inverse limit of product

**KaKa** · November 21st 2010, 12:53 AM

I know that $\mathbb{Z}/(6)^i\mathbb{Z}\cong \mathbb{Z}/(2)^i\mathbb{Z}\times \mathbb{Z}/(3)^i\mathbb{Z}$ for any $i$ .

But $\varprojlim_{i\in I} \mathbb{Z}/(6)^i\mathbb{Z}\cong \varprojlim_{i\in I} \mathbb{Z}/(2)^i\mathbb{Z}\times \varprojlim_{i\in I} \mathbb{Z}/(3)^i\mathbb{Z}??$

Please give an explanation if it's right.

**tonio** · November 21st 2010, 03:08 AM

Originally Posted by KaKa

I know that $\mathbb{Z}/(6)^i\mathbb{Z}\cong \mathbb{Z}/(2)^i\mathbb{Z}\times \mathbb{Z}/(3)^i\mathbb{Z}$ for any $i$ .

But $\varprojlim_{i\in I} \mathbb{Z}/(6)^i\mathbb{Z}\cong \varprojlim_{i\in I} \mathbb{Z}/(2)^i\mathbb{Z}\times \varprojlim_{i\in I} \mathbb{Z}/(3)^i\mathbb{Z}??$

Please give an explanation if it's right.

To have an inverse limit you must have, besides a set of groups, a corresponding directed set of indexes and a

corresdponding set of homomorphisms between those indexed groups with some properties.

You must check these are fulfilled in this case.

Tonio

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