}, \dots, m_{s}}$ is also an $R$ -linearly independent set of elements in $M$ . Indeed, any relation $r_{1} m_{1} + \dots + r_{s} m_{s} = 0_{M}$ immediately yields a relation $r_{1} (m_{1} + N) + \dots + r_{s} (m_{s} + N) = 0_{M} + N = 0_{M / N}$ . It follows that the rank of $M / N$ is no more than the rank $k$ of $M$ . In particular, the rank of $M / N$ is finite.

Let $s$ be the rank of $M / N$ and let ${m_{1} + N, \dots, m_{s} + N}$ be an $R$ -linearly independent set. By the previous comment, the set ${m_{1}, \dots, m_{s}}$ is then $R$ -linearly independent in $M$ . Let ${n_{1}, \dots, n_{l}}$ be an $R$ -linearly independent set in $N$ . We claim the combined set $S = {m_{1}, \dots, m_{s}, n_{1}, \dots, n_{l}}$ is an $R$ -linearly independent set in $M$ . To see this, suppose we have a relation

$r_{1} m_{1} + \dots + r_{s} m_{s} + r_{1}^{'} n_{1} + \dots + r_{l}^{'} n_{l} = 0_{M} .$

In the quotient module $M / N$ we have

$\begin{aligned} (r_{1} m_{1} + \dots + r_{s} m_{s} + r_{1}^{'} n_{1} + \dots + r_{l}^{'} n_{l}) + N & = (r_{1} m_{1} + \dots + r_{s} m_{s}) + N \\ = r_{1} (m_{1} + N) + \dots + r_{s} (m_{s} + N) . \end{aligned}$

Our earlier relation then gives the relation

$r_{1} (m_{1} + N) + \dots + r_{s} (m_{s} + N) = 0_{M / N} .$

Since the set ${m_{1} + N, \dots, m_{s} + N}$ is $R$ -linearly independent in $M / N$ this implies all $r_{i} = 0_{R}$ . Our earlier relation then simplifies to

$r_{1}^{'} n_{1} + \dots + r_{l}^{'} n_{l} = 0_{M} = 0_{N} .$

Since the set ${n_{1}, \dots, n_{l}}$ is $R$ -linearly independent in $N$ this implies all $r_{j}^{'} = 0_{R}$ . Thus our set $S = {m_{1}, \dots, m_{s}, n_{1}, \dots, n_{l}}$ is $R$ -linearly independent. This proves that the rank $k$ of $M$ is at least $s + l$ .

To prove the rank $k$ of $M$ is exactly $s + l$ , we need only show that $M / P$ is torsion, where $P$ is the free submodule of $M$ generated by our set $S$ . Take any element $m + P$ in the quotient module $M / P$ . The set ${m_{1} + N, \dots, m_{s} + N, m + N}$ must be $R$ -linearly dependent in $M / N$ , so there are some $r_{i} \in R$ (not all zero) such that

$r_{1} (m_{1} + N) + \dots + r_{s} (m_{s} + N) + r_{s + 1} (m + N) = 0_{M / N} .$

Note that $r_{s + 1}$ must be nonzero since the set ${m_{1} + N, \dots, m_{s} + N}$ is $R$ -linearly independent. The above equality implies

$r_{1} m_{1} + \dots + r_{s} m_{s} + r_{s + 1} m \in N .$

If we add this element to the set ${n_{1}, \dots, n_{l}}$ then we must obtain an $R$ -linearly dependent set, which means there are $r_{j}^{'} \in R$ (not all zero) such that

$r_{1}^{'} n_{1} + \dots + r_{l}^{'} n_{l} + r_{l + 1}^{'} (r_{1} m_{1} + \dots + r_{s} m_{s} + r_{s + 1} m) = 0_{N} .$

Once again, the coefficient $r_{l + 1}^{'}$ must be nonzero since the set ${n_{1}, \dots, n_{l}}$ is $R$ -linearly independent. We can then use the above equation to obtain the equality

