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- Jun 22, 2012

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I am currently studying Example 2, page 363 (see attachment) and I am trying to

*relate the example to Theorem 8 and the D&F text on extension of the scalars preceding Theorem 8 on pages 359-362)*

**closely**In Example 2 (see attachment) we have

\(\displaystyle R = \mathbb{Z}, \ S = \mathbb{Q}, \text{ and } A

\) is a finite abelian group.

Since A is an abelian group, it is the \(\displaystyle \mathbb{Z} \)-module N in the notation of Theorem 8.

Given the above and following the construction of D&F pages 359-362 (and Theorem 8) - we can form the tensor product of S and N over R, so doing this we have:

\(\displaystyle S \otimes_R N = \mathbb{Q} \otimes_\mathbb{Z} A \)

That is fine ... BUT ..

D&F say the following:

"In this case the \(\displaystyle \mathbb{Q} \)-module \(\displaystyle \mathbb{Q} \otimes_\mathbb{Z} A \) is 0."

But D&F in saying this are calling the tensor product of S and N (in this case \(\displaystyle \mathbb{Q} \) and A) the extension of the scalars from N!

However, after some careful study and considerable generous help, I had formed the view that the extension of the scalars in Theorem 8 was, in fact, the S_module L and not the S-module \(\displaystyle S \otimes_R N \)

**Can someone please clarify this for me?**

Further to the above, D&F go on to write:

"By Theorem 8, we see again that any homomorphism from a finite abelian group into a rational vector space is the zero map"

Is this statement referring to the map \(\displaystyle \phi \ : \ N \to L \)? Why?

If not, then what map in Theorem 8 is it referring to? Why?

**Hoping someone can help.**

Peter