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In Advanced Modern Algebra, Rotman gives the following construction of a module that he denotes by $V^T$: Let $V$ be a vector space over a field $k$ and let $T:V \rightarrow V$ be a linear operator on $V$. Then, for a given polynomial $f(x) = \sum_i c_ix^i \in k[x]$ and $v \in V$ define scalar multiplication according to $$ \cdot :k[x] \times V \rightarrow V : f \cdot v \mapsto \sum_i c_i T^i(v) $$ where $T^i$ denotes the $i$-fold composition of $T$ with itself. It is clear that $V$ together with the indicated scalar multiplication is a $k[x]$-module.

My question is the following: Is this construction important enough to have a name and, if so, what is it called and how does it fit into the bigger picture? What is the motivation for this construction?

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This is just an unpacking of the definition of a $k[x]$-module. Every $k[x]$-module looks like this. –  Qiaochu Yuan May 1 '12 at 17:57
    
Umm..ok. Well, I guess every $k[x]$ module would probably have to look something like this but that doesn't really get at the heart of my question nor does it address the role of $T$ which makes the construction a little more interesting. One can consider modules over any ring and since $k[x]$ is a ring it seems reasonable that one could consider modules over $k[x]$, but I'm trying to understand the motivation for this particular construction. Surely, there is something more than "this just follows from the definitions" –  ItsNotObvious May 1 '12 at 18:09
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Every $k[x]$-module has to look exactly like this, and it is an unpacking of the definition that this is true. If this is not clear to you it is worth thinking about, and if you still have questions I can expand my comment into an answer. –  Qiaochu Yuan May 1 '12 at 18:10
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More often than not the motivation of this module action is that we can say a lot of things about the linear mapping $T$ (or about all linear mappings) by using the fact that this is a module over $k[x]$. The polynomial ring being a PID gives us a number of useful tools. –  Jyrki Lahtonen May 1 '12 at 18:22
    
One famous example of Jyrki's comment is that the Jordan decomposition of a linear operator is a corollary of the structure theorem of finitely generated modules over a PID. –  Michael Joyce May 1 '12 at 21:15
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1 Answer 1

Let $R$ be a ring and $M$ an abelian group. To give $M$ the structure of a left $R$-module is precisely to give a ring homomorphism $R \to \text{End}(M)$, where $\text{End}(M)$ denotes the ring of endomorphisms $M \to M$.

If $R = k$ is a field, giving a ring homomorphism $k \to \text{End}(M)$ equips $M$ with a notion of scalar multiplication by elements of $k$, so in fact a $k$-module is the same thing as a $k$-vector space.

If $R = k[x]$, then a ring homomorphism $\phi : k[x] \to \text{End}(M)$ gives by precomposition a ring homomorphism $k \to k[x] \to \text{End}(M)$, so $M$ is in particular a $k$-vector space. In addition, we get an additive map $\phi(x) \in \text{End}(M)$, that is, one satisfying $$\phi(x)(m + n) = \phi(x)m + \phi(x)n.$$

Since $x$ commutes with all scalars in $k$, we have in fact that $\phi(x)$ is an endomorphism of $M$ as a $k$-vector space; moreover, the image of any element of $k[x]$ is determined by the structure of $M$ as a $k$-vector space and by $\phi(x)$. So we conclude that

A $k[x]$-module is precisely a $k$-vector space together with a linear operator.

So I do not really know how to answer your questions as they are phrased. Rotman is just unpacking the definition of a $k[x]$-module. To me the question you are asking is roughly analogous to the following question:

My calculus teacher gave me the following construction of the derivative of a polynomial: to a polynomial $\sum a_i x^i$ you associate the polynomial $\sum i a_i x^{i-1}$. Is this construction important enough to have a name and, if so, what is it called and how does it fit into the bigger picture? What is the motivation for this construction?

Your calculus teacher is just unpacking the definition of differentiation.

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