# Tensor Products - D&F page 369 Example 2

#### Peter

##### Well-known member
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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.

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

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#### ThePerfectHacker

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$$r\otimes 1 = (r\cdot 1)\otimes 1 = r(1\otimes 1)$$

#### Peter

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MHB Site Helper
$$r\otimes 1 = (r\cdot 1)\otimes 1 = r(1\otimes 1)$$
Thanks, ThePerfecthacker

Obvious when you see it

Peter

#### Peter

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Thanks, ThePerfecthacker

Obvious when you see it

Peter
A further question on this example:

It appears to me that the relations:

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

are quite general depending only on the properties of the tensor product (and not on the particular modules involved - we do not draw anything from the nature of $$\displaystyle \mathbb{Z} / m \mathbb{Z}$$ or $$\displaystyle \mathbb{Z} / n \mathbb{Z}$$?

Am I correct in this?

If I am then the relations:

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

are true for all modules and all tensor products are cyclic groups?

But is this the case????

Peter

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#### Peter

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$$r\otimes 1 = (r\cdot 1)\otimes 1 = r(1\otimes 1)$$
Sorry ThePerfectHacker ... I was far too quick by far in my response to your post pointing out that

$$r\otimes 1 = (r\cdot 1)\otimes 1 = r(1\otimes 1)$$

I thought I 'saw' the situation but now after checking I am perplexed as to why

$$(r\cdot 1)\otimes 1 = r(1\otimes 1)$$

Peter

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#### ThePerfectHacker

##### Well-known member
I thought I 'saw' the situation but now after checking I am perplexed as to why

$$(r\cdot 1)\otimes 1 = r(1\otimes 1)$$
That is a property of tensor products. Look at the properties that it satisfies.

are true for all modules and all tensor products are cyclic groups?
Consider the module $\mathbb{Z}\otimes \mathbb{Z}$. As you said $a\otimes b = ab(1\otimes 1)$. A typical element in $\mathbb{Z}\otimes \mathbb{Z}$ has the form $a_1\otimes b_1 + ... + a_n \otimes b_n$ which we can rewrite as $(a_1b_1 + ... + a_nb_n)(1\otimes 1)$. However, here you are using the fact that $\mathbb{Z}$ has a generating element (cyclic), in general a module does not have a generating element.

#### Peter

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That is a property of tensor products. Look at the properties that it satisfies.

Consider the module $\mathbb{Z}\otimes \mathbb{Z}$. As you said $a\otimes b = ab(1\otimes 1)$. A typical element in $\mathbb{Z}\otimes \mathbb{Z}$ has the form $a_1\otimes b_1 + ... + a_n \otimes b_n$ which we can rewrite as $(a_1b_1 + ... + a_nb_n)(1\otimes 1)$. However, here you are using the fact that $\mathbb{Z}$ has a generating element (cyclic), in general a module does not have a generating element.
You write that

$$\displaystyle (r\cdot 1)\otimes 1 = r(1\otimes 1)$$ ... ... ... ... (1)

is a property of tensor products ... but it does not seem to follow any of the relations of 10-7 (see D&F page 364 attached) ... as the only one that seems close is

$$\displaystyle mr \otimes n = m \otimes rn$$

which does not seem to cover (1) above.

Can you be specific as to what property you are referring to and indicate why the property is the case?

Peter

#### Peter

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Well yes, the relationship is definitely there in Wikipedia, as you say.

Why D&F only state $$\displaystyle mr \otimes n = m \otimes rn$$ and do omit the $$\displaystyle = r (m \otimes n)$$ I have no idea.

Wikipedia also seem to have stated the relationship as

$$\displaystyle rm \otimes n = m \otimes rn = r (m \otimes n)$$ - that is use $$\displaystyle rm \otimes n$$ instead of $$\displaystyle mr \otimes n$$ ... ...

Also Wikipedia are dealing with vector spaces not modules ... but whether all this makes any difference here, I am not sure ...

I am trying to gain a sense of why the relationship follows:

In Dummit and Foote, on page 364 (see attached) D&F state the following:

============================================

" Suppose then that N is a left R-module and that M is a right R-module. The quotient of the free $$\displaystyle \mathbb{Z}$$-module on the set $$\displaystyle M \times N$$ by the subgroup generated by all elements of the form;

$$\displaystyle (m_1 + m_2, n) - (m_1, n) - (m_2, n)$$ ... ... (10.6.1)

$$\displaystyle (m, n_1 + n_2) - (m, n_1) - (m, n_2)$$ ... ... (10.6.2)

$$\displaystyle (mr, n) - (m, rn)$$ ... ... (10.6.3)

for $$\displaystyle m, m_1, m_2 \in M, n, n_1, n_2 \in N, r \in R$$

==========================================

Now I can see that if you are taking the quotient of the subgroup generated by all the above elements you will have

$$\displaystyle (mr, n) - (m, rn) = 0$$ in the cosets so

$$\displaystyle mr \otimes n = m \otimes rn$$

I am not completely certain I am expressing this absolutely correctly but if I am correct there is some logic for accepting that

$$\displaystyle mr \otimes n = m \otimes rn$$ ... ...

