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Let $k$ be a field. There are several ways to make $k$ into a $k[X]$-module; for example, by fixing $\alpha \in k$ and taking $P(X) \cdot t := P(\alpha)t$ for all $t \in k$. Are there other obvious ways to make $k$ into a $k[X]$-module? Can we actually give a nice descriptions of all possible such structures?

What can be said about the $k[X_1,X_2,\cdots,X_n]$-module structures on $k$ for $n \geq 2$?

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What can be said about background and context? –  Phira Oct 28 '11 at 23:23
    
It's a logical question to ask... there is not much more to say, really. –  Evariste Oct 28 '11 at 23:26
    
You really don't think that "obvious" does not depend on what you already know? –  Phira Oct 28 '11 at 23:28
    
Of course it does, but... I don't see your point, sorry. –  Evariste Oct 28 '11 at 23:29

1 Answer 1

up vote 4 down vote accepted

All interesting $k[x]$-module structures on $k$ can be given by a pair $(\phi,\alpha)$ where $\phi$ is an automorphism of $k$ and $\alpha\in k$. Then the module action is determined by:

$$P(x)\cdot t = \phi(P(\phi^{-1}(\alpha)))t$$

If you want $k[x]$ to act on $k$ consistent with the "obvious" action of $k$ on $k$, then $\phi = id_k$ and arbitrary $\alpha$ give you all the possible solutions.

This doesn't give all module structures. Let $k_0$ be the least subfield of $k$ containing $0$ and $1$. So $k_0\cong\mathbb Q$ or $k_0\cong\mathbb Z_p$. Let $T$ be an invertible linear operator on the field $k$ when viewed as a vector space over $k_0$. Then you can get an even more general operator given $(\phi,\alpha,T)$:

$$P(x)\cdot t = T^{-1} \phi(P(\phi^{-1}(\alpha)))(Tt)$$

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Hm, interesting... –  Evariste Oct 28 '11 at 23:41
    
Yes, I see where your operator $T$ comes from! –  Evariste Oct 28 '11 at 23:47

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