## Dual Modules

**Point of Post: **In this post we discuss the notion of dual modules, and give a special treatment showing that for vector spaces, the dual space is isomorphic to the original space if and only in the case of finite dimension.

*Motivation*

Just as in the case for vector spaces, one is naturally inclined in module theory to look at the set of linear functionals from module to the ground ring, or using the notation of homomorphism groups . The idea of a dual module are vital in many ares of mathematics, including such non-algebraic subjects as analysis (where the dual space is vital for defining generalizations of integration, etc.). For us though, we are interested in the purely algebraic traits of dual modules and what they can tell us about the module itself. We shall be particularly interested in reminding ourselves of the following very interesting fact from linear algebra, which says that for a vector space the dual space of is isomorphic to precisely when is finite dimensional.

*Dual Modules*

Let be a ring and a left -module. Then, we define the *dual module *of to be the set of all -maps (such maps are called *linear functionals*). We denote the dual space of as either or , the former being more common when is a field.

As a particular case of the study of homomorphism groups we know that is an abelian group always, and a left -module when is commutative.

Now, note that one must really be careful when dealing with modules. From linear algebra, one might be prone to say that the dual module is “bigger” than in the sense that there is some embedding . While this is always true if is a field, this is far from true otherwise. For example, looking at -modules it’s not hard to see that where is any finite abelian group. Indeed, clearly there exists no nonzero homomorphisms since given any one has that for all . Now, since is divisible and is not–put more directly if then one has that for every , and so for all so that . That said, the following result, which should be familiar from linear algebra, is true:

**Theorem: ***Let be any commutative unital ring, then as left -modules for all . *

**Proof:** Define the functions defined by . It’s not hard to see that is a linear functional. We then define the map given by . Clearly is a group homomorphism and -morphism when is commutative, what we now claim is that is bijective. Indeed, if , then one has by assumption that (where denotes the canonical injection) and so is trivial. To see that is surjective let be arbitrary, we claim that . Indeed, by pure computation:

from where the conclusion follows.

*Remark: *Of course the above should generalize to modules that have finite “bases” (i.e. free modules of finite rank)–we shall see this when we define the appropriate concepts later.

*Dual Spaces of Vector Spaces*

What we’d now like to do is prove a classic theorem from linear algebra, something that shall be useful in the coming theory. What it says, roughly, is that for a vector space over a field one has that is equivalent to . The argument we give here is essentially that of Richard Foote (yes, that Foote).

**Theorem: ***Let be a field and a -space. Then, if and only if .*

**Proof: **From basic linear algebra we know that if then and so (where we used the previous theorem).

Conversely, suppose that is infinite dimensional. Let be a basis for , we then have from first principles that , and evidently . Let be the prime subfield of , we claim that

Indeed, the first equality holds since, by first principle, both have dimension . To see the rightmost inequality we take a basis for . We note that since each is naturally an element of , we claim that is linearly independent over . Indeed, since is a vector space over we can find a basis for over . Suppose then that were such that . We have that there exists scalars so that (where we have made a little bit of a notational shortcut, by taking the set of all vectors in to express each of the and enumerating them from the start) and so . We see then, choosing some this tells us that and since we may conclude that . Since this was true for all we must have that and so by assumption . From there linear independence of the ‘s over and thus the dimension inequality follows. To see the central inequality we let we note then, using basic cardinal arithmetic and a common set theoretic fact, that

Yet, since has at least two elements (as every field does) we know that contains and, by Cantor’s theorem, has cardinality strictly greater than . Thus, since obviously the (non-strict) inequality holds, and equality can’t hold otherwise contradictory to what we just proved. The conclusion follows.

**References:**

[1] Dummit, David Steven., and Richard M. Foote. *Abstract Algebra*. Hoboken, NJ: Wiley, 2004. Print.

[2] Rotman, Joseph J. *Advanced Modern Algebra*. Providence, RI: American Mathematical Society, 2010. Print.

[3] Blyth, T. S. *Module Theory.* Clarendon, 1990. Print.

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