How do you define modules for polynomial algebras?

Question

I am currently reading this book and I try to understand p. 33.

I understand Definition 2.7, but Proposition 2.9. seems weird to me. How is $\alpha : V\to V, \, v\mapsto X\cdot v$ a function from $V$ to $V$?

Thanks in advance!

Answer 1

In the proposition, we assume that $V$ is a $K[X]$-module, which means that we know how polynomials act on vectors. This means that for any $g \in K[X]$, the linear map $V \to V$ defined by $v \mapsto g \cdot v$ is specified. In particular, we're interested in the polynomial is $g = X \in K[X]$, and we define the linear map $\alpha: V \to V$ by it.

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Added in response to comments:

There's a natural correspondence \begin{align} \left\{ \begin{matrix} \text{linear maps} \\ \alpha: V \to V \end{matrix} \right\} &\longleftrightarrow \left\{ \begin{matrix} \text{module actions} \\ \lambda: K[X] \times V \to V \end{matrix} \right\} \end{align} which we can spell out explicitly.

Beginning with a linear map $\alpha: V \to V$, we can construct a module action $\lambda: K[X] \times V \to V$ as follows. Define $\lambda(X, v) = \alpha(v)$ for all $v \in V$. This is what we mean by the action of $X$ is determined by $\alpha$. Now it extends uniquely to a linear action over all of $K[X]$ by $$ \lambda\biggl( \sum_i c_i X^i,\, v \biggr) = \sum_i c_i \,\alpha(v)^i $$ You can verify that this definition satisfies all of the module axioms.

In the other direction, beginning with a linear action $\lambda: K[X] \times V \to V$, we can construct uniquely a linear map $\alpha: V \to V$ by $\alpha = \lambda(X, -)$, in other words, $$ \alpha(v) = \lambda(X, v) $$ for all $v \in V$. You can check that this must be a linear map.

It remains to see that these correspondences are mutually inverse. If you want to be really explicit, give names to these bijections, say $F$ goes from left to right and $G$ goes from right to left. Then, we would like to verify that $$ G\bigl( F(\alpha) \bigr) = \alpha $$ for any linear $\alpha: V \to V$ and that $$ F\bigl( G(\lambda) \bigr) = \lambda $$ for any $\lambda: K[X] \times V \to V$. Can you see how this goes?