Homework 1

Problem 1

Suppose $R$ is a ring, $M$ is a left $R$ -module, and $r \in R$ is an element for which there exists a nonzero $m \in M$ such that $r m = 0$ . Prove $r$ does not have a left inverse.

Solution

Suppose, towards a contradiction, that $r$ had a left inverse. Let $s \in R$ be a left inverse of $r$ in $R$ , so that $s r = 1_{R}$ . On the one hand, we then have $s (r m) = s \cdot 0 = 0$ ; on the other hand we also have $s (r m) = (s r) m = 1_{R} \cdot m = m$ . Since $m$ was assumed to be nonzero, this is a contradiction.

Problem 2

Suppose $R$ is a ring, $I \subseteq R$ is a left ideal, and $M$ is a left $R$ -module. Let $I M \subseteq M$ denote the subset of all finite $I$ -linear combinations in $M$ , i.e., $I M = {\sum_{finite} i_{k} m_{k} ∣ i_{k} \in I, m_{k} \in M} .$ Prove $I M$ is a submodule of $M$ .

Solution

It is clear (always dangerous to say!) that $I M$ is a subgroup of the abelian group $M$ , so it only remains to verify $I M$ is closed under scaling, i.e., the action of $R$ . To see that, suppose we have an element of $\sum_{k} i_{k} m_{k}$ of $I M$ and observe that $r \cdot \sum_{k} i_{k} m_{k} = \sum_{k} r (i_{k} m_{k}) = \sum_{k} (r i_{k}) m .$

Since $I$ is a left ideal of $R$ , we have $r i_{k} \in R$ for every $k$ and so the final sum is once again another element of the set $I M$ .

Problem 3

Suppose $R$ is a ring and $M$ is a left $R$ -module. An element $m \in M$ is called a torsion element if $r m = 0$ for some nonzero $r \in R$ . The set of torsion elements in $M$ is denoted $Tor (M)$ .

Prove that if $R$ is an integral domain then $Tor (M)$ is a submodule of $M$ .
Give an example of a ring $R$ and $R$ -module $M$ such that $Tor (M)$ is not a submodule. (Hint: Consider torsion elements in the $R$ -module $R$ for some specific ring $R$ .)
Suppose $R$ has a (nonzero) zero divisor and $M$ is nontrivial. Prove that $M$ has nonzero torsion elements.
Suppose $ϕ : M \to N$ is an $R$ -module morphism. Prove that $ϕ (Tor (M)) \subseteq Tor (N)$ .

Solution

First observe that $1_{R} \cdot 0 = 0$ so $0 \in Tor (M)$ . Next suppose $m, n \in Tor (M)$ , and let $r, s \in R$ be nonzero elements such that $r m = 0$ and $s n = 0$ . We claim that $r s$ is nonzero and $(r s) (m - n) = 0$ . The first statement follows from the fact that we assumed $R$ is an integral domain; for the second, observe that $(r s) (m - n) = (r s) m - (r s) n = (s r) m - (r s) n = s (r m) - r (s n) = s \cdot 0 - r \cdot 0 = 0 - 0 = 0.$ Note that we used the commutativity of $R$ (the "integral" part of "integral domain") to switch from $(r s) m$ to $(s r) m$ . > 2. In light of part 1, we need to look at rings $R$ that are not integral domains. One such ring is $Z_{6}$ . When considered as an module over itself, the elements $2$ and $3$ are torsion elements, since $3 \cdot 2 = 0$ and $2 \cdot 3 = 0$ . On the other hand, the element $2 + 3 = 5$ is not torsion, as one can quickly verify that $r \cdot 5 \neq 0$ for each of the five nonzero elements of $Z_{6}$ .
Let $r \in R$ be a nonzero zero divisor, so that $r s = 0_{R}$ for some nonzero $s \in R$ . Let $m \in M$ be any nonzero element, and consider the element $s m$ . If $s m = 0$ , then $m$ is torsion; if $s m \neq 0$ , then $r (s m) = (r s) m = 0_{R} \cdot m = 0$ , and so $s m$ is torsion. So in either case there exists a nonzero torsion element in $M$ .
Take any torsion element $m \in M$ and let $r \in R$ be a nonzero element such that $r m = 0_{M}$ . Then observe that $r \cdot ϕ (m) = ϕ (r m) = ϕ (0_{M}) = 0_{N}$ . This prove $ϕ (m)$ is a torsion element in $N$ .

Problem 4

Suppose $R$ is a ring and $M$ is a left $R$ -module.

For each submodule $N$ on $M$ , the annihilator of $N$ in $R$ is defined to be the set of elements $r \in R$ such that $r n = 0$ for all $n \in N$ . Prove that the annihilator of $N$ in $R$ is a 2-sided ideal of $R$ .
For each right ideal $I$ of $R$ , the annihilator of $I$ in $M$ is defined to be the set of all elements $m \in M$ such that $i m = 0$ for all $i \in I$ . Prove that the annihilator of $I$ in $M$ is a submodule of $M$ .
Consider the $Z$ -module $M = Z_{24} \times Z_{15} \times Z_{50}$ and ideal $I = 2 Z$ . Determine the annihilator of $M$ in $Z$ and the annihilator of $I$ in $M$ .

Solution

We first verify the annihilator of $N$ in $R$ is an abelian subgroup. First recall that $0_{R} \cdot n = 0_{M}$ for every $n\in $N$ . Take any $r_{1}, r_{2}$ in the annihilator of $N$ and observe that for all $n \in N$ we have $(r_{1} - r_{2}) n = r_{1} n - r_{2} n = 0_{N} - 0_{N} = 0_{N}$ , and so $r_{1} - r_{2}$ is in the annihilator of $N$ . By the Subgroup Criterion, we've thus proven the annihilator is a subgroup of $R$ under addition. Now fix any element $r$ of the annihilator and let $s \in R$ be an arbitrary element. Then for every $n \in N$ observe that $(s r) n = s (r n) = s \cdot 0_{M} = 0_{M}$ , where for the last equality we once again used the fact that zero acts trivially. This proves $s r$ is in the annihilator for every $s \in R$ and hence the annihilator is a left ideal of $R$ . Similarly, we have $(r s) n = r (s n) = r \cdot 0_{M} = 0_{M}$ , where for the second equality we used the fact that $N$ is a submodule so $s n \in N$ and hence it is annihilated by $r$ . This proves $r s$ is in the annihilator for every $s \in R$ , and hence the annihilator is a right ideal of $R$ .
This will prove very similar to the argument above. First note once again that $i \cdot 0_{M} = 0_{M}$ for every $i \in I$ , and so $0_{M}$ is the annihilator of $I$ in $M$ . Next suppose $m_{1}, m_{2}$ are in the annihilator of $I$ in $M$ . Then for every $i \in I$ we have $i (m_{1} - m_{2}) = i m_{1} - i m_{2} = 0_{M} - 0_{M} = 0_{M}$ , and so $m_{1} - m_{2}$ is in the annihilator of $I$ in $M$ . We've thus proven the annihilator of $I$ in $M$ is a subgroup of $M$ . Finally, take any $r \in R$ and $m$ in the annihilator of $I$ in $M$ . Then $i (r m) = (i r) m = i^{'} m = 0_{M}$ , since $i^{'} \in I$ (as $I$ is a right ideal) and $m$ annihilates everything in $I$ . Thus $r m$ is in the annihilator of $I$ in $M$ , and hence that annihilator is a left $R$ -submodule.
By the definition, the annihilator of $M$ in $Z$ is the collection of all integers $n$ such that $n m = 0$ for every $m \in M$ . The module $M$ is a Cartesian product, which means that its operation is component-wise (addition) and that its zero element is the triple $(0, 0, 0)$ . If we denote the general element in $M$ as $(m_{1}, m_{2}, m_{3})$ then $n \cdot (m_{1}, m_{2}, m_{3}) = (n m_{1}, n m_{2}, n m_{3})$ , and so $n$ annihilates that element exactly when $n m_{1} = 0$ in $Z_{24}$ , $n m_{2} = 0$ in $Z_{15}$ , and $n m_{3} = 0$ in $Z_{50}$ . In other words, the desired annihilator is $A_{1} \cap A_{2} \cap A_{3}$ , where each $A_{i}$ is the annihilator in $Z$ of the corresponding factor group $M_{i}$ of $M$ . By our basic knowledge of the cyclic groups $Z_{k}$ , we know these annihilators are the ideals $A_{1} = 24 Z$ , $A_{2} = 15 Z$ , and $A_{3} = 50 Z$ . The intersection of those three ideals in $Z$ is the ideal generated by their three generators, which (by definition!) is the least common multiple $lcm (24, 15, 50) = 600$ . So in summary, the annihilator of $M$ in $Z$ is the ideal $600 Z$ .

