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Let $a,b\in \mathbb N,$ then

$$\mathbb Z_p[T]/(T^a,p^b)\cong\mathbb Z_p[[T]]/(T^a,p^b)$$

1.What is this isomorphism ?

2.How to prove that $|\mathbb Z_p[[T]]/(T,p)^t|=p^{t(t+1)/2}$

Now Let $X$ be a $\mathbb Z_p[[T]]-$module.

3.we can view $X$ as a $\mathbb Z_p-$module ?

Community
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Med
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1 Answers1

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Hint for 1: What is the most natural homomorphism, from $\Bbb Z_p[T] \rightarrow \Bbb Z_p[[T]]$? Your isomorphism is induced from that homomorphism.

Hint for 2: A similar argument to 1. shows $\mathbb Z_p[[T]]/(T,p)^t \cong \mathbb Z_p[T]/(T,p)^t$. Note $\mathbb Z_p/(p^t) \cong \mathbb Z/p^t\mathbb Z$, and now you can just work out what $\mathbb Z_p[T]/(T,p)^t$ is.

Hint for 3:For any commutative rings $R$ and $S$, a homomorphism of rings $\phi : R\rightarrow S$ makes every $S$-module a $R$-module in a natural way.

PVAL-inactive
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  • obviously the natural injection :) – Med Dec 19 '13 at 02:25
  • thank you for 1 can you explain more – Med Dec 19 '13 at 02:30
  • I think $\mathbb Z_p[[T]]/(T,p)^t\cong \mathbb Z_p[T]/(T,p)^t$ is not just!! – Med Dec 19 '13 at 03:05
  • The induced map from the natural injection is clearly injective. It is surjective since we can write every element of the quotient ring of the power series with a polynomial representative, since $T^t \in (T,p)^t$. – PVAL-inactive Dec 19 '13 at 03:12