Homework 1
Problem 1
Suppose
Suppose, towards a contradiction, that
Problem 2
Suppose
It is clear (always dangerous to say!) that
Since
Problem 3
Suppose
- Prove that if
is an integral domain then is a submodule of . - Give an example of a ring
and -module such that is not a submodule. (Hint: Consider torsion elements in the -module for some specific ring .) - Suppose
has a (nonzero) zero divisor and is nontrivial. Prove that has nonzero torsion elements. - Suppose
is an -module morphism. Prove that .
- First observe that
so . Next suppose , and let be nonzero elements such that and . We claim that is nonzero and . The first statement follows from the fact that we assumed is an integral domain; for the second, observe that Note that we used the commutativity of (the "integral" part of "integral domain") to switch from to . > 2. In light of part 1, we need to look at rings that are not integral domains. One such ring is . When considered as an module over itself, the elements and are torsion elements, since and . On the other hand, the element is not torsion, as one can quickly verify that for each of the five nonzero elements of . - Let
be a nonzero zero divisor, so that for some nonzero . Let be any nonzero element, and consider the element . If , then is torsion; if , then , and so is torsion. So in either case there exists a nonzero torsion element in . - Take any torsion element
and let be a nonzero element such that . Then observe that . This prove is a torsion element in .
Problem 4
Suppose
- For each submodule
on , the annihilator of in is defined to be the set of elements such that for all . Prove that the annihilator of in is a 2-sided ideal of . - For each right ideal
of , the annihilator of in is defined to be the set of all elements such that for all . Prove that the annihilator of in is a submodule of . - Consider the
-module and ideal . Determine the annihilator of in and the annihilator of in .
-
We first verify the annihilator of
in is an abelian subgroup. First recall that for every $n\in . Take any in the annihilator of and observe that for all we have , and so is in the annihilator of . By the Subgroup Criterion, we've thus proven the annihilator is a subgroup of under addition. Now fix any element of the annihilator and let be an arbitrary element. Then for every observe that , where for the last equality we once again used the fact that zero acts trivially. This proves is in the annihilator for every and hence the annihilator is a left ideal of . Similarly, we have , where for the second equality we used the fact that is a submodule so and hence it is annihilated by . This proves is in the annihilator for every , and hence the annihilator is a right ideal of . -
This will prove very similar to the argument above. First note once again that
for every , and so is the annihilator of in . Next suppose are in the annihilator of in . Then for every we have , and so is in the annihilator of in . We've thus proven the annihilator of in is a subgroup of . Finally, take any and in the annihilator of in . Then , since (as is a right ideal) and annihilates everything in . Thus is in the annihilator of in , and hence that annihilator is a left -submodule. -
By the definition, the annihilator of
in is the collection of all integers such that for every . The module is a Cartesian product, which means that its operation is component-wise (addition) and that its zero element is the triple . If we denote the general element in as then , and so annihilates that element exactly when in , in , and in . In other words, the desired annihilator is , where each is the annihilator in of the corresponding factor group of . By our basic knowledge of the cyclic groups , we know these annihilators are the ideals , , and . The intersection of those three ideals in is the ideal generated by their three generators, which (by definition!) is the least common multiple . So in summary, the annihilator of in is the ideal .Turning things around, the annihilator of the ideal
in the given module is the collection of all elements such that for every . Given the description of the ideal , the annihilator are those triples such that for every integer we have in , in , and in . These conditions are satisfied exactly by , in , and . The annihilator of in is therefore the submodule , which is isomorphic as an abelian group to .
Problem 5
Give an example of a ring
There are many possible such examples. One familiar examples that of complex conjugation. Let
Problem 6
Suppose
Bonus challenge: Is your isomorphism natural in
First recall that
We will now define a set map