Turning things around, the annihilator of the ideal $I = 2 Z$ in the given module $M$ is the collection of all elements $(m_{1}, m_{2}, m_{3})$ such that $(i m_{1}, i m_{2}, i m_{3}) = (0, 0, 0)$ for every $i \in I$ . Given the description of the ideal $I$ , the annihilator are those triples such that for every integer $k$ we have $2 k m_{1} = 0$ in $Z_{24}$ , $2 k m_{2} = 0$ in $Z_{15}$ , and $2 k m_{3} = 0$ in $Z_{50}$ . These conditions are satisfied exactly by $m_{1} \in 12 Z_{24}$ , $m_{2} = 0$ in $Z_{15}$ , and $m_{3} \in 25 Z_{50}$ . The annihilator of $I$ in $M$ is therefore the submodule $(12 Z_{24}) \times 0 \times (25 Z_{50})$ , which is isomorphic as an abelian group to $Z_{2} \times Z_{2}$ .

Problem 5

Give an example of a ring $R$ , two $R$ -modules $M$ and $N$ , and a set map $f : M \to N$ such that $f$ is a group morphism but not an $R$ -module morphism.

Solution

There are many possible such examples. One familiar examples that of complex conjugation. Let $R$ be the field of complex numbers, let $M$ and $N$ both be the abelian group of complex numbers (under addition), and let $f$ be complex conjugation. The familiar fact that complex conjugation is compatible with addition (i.e., $\overset{―}{z + w} = \overset{―}{z} + \overset{―}{w}$ ) is really the statement that the map $f : C \to C$ is a morphism of abelian groups (i.e., $f (z + w) = f (z) + f (w)$ .) But conjugation is not compatible with complex scaling. For instance, if $z = 1 + 2 i$ and $r = i$ , then $\overset{―}{r z} = \overset{―}{i (1 + 2 i)} = \overset{―}{i - 2} = - i - 2$ , while $r \overset{―}{z} = i \cdot \overset{―}{1 + 2 i} = i (1 - 2 i) = i + 2$ . In our notation, this exactly says that $f (r z) \neq r f (z)$ , and so our map $f$ is not an $R$ -module morphism.

Problem 6

Suppose $R$ is a commutative ring and $M$ is left $R$ -module. Prove that ${Hom}_{R} (R, M)$ and $M$ are isomorphic as left $R$ -modules.

Bonus challenge: Is your isomorphism natural in $M$ ?

Solution

First recall that $R$ is being viewed as a module over itself by way of left multiplication. Next, recall that the set ${Hom}_{R} (R, M)$ always has the structure of an abelian group, and since $R$ is commutative here it also has the structure of an $R$ -module.

We will now define a set map ${Hom}_{R} (R, M) \to M$ and eventually prove it is a module isomorphism. How can we come up with a map? Suppose $f : R \to M$ is a module morphism. What do we know about $f$ ? At first glance, not much. By the group morphism property we must have $f (0_{R}) = 0_{M}$ , so that doesn't tell us much about the map $f$ . But our ring $R$ has another distinguished element, its multiplicative identity $1_{R}$ . That element gets mapped to some element $f (1_{R}) \in M$ . We claim that this element (the image of $1_{R}$ in $M$ ) completely determines the map $f$ as a module morphism. To see this, suppose $r \in R$ is any element. Since $f$ is a module morphism, we must have $f (r) = f (r \cdot 1_{R}) = r f (1_{R})$ . In other words, the image of $r$ is determined by $f (1_{R})$ . We now have our candidate for a map from ${Hom}_{R} (R, M) \to M$ , which we will now name ${ev}_{1_{R}}$ map ("evaluate at $1_{R}$ "). We claim this map is an $R$ -module isomorphism.

First we show it is group morphism, so suppose $f, g \in {Hom}_{R} (R, M)$ . Then ${ev}_{1_{R}} (f + g) = (f + g) (1_{R}) = f (1_{R}) + g (1_{R}) = {ev}_{1_{R}} (f) + {ev}_{1_{R}} (g)$ , where the second equality holds by the definition of the group structure on ${Hom}_{R} (R, M)$ .

We next verify ${ev}_{1_{R}}$ an $R$ -module morphism. Take any $r \in R$ and $f \in {Hom}_{R} (R, M)$ , and observe that ${ev}_{1_{R}} (r f) = (r f) (1_{R}) = r \cdot f (1_{R}) = r \cdot {ev}_{1_{R}} (f)$ , where the second equality holds by the definition of the $R$ -action on ${Hom}_{R} (R, M)$ .