But what is the logic for accepting the last equality of

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

Can you help? I am seeking some justification and some sense of why the relationship in question holds.

Peter

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#### Peter

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That is a property of tensor products. Look at the properties that it satisfies.

Consider the module $\mathbb{Z}\otimes \mathbb{Z}$. As you said $a\otimes b = ab(1\otimes 1)$. A typical element in $\mathbb{Z}\otimes \mathbb{Z}$ has the form $a_1\otimes b_1 + ... + a_n \otimes b_n$ which we can rewrite as $(a_1b_1 + ... + a_nb_n)(1\otimes 1)$. However, here you are using the fact that $\mathbb{Z}$ has a generating element (cyclic), in general a module does not have a generating element.
Thanks once again for your help ...

Still thinking over the issue that the relations:

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

are true for all modules and thus all tensor products are cyclic groups ...???

Still struggling with this issue ... reflecting further on what you have said ...

Peter

#### ThePerfectHacker

##### Well-known member
Why D&F only state $$\displaystyle mr \otimes n = m \otimes rn$$ and do omit the $$\displaystyle = r (m \otimes n)$$ I have no idea.
Because $M$ is a right module and so you multiply on the right instead of usual left.

I think it is best not to follow DF here as it seems they treat modules way too generally. It is better to learn it elsewhere. I think what they do is a little non-standard.

Also Wikipedia are dealing with vector spaces not modules ... but whether all this makes any difference here, I am not sure ...
A module over a ring is more general than a vector space over a field. The difference between module and vector space is that a module is an abelian group over a ring and a vector space is an abelian group over a field. So all vector spaces are modules but not all modules are vector spaces.

.. are all tensor products cyclinc?
$$\mathbb{Z}\otimes (\mathbb{Z}\oplus \mathbb{Z}) \simeq (\mathbb{Z} \otimes \mathbb{Z})\oplus (\mathbb{Z}\otimes \mathbb{Z}) \simeq \mathbb{Z}\oplus \mathbb{Z}$$

#### Peter

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MHB Site Helper
Because $M$ is a right module and so you multiply on the right instead of usual left.

I think it is best not to follow DF here as it seems they treat modules way too generally. It is better to learn it elsewhere. I think what they do is a little non-standard.

A module over a ring is more general than a vector space over a field. The difference between module and vector space is that a module is an abelian group over a ring and a vector space is an abelian group over a field. So all vector spaces are modules but not all modules are vector spaces.

$$\mathbb{Z}\otimes (\mathbb{Z}\oplus \mathbb{Z}) \simeq (\mathbb{Z} \otimes \mathbb{Z})\oplus (\mathbb{Z}\otimes \mathbb{Z}) \simeq \mathbb{Z}\oplus \mathbb{Z}$$

Thanks ThePerfectHacker.

Now just a further note ... as will be very clear to MHB members who have read the above posts, I have been very perplexed as to why, given the D&F construction/definitions of tensor products of modules that the following relation holds:

$$\displaystyle r(m \otimes n) = rm \otimes n$$ ... ... ... ... (1)

for tensor products.

Well ... have been reflecting ...

... ... I think that so long as we are careful about the nature of the elements of the tensor product, (1) above can be justified in terms of the D&F definitions/construction of tensor products of modules ... ...

If, in the construction of the tensor product we take N as a left R-module and M as an (S, R)-bimodule and the define the action f the left S-module $$\displaystyle M \otimes_R N$$ as in (10.8) D&F page 367 (see attachment), that is as follows:

$$\displaystyle s ( \sum_{Finite} m_i \otimes n_i = \sum_{Finite} (sm_i) \otimes n_i$$ ... ... ... (10.8)

Given the above, if, in (10.8) we take the elementary tensor $$\displaystyle m \otimes n$$ - that is a 'sum' of one tensor, we have:

$$\displaystyle s(m \otimes n) = sm \otimes n$$

But, further, if we take M as a standard R-module (that is an (R,R)-bimodule with rm = mr we can write (as D&F do near the bottom of page 367 (see attachment):

$$\displaystyle r(m \otimes n) = rm \otimes n = mr \otimes n = m \otimes rn$$

Can someone please confirm that the above is the correct (or otherwise) reasoning for the justification of

$$\displaystyle r ( m \otimes n) = rm \otimes n$$

Peter

#### Deveno

##### Well-known member
MHB Math Scholar
As I have remarked before, the tensor product needs "a ring in the middle" (and this is why we do algebra over commutative rings FIRST).

What is actually happening with extension of scalars is this:

($S,R$)-bimodule tensor