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I'm reading a remark in Atiyah, Macdonald - Introduction to Commutative Algebra and I can't really see what is going on.

After giving the definition and a few examples of $R$-Modules (for a ring $R$), the following statement appears:

If $A = k[x]$, where $k$ is a field, then an $A$-module is a $k$-vector space with a linear transformation.

I'm not sure how to "precisify" this statement. Certainly $A$ is a ring in this case so the concept of an $A$-module is well defined.

I'm certainly open to even a vague hint; I think I would benefit from working out the details and I need to rapidly get familiar these ideas.

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You have a natural inclusion of $k$ into $k[x]$ so you can invoke restriction of scalars to get a $k$-module structure on any $k[x]$-module. More basically, $k$ is a subring of $k[x]$, so any $k[x]$-module has a $k$-module structure. Relative to the $k$-module structure, multiplication by $x$ determines a $k$-linear transformation. This may be what he is driving at, but I'm not sure. – Chris Leary Jan 11 '12 at 3:33
up vote 5 down vote accepted

Let $M$ be a left $A$-module. Then $M$ is also a module over the subring of constant polynomials, which is isomorphic to $k$. This makes $M$ a vector space over $k$. Now let $T:M\to M$ be defined by $T(m)=x\cdot m$. Then $T$ is a linear transformation on $M$. The action of $A$ on $M$ is given by $p(x)\cdot m =p(T)(m)$.

Conversely, if $V$ is a $k$-vector space and $T$ is a linear transformation on $V$, then defining $p(x)\cdot v=p(T)(v)$ makes $V$ an $A$-module.

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Thank you very much. This clarifies a lot (even more than what I asked). Very appreciated. – roo Jan 11 '12 at 3:53
You're welcome. – Jonas Meyer Jan 11 '12 at 3:55
what is meant by $p(T)(v)$ ? – Epsilon May 19 '13 at 14:15
@André: If $T$ is a linear transformation on a $k$ vector space $V$, and $p(x)$ is a polynomial in $k[x]$, then the evaluation $p(T)$ is a linear transformation on $V$. If $v\in V$, then $p(T)(v)$ means the result of applying the linear transformation $p(T)$ to the vector $v$. E.g., if $p(x)=1-3x+4x^2$, then $p(T)=I-3T+4T^2$, and $p(T)(v)=v-3T(v)+4T(T(v))$. – Jonas Meyer May 19 '13 at 18:09

First of all, if $S$ is a subring of $R$ and $M$ is an $R$-module, then $M$ is automatically an $S$-module. [To make this concrete just think of something like $\mathbb{R}^2$. It's an $\mathbb{R}$-module (a real vector space), but it's also a $\mathbb{Z}$-module (since you can scale by real numbers, you can also scale by integers).]

Thus any $k[x]$-module is a $k$-module (thus a $k$-vector space). Next, just consider module axioms. Can you show that the action of "$x$" is a linear transformation?

Conversely, if you take a vector space with linear transformation (call it "$T$"). Then let $x$ act like $T$: $x \cdot x \cdot v =T(T(v))$ etc. Poof! You get a $k[x]$ action on your vector space.

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Thank you! So if we take $B = k$, then a $B$-module would be a $k$-vector space as well, correct? – roo Jan 11 '12 at 3:41
Yes. A module over a field is a vector space. – Bill Cook Jan 11 '12 at 4:38

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