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If $V$ is a $\mathbb{C}$-vector space, and $a\in End \ V $, then we let $V_a$ be the $\mathbb{C}[t]$-module with ground set $V$ and scaling defined by: $Pv=P(a)v, \ \ (P\in \mathbb{C}[t], \ v\in V)$

i) Suppose $M$ is a $\mathbb{C}[t]$-module. Then there exist $V,a$ and a $\mathbb{C}[t]$-module isomorphism $f:M\rightarrow V_{a}$.

ii) Suppose $R= \mathbb{C}[t]/t^{2}\mathbb{C}[t]$ and let $M$ be a $R$-module. Then there exist $V,a$ so that $a^{2}=0$ and also a $R$-module isomorphism $f:M\rightarrow V_{a}$.

i) To show that this is true, we must find a module homomorphism which is one-to-one and onto. If $P(t)= b_{n}t^{n}+b_{n-1}t^{n-1}+...b_{1}t+b_{0}$ and one puts: $P(a)=b_{n}a^{n}+b_{n-1}a^{n-1}+...+b_{1}a+b_{0}I$, then $a^{n}$, where I is the identity and $a^{n}$ is the composition of a with itself ($va^{n}=(...(va)a...)a)$ This is then made into a $\mathbb{C}[t]$ module by: $P(t)v = P(a)v$.

How to construct the isomorphism $f: M\rightarrow V_{a}$?

ii) I know that there must be a vector space with $a^{2}=0$ because the factorial ring contains $t^{2}\mathbb{C}[t]$, but I don't see how to show that it exists and also not how to construct a module isomorphism?

Would be glad if somebody could answer my questions.

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I don't understand your remarks, so forgive me for not addressing them. It does seem like you know how the $V_a$ are constructed, which is good. We'll need that.

For both of these, note that $M$ is already a vector space over $\mathbf{C}$. The next thing to think about is whether the map $M \to M$ given by $x \mapsto tx$ is $\mathbf{C}$-linear. If it is, then I think you have found your $a$. The rest is just filling in details, but of course this can also be tricky—I'm certainly willing to say more.

Added. Let $V$ be $M$, viewed as a $\mathbf{C}$-vector space. View $a$ as a $\mathbf{C}$-linear map $V \to V$. We obtain a $\mathbf C[t]$-module $V_a$ by the construction you describe. You want to show that the map $M \to V_a$ which is the identity as a map of sets is actually $\mathbf C[t]$-linear; so check that for $P(t) \in \mathbf{C}[t]$, $x \in M = V_a$ we have \[ P(t)x = P(a)x. \] This is really close to being a tautology!

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    @Tashi Hm. Trying to invite someone to$a$room seems impossible at the moment, so just drop into the main room sometime and say hello if you want to chat. I'm around pretty often.2011-12-05