Abstract Nonsense

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.

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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 $\text{Hom}_R(M,R)$. 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 $V$ the dual space of $V$ is isomorphic to $V$ precisely when $V$ is finite dimensional.

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Dual Modules

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Let $R$ be a ring and $M$ a left $R$-module. Then, we define the dual module of $M$ to be the set $\text{Hom}_R(M,R)$ of all $R$-maps $M\to R$ (such maps are called linear functionals). We denote the dual space of $M$ as either $M^{\vee}$ or $M^\ast$, the former being more common when $R$ is a field.

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

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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 $M^\vee$ is “bigger” than $M$ in the sense that there is some embedding $M\hookrightarrow M^\vee$. While this is always true if $R$ is a field, this is far from true otherwise. For example, looking at $\mathbb{Z}$-modules it’s not hard to see that $A^\vee=\mathbb{Q}^\vee=0$ where $A$ is any finite abelian group. Indeed, clearly there exists no nonzero homomorphisms $A\to\mathbb{Z}$ since given any $f\in A^\vee$ one has that $|A|f(x)=f(|A|x)=f(0)=0$ for all $x\in A$. Now, $\mathbb{Q}^\vee=0$ since $\mathbb{Q}$ is divisible and $\mathbb{Z}$ is not–put more directly if $f\in\mathbb{Q}^\vee$ then one has that $f(x)=nf(n^{-1}x)$ for every $n\in\mathbb{N}$, and so $n\mid f(x)$ for all $n\in\mathbb{N}$ so that $f(x)=0$. That said, the following result, which should be familiar from linear algebra, is true:

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Theorem: Let $R$ be any commutative unital ring, then $\left(R^n\right)^\vee\cong R^n$ as left $R$-modules for all $n\in\mathbb{N}$

Proof: Define the functions $\delta_j:R^n\to R$ defined by $\delta_j(r_1,\cdots,r_n)=r_j$. It’s not hard to see that $\delta_j$ is a linear functional. We then define the map $f:R^n\to (R^n)^\vee$ given by $(r_1,\cdots,r_n)\mapsto r_1\delta_1+\cdots+r_n\delta_n$. Clearly $f$ is a group homomorphism and $R$-morphism when $R$ is commutative, what we now claim is that $f$ is bijective. Indeed, if $r_1\delta_1+\cdots+r_n\delta_n=0$, then  one has by assumption that $0=(r_1\delta_1+\cdots+r_n\delta_n)(\iota_j(1))=r_j$ (where $\iota_j$ denotes the canonical injection) and so $\ker f$ is trivial. To see that $f$ is surjective let $g\in (R^n)^\vee$ be arbitrary, we claim that $g=g(\iota_1(1))\delta_1+\cdots+g(\iota_n(1))\delta_n$. Indeed, by pure computation:

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\begin{aligned}g(r_1,\cdots,r_n) &=g(r_1\iota_1(1)+\cdots+r_n\iota_n(1))\\ &=r_1g(\iota_1(1))+\cdots+r_ng(\iota_n(1))\\ &=\delta_1(r_1,\cdots,r_n)g(\iota(1))+\cdots+\delta_n(r_1,\cdots,r_n)g(\iota_n(1))\end{aligned}

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from where the conclusion follows. $\blacksquare$

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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.

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Dual Spaces of Vector Spaces

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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 $V$ over a field $k$ one has that $V\cong V^\ast$ is equivalent to $\dim_k V<\infty$. The argument we give here is essentially that of Richard Foote (yes, that Foote).

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Theorem: Let $k$ be a field and $V$ a $k$-space. Then, $V\cong V^\ast$ if and only if $\dim_k V<\infty$.

Proof: From basic linear algebra we know that if $\dim_k V=n\in\mathbb{N}$ then $V\cong k^n$ and so $V^\ast\cong (k^n)^\ast\cong k^n\cong V$ (where we used the previous theorem).

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Conversely, suppose that $V$ is infinite dimensional. Let $\mathcal{B}$ be a basis for $V$, we then have from first principles that $V\cong k^{\oplus\mathcal{B}}$, and evidently $V^\ast\cong k^{\mathcal{B}}$. Let $k_0$ be the prime subfield of $k$, we claim that

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$\dim_k k^{\oplus\mathcal{B}}=\dim_{k_0}k_0^{\oplus\mathcal{B}}<\dim_{k_0}k_0^{\mathcal{B}}\leqslant \dim_k k^{\mathcal{B}}$

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Indeed, the first equality holds since, by first principle, both have dimension $\#(\mathcal{B})$. To see the rightmost inequality we take a basis $\left\{f_\alpha\right\}$ for $k_0^{\mathcal{B}}$. We note that since $k_0\subseteq k$ each $f_\alpha$ is naturally an element of $k^{\mathcal{B}}$, we claim that $\left\{f_\alpha\right\}$ is linearly independent over $k$. Indeed, since $k$ is a vector space over $k_0$ we can find a basis $\{v_\beta\}_{\beta\in\mathcal{B}}$ for $k$ over $k_0$. Suppose then that $c_i\in k$ were such that $\displaystyle \sum_i c_i f_i=0$. We have that there exists scalars $d_{i,1},\cdots,d_{i,m_i}$ so that $\displaystyle c_i=\sum_j d_{i,j}v_{\beta_j}$ (where we have made a little bit of a notational shortcut, by taking the set of all vectors in $\{v_\beta\}$ to express each of the $c_i$ and enumerating them from the start) and so $\displaystyle 0=\sum_i\sum_j d_{i,j}v_{\beta_j}f_i$.  We see then, choosing some $v\in\mathcal{B}$ this tells us that $\displaystyle \sum_j\left(\sum_i d_{i,j}f_i(v)\right)v_{\beta_j}=0$ and since $\displaystyle \sum_i d_{i,j}f_i(v)\in k_0$ we may conclude that $\displaystyle \sum_i d_{i,j}f_i(v)=0$. Since this was true for all $v\in\mathcal{B}$ we must have that $\displaystyle \sum_i d_{i,j}f_i=0$ and so by assumption $d_{i,j}=0$. From there linear independence of the $f_i$‘s over $k$ and thus the dimension inequality follows. To see the central inequality we let $\mathcal{S}=\left\{S\subseteq\mathcal{B}:\#(S)<\infty\right\}$ we note then, using basic cardinal arithmetic and a common set theoretic fact, that

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$\displaystyle \#(k_0^{\oplus\mathcal{B}})=\#\left(\bigcup_{S\in\mathcal{S}}k_0^S\right)\leqslant \sum_{S\in\mathcal{S}}\aleph_0=\#(\mathcal{S})\aleph_0=\#(\mathcal{B})\aleph_0=\#(\mathcal{B})$

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Yet, since $k$ has at least two elements (as every field does) we know that $k^{\mathcal{B}}$ contains $2^\mathcal{B}$ and, by Cantor’s theorem,  has cardinality strictly greater than $\#(\mathcal{B})$. Thus, $\dim_{k_0}k_0^{\oplus\mathcal{B}}<\dim_{k_0}k_0^{\mathcal{B}}$ since obviously the (non-strict) inequality holds, and equality can’t hold otherwise $\#(k_0^{\oplus\mathcal{B}})=\#(k_0^{\mathcal{B}})$ contradictory to what we just proved. The conclusion follows. $\blacksquare$

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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.

November 12, 2011 -

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