MHB Tensor Products - D&F page 369 Example 3 - The map phi

[Math Amateur]

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: (see attachment)

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

In general,

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

To see this observe first that

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

... ... ... etc etc ...

... The map

$$ \phi \ : \ \mathbb{Z} / m \mathbb{Z} \times_\mathbb{Z} \mathbb{Z} / n \mathbb{Z} \to \mathbb{Z} / d \mathbb{Z} $$

defined by

$$ \phi (a mod \ m , b mod \ n ) = ab mod \ d $$

is well defined since d divides both m and n. ... ...

... ...

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

Can someone please help with the following issue:

What is meant by the map $$ \phi $$ being 'well defined' and why is d dividing both m and n important in this matter?

I would appreciate some help.

Peter

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

 

[Deveno]

This is pretty basic.

When one defines a mapping on a quotient object (which is typically induced by some equivalence relation) one has to make sure the mapping depends only on the equivalence classes themselves, and not "representative elements".

For example, the map:

$f:\Bbb Z_5 \to \Bbb Z_4$ "defined" by $f(a\text{ (mod }5)) = a\text{ (mod }4)$

is not well-defined, because:

$[3]_5 = [8]_5$ but $f([3]_5) = [3]_4$ whereas $f([8]_5) = [0]_4$.

So to prove that $\phi$ is well-defined, we need to show that if:

$a \equiv a'\text{ (mod }m)$
$b \equiv b'\text{ (mod }n)$

that:

$ab \equiv a'b'\text{ (mod }d)$.

Now the first condition means that:

$a - a' = km$ for some $k \in \Bbb Z$, and the second condition means that:
$b - b' = sn$ for some $s \in \Bbb Z$.

Thus:

$ab - a'b' = ab - a'b + a'b - a'b' = (a - a')b + a'(b - b') = kmb + a'sn$.

Since $d|m$ and $d|n$, we have:

$m = td$
$n = ud$ for some (positive) integers $t,u$.

Hence:

$ab - a'b' = kmb + a'sn = k(td)b + a's(ud) = (ktb + a'su)d$

which shows that:

$ab \equiv a'b'\text{ (mod }d)$.

 

[ThePerfectHacker]

... The map

$$ \phi \ : \ \mathbb{Z} / m \mathbb{Z} \times_\mathbb{Z} \mathbb{Z} / n \mathbb{Z} \to \mathbb{Z} / d \mathbb{Z} $$

defined by

$$ \phi (a mod \ m , b mod \ n ) = ab mod \ d $$

is well defined since d divides both m and n. ... ...

This is how the map is being define $\varphi: \mathbb{Z}_m \times \mathbb{Z}_n \to \mathbb{Z}_d$, for $(\alpha,\beta) \in \varphi: \mathbb{Z}_m \times \mathbb{Z}_n \to \mathbb{Z}_d$ we choose representatives $\alpha = [a]_n$ and $\beta = _m$. In other words, $\alpha,\beta$ are classes mod $n$ and $m$ and can be represented by some integer. Then we define $\alpha \beta\in \mathbb{Z}_d$ to be the class mod $d$ that is represented by the integer $ab$. The question here is how do we know that if we picked different representatives in the beginning that we end up in the same class in $\mathbb{Z}_d$?

The next step is to check that this map $\varphi: \mathbb{Z}_m \times \mathbb{Z}_n \to \mathbb{Z}_d$ is bilinear. From here it will follow from the universal property of tensor product that there is a morphism $\psi : \mathbb{Z}_m \otimes \mathbb{Z}_n \to \mathbb{Z}_d $ which is defined by $\psi (a\otimes b) = ab$.