# $\mathbb{C}[t]$ module , module isomorphisms

If $V$ is a $\mathbb{C}$ vector space and $a\in End \ V$. Then let $V=V_{a}$ be the $\mathbb{C}[t]$-module 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?

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"How to find a $\mathbb C\left[t\right]$ module which isomorphic?" What does this question mean? – darij grinberg Dec 1 '11 at 0:55
Also, your remark about ii) is not really understandable. – darij grinberg Dec 1 '11 at 0:56
I corrceted my remarks, thanks for yours. – Tashi Dec 1 '11 at 15:31

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!
Thank you Dylan Moreland. The map given by $x \mapsto tx$ is C-linear because $t(zx) = z(tx)$ and $t(x+y)=t(x)+t(y)$. I can't undrstand where you are going with this. I would be very glad if you could show me. – Tashi Dec 1 '11 at 15:26
@Tashi The plan is to have $V = M$ with this $\mathbf{C}$-linear structure, and to have $a$ as above: $a(x) = tx$. Can you check that this works? – Dylan Moreland Dec 1 '11 at 18:13
Thank you @Dylan Moreland: If $P(t) = a_{n}t^{n}+....a_{0}$, then $P(a(x))=a_{n}(tx)^{n}+...+a_{0}$. Then by $P(t)v=P(a)v$ this is again a $\mathbb{C}[t]$ module. I don't think this is what you had in mind... – Tashi Dec 1 '11 at 18:34
But a isn't a homomorphism (so also not a isomorphism): Since a(x+y)= tx+ty = a(tx)+a(ty) but $txy = a(xy)\ne a(x)a(y) = txty$?? So $a(x)=tx$ won't work...? – Tashi Dec 1 '11 at 20:00
@Tashi It needs to be a homomorphism of modules, and really $M$ doesn't come with a notion of multiplication with itself. – Dylan Moreland Dec 1 '11 at 20:09