Finally, we verify ${ev}_{1_{R}}$ is an isomorphism. We could check it is bijective, but I prefer to instead produce the inverse $R$ -module morphism. To that end, define a map $g : M \to {Hom}_{R} (R, M)$ as follows. For each element $m \in M$ , let $f_{m} : R \to M$ be the map defined by $f_{m} (r) = r m$ . Note that $f_{m} (1_{R}) = 1_{R} \cdot m = m$ , as one would hope. Also observe that $f_{m}$ is indeed an $R$ -module morphism, since $f_{m} (r + s) = (r + s) m = r m + s m = f_{m} (r) + f_{m} (s)$ and $f_{m} (s r) = (s r) m = s (r m) = s f_{m} (r)$ . We have therefore indeed defined a set map $g : M \to {Hom}_{R} (R, M)$ . This map is a group morphism since $f_{m_{1} + m_{2}} (r) = r (m_{1} + m_{2}) = r m_{1} + r m_{2} = f_{m_{1}} (r) + f_{m_{2}} (r)$ , i.e., $g (m_{1} + m_{2}) = g (m_{1}) + g (m_{2})$ . This map is an $R$ -module morphism since $f_{s m} (r) = r (s m) = (r s) m = (s r) m = s (r m) = s \cdot f_{m} (r)$ , i.e., $g (s m) = s \cdot g (m)$ . (Note that we used the commutativity of $R$ for the third equality.) Lastly, this $R$ -module morphism $g$ is indeed inverse to ${ev}_{1_{R}}$ , since $({ev}_{1_{R}} \circ g) (m) = {ev}_{1_{R}} \circ f_{m} = f_{m} (1_{R}) = 1_{R} \cdot m = m$ , i.e., ${ev}_{1_{R}} \circ g$ is the identity module morphism on $R$ . (You can also verify that $g \circ {ev}_{1_{R}}$ is the identity module morphism on ${Hom}_{R} (R, M)$ ).

Curious about the bonus challenge? Naturality in this case means that for every $R$ -module morphism $g : M \to N$ we should have a commutative diagram as below:

Chasing the diagram around, suppose we have an element in the module on the top left, i.e., an $R$ -module morphism $f : R \to M$ . Going over and down gives $g (f (1_{R}))$ , while going down and then over gives $(g \circ f) (1_{R})$ , which by the definition of function composition does indeed agree with $g (f (1_{R}))$ . So we have indeed defined a natural isomorphism.

Problem 7

Suppose $R$ is a commutative ring. Prove that ${Hom}_{R} (R, R)$ and $R$ are isomorphic as rings.

Solution

We know that when $R$ is a commutative ring and $M$ is an $R$ -module, the set ${Hom}_{R} (R, M)$ is an $R$ -module and is isomorphic to $M$ . So we already know that in the special case $M = R$ (when $R$ is a viewed as an $R$ -module) the map ${ev}_{1_{R}} : {Hom}_{R} (R, R) \to R$ is an $R$ -module isomorphism. We thus only need to verify it's actually a ring morphism. To see that, suppose $f_{1}, f_{2} \in {Hom}_{R} (R, R)$ and recall that the product in the ring ${Hom}_{R} (R, R)$ is given by composition (or $R$ -module morphisms). Then note that
${ev}_{1_{R}} (f_{1} \cdot f_{2}) = (f_{1} \cdot f_{2}) (1_{R}) = f_{1} (f_{2} (1_{R})) .$
Let $r^{'} = f_{2} (1_{R}) \in R$ . Since $f_{1}$ is an $R$ -module morphism, we must have $f_{1} (r^{'}) = r^{'} \cdot f_{1} (1_{R}) = r^{'} \cdot {ev}_{1_{R}} (f_{1})$ . Since $r^{'} = f_{2} (1_{R}) = {ev}_{1_{R}} (f_{2})$ , the above equality becomes
${ev}_{1_{R}} (f_{1} \cdot f_{2}) = {ev}_{1_{R}} (f_{2}) {ev}_{1_{R}} (f_{1}) .$
The right-hand side of the above equality is a product in the commutative ring $R$ , so that commutativity allows us to conclude
${ev}_{1_{R}} (f_{1} \cdot f_{2}) = {ev}_{1_{R}} (f_{1}) {ev}_{1_{R}} (f_{2}) .$