$(- r_{l + 1}^{'} r_{s + 1}) m = (r_{1} r_{l + 1}^{'}) m_{1} + \dots (r_{s} r_{l + 1}^{'}) m_{s} + r_{1}^{'} n_{1} + \dots + r_{l}^{'} n_{l} .$

If we let $r = - r_{l + 1}^{'} r_{s + 1}$ , then $r$ is nonzero (since $R$ is an integral domain) and the above equality implies $r m \in P$ . Thus, the coset $r (m + P) = r m + P = P = 0_{M / P}$ and hence the coset $m + P$ is torsion. This proves that the quotient module $M / P$ is torsion and hence, by our earlier discussion, the rank of $M$ is exactly $s + l$ .

Problem 4

Suppose $R$ is an integral domain and $M$ is a $R$ -module.

Show that if $m$ is a nonzero torsion element in $M$ , then the set ${m}$ is $R$ -linearly dependent. Conclude that the rank of $Tor (M)$ is 0.
Show that the rank of $M$ is the same as the rank of the quotient $M / Tor (M)$ .

Solution

Suppose $m \in M$ is a nonzero torsion element, so $r m = 0_{M}$ for some nonzero $r \in R$ . The equation $r m = 0_{M}$ implies (by definition!) that the set ${m}$ is $R$ -linearly dependent. Also note that $1_{R} \cdot 0_{M} = 0_{M}$ , so even the set ${0_{M}}$ is $R$ -linearly dependent.

Now suppose $S$ is any nonempty of $Tor (M)$ . If $S = {0_{M}}$ , then $S$ is $R$ -linearly dependent. If $S$ contains any nonzero element $m$ , then $m$ is a nonzero torsion element so by the above argument the set ${m}$ is $R$ -linearly dependent and hence so is $S$ .

It follows that the only $R$ -linearly independent subset of $Tor (M)$ is the empty set, and hence the rank of $Tor (M)$ is 0.
For a quick solution, we can use the relation between ranks and quotients. In this case, the submodule $Tor (M) \subseteq M$ is a submodule of rank 0 (by above), so rank of $M$ is equal to the rank of $M / Tor (M)$ .

We can also give a standalone proof. First suppose ${m_{1}, \dots, m_{k}}$ is any $R$ -linearly independent subset of $M$ . We claim that the image of this set is an $R$ -linearly independent subset of $M / Tor (M)$ . To see this, suppose we have a relation

$r_{1} (m_{1} + Tor (M)) + \dots + r_{k} (m_{k} + Tor (M)) = 0.$

By the definition of coset operations, the coset on the left is represented by $r_{1} m_{1} + \dots + r_{k} m_{k}$ , and this element represents the zero coset in $M / Tor (M)$ exactly when $r_{1} m_{1} + \dots + r_{k} m_{k} \in Tor (M)$ . This implies there is some nonzero $r \in R$ such that

$r (r_{1} m_{1} + \dots + r_{k} m_{k}) = 0_{M},$

which in turn implies

$(r r_{1}) m_{1} + \dots + (r r_{k}) m_{k} = 0_{M} .$

Since the set ${m_{1}, \dots, m_{k}}$ is $R$ -linearly independent, the above equality implies $r r_{i} = 0_{R}$ for every $i$ . Since $r$ is nonzero and $R$ is an integral domain, this implies $r_{i} = 0$ for every $i$ . We've therefore proven the set ${m_{1} + Tor (M), \dots, m_{k} + Tor (M)}$ is an $R$ -linearly independent subset of $M / Tor (M)$ . This proves that the rank of $M / Tor (M)$ is at least as large as the rank of $M$ .

Conversely, suppose ${m_{1} + Tor (M), \dots, m_{k} + Tor (M)}$ is an $R$ -linearly independent set in $M / Tor (M)$ . Suppose there is an $R$ -linear relation among the elements $m_{1}, \dots, m_{k}$ in $M$ , i.e., there are $r_{i} \in R$ such that

$r_{1} m_{1} + \dots + r_{k} m_{k} = 0_{M} .$

Since $0_{M} \in Tor (M)$ , we then have the coset equality

$(r_{1} m_{1} + \dots + r_{k} m_{k}) + Tor (M) = 0_{M / Tor (M)};$

i.e.,

$r_{1} (m_{1} + Tor (M)) + \dots + r_{k} (m_{k} + Tor (M)) = 0$

in $M / Tor (M)$ . Since the set ${m_{1} + Tor (M), \dots, m_{k} + Tor (M)}$ is $R$ -linearly independent, the above equality implies every $r_{i} = 0_{R}$ . Thus, the set ${m_{1}, \dots, m_{k}}$ is also $R$ -linearly independent in $M$ . This proves that the rank of $M$ is at least as large as the rank of $M / Tor (M)$ .

We've therefore proven $M$ and $M / Tor (M)$ have the same rank.

Problem 5

Let $R$ be an integral domain and $I \subseteq R$ be a non-principal ideal. Prove that $I$ is torsion free of rank 1, but $I$ is not a free $R$ -module.

Hints

To show that the rank of $I$ is no more than 1, observe that for any pair of nonzero elements $a, b \in I$ one has $b \cdot a + (- a) \cdot b = 0$ and so the set ${a, b}$ is $R$ -linearly dependent.

Solution

First note that $I$ is contained in the integral domain $R$ , so it is automatically torsion free. (If $r i = 0$ for some $i \in I$ and nonzero $r \in R$ , then the fact that $R$ is an integral domain implies $i = 0$ .) Also note that $I$ is nonzero, since the zero ideal is principal.

Next observe that for any pair of nonzero elements $a, b \in I$ one has $b \cdot a + (- a) \cdot b = 0$ and so the set ${a, b}$ is $R$ -linearly dependent. This implies that the rank of $I$ must be less than 2. On the other hand, for any nonzero element $a \in A$ the set ${a}$ is $R$ -linearly independent (since the only possible relations are of the form $r a = 0$ , which for nonzero $a$ yield only $r = 0$ ). Thus, the rank of $I$ is exactly 1.

Finally, we claim $I$ is not free as an $R$ -module. To see why, suppose to the contrary that $I$ were free as an $R$ -module. We would then have an $R$ -module isomorphism $I ≃ R^{n}$ for some $n \geq 1$ . The rank of $R^{n}$ is $n$ (and rank is invariant under isomorphism), so we must then have $I ≃ R$ . Such an $R$ -module isomorphism corresponds to the choice of an element $a \in I$ that generates $I$ as an $R$ -module. But this implies $I$ is a principal ideal, contrary to our assumption $I$ is non-principal.

Problem 6

Let $R$ be a PID and $M$ be a torsion $R$ -module. Suppose $p m = 0$ for some nonzero $m \in M$ and prime element $p \in R$ . Prove that $Ann (M) \subseteq ⟨ p ⟩$ .

Hints

The ideal $Ann (M) \subseteq R$ is principal. Also, since $p \in R$ is prime and $R$ is a PID, for every element $a \in R$ we either have that $p$ divides $a$ (i.e., $a \in ⟨ p ⟩$ ), or $a$ and $p$ are relatively prime (in which case we can write $r a + s p = 1_{R}$ for some $r, s \in R$ ).

Solution

First recall that $Ann (M)$ is an ideal in $R$ and hence (since $R$ is a PID) is generated by some element $a \in R$ . Our goal, then, is to prove $a \in ⟨ p ⟩$ ; i.e., $p$ divides $a$ . To that end, note that either $p$ divides $a$ , or else $p$ and $a$ are relatively prime. If the former, then we are done. If the latter, then we can write we can write $r a + s p = 1_{R}$ for some $r, s \in R$ . Then

$m = 1_{R} \cdot m = (r a + s p) m = r a m + s p m = 0 + 0 = 0,$

contradicting the assumption that $m$ is nonzero.