- Thread starter
- #1

- Jun 22, 2012

- 2,918

I am reading Dummit and Foote, Section 10.4: Tensor Products of Modules. I am currently studying Example 3 on page 369 (see attachment).

Example 3 on page 369 reads as follows:

-------------------------------------------------------------------------------

In general,

\(\displaystyle \mathbb{Z} / m \mathbb{Z} \otimes_\mathbb{Z} \mathbb{Z} / n \mathbb{Z} \cong \mathbb{Z} / d \mathbb{Z}\) where d is the g.c.d. of the integers m and n.

To see this observe first that

\(\displaystyle a \otimes b = a \otimes (b \cdot 1) = (ab) \otimes 1 = ab(1 \otimes 1) \)

... ... ... etc etc

-----------------------------------------------------------------------------

I can see how one of the relations (equalities) in

\(\displaystyle a \otimes b = a \otimes (b \cdot 1) = (ab) \otimes 1 = ab(1 \otimes 1) \)

is justified by the relations in 10.7 (see D&F page 364 - attached) that flow from the form of the quotient taken of the the free \(\displaystyle \mathbb{Z} \)-module on the set \(\displaystyle M \times N \).

For example

\(\displaystyle a \otimes (b \cdot 1) = (ab) \otimes 1 \)

follows the 'rule' (see 10.7 on page 364 attached)

\(\displaystyle m \otimes r n = mr \otimes n \)

Further \(\displaystyle a \otimes b = a \otimes (b \cdot 1) \) is trivially true.

Why, exactly, is this the case? Further, can someone give me a sense of why it should be the case?

Peter

Example 3 on page 369 reads as follows:

-------------------------------------------------------------------------------

In general,

\(\displaystyle \mathbb{Z} / m \mathbb{Z} \otimes_\mathbb{Z} \mathbb{Z} / n \mathbb{Z} \cong \mathbb{Z} / d \mathbb{Z}\) where d is the g.c.d. of the integers m and n.

To see this observe first that

\(\displaystyle a \otimes b = a \otimes (b \cdot 1) = (ab) \otimes 1 = ab(1 \otimes 1) \)

... ... ... etc etc

-----------------------------------------------------------------------------

I can see how one of the relations (equalities) in

\(\displaystyle a \otimes b = a \otimes (b \cdot 1) = (ab) \otimes 1 = ab(1 \otimes 1) \)

is justified by the relations in 10.7 (see D&F page 364 - attached) that flow from the form of the quotient taken of the the free \(\displaystyle \mathbb{Z} \)-module on the set \(\displaystyle M \times N \).

For example

\(\displaystyle a \otimes (b \cdot 1) = (ab) \otimes 1 \)

follows the 'rule' (see 10.7 on page 364 attached)

\(\displaystyle m \otimes r n = mr \otimes n \)

Further \(\displaystyle a \otimes b = a \otimes (b \cdot 1) \) is trivially true.

**BUT ... ... ...**how is the relationship \(\displaystyle (ab) \otimes 1 = ab(1 \otimes 1) \) justified?Why, exactly, is this the case? Further, can someone give me a sense of why it should be the case?

Peter

Last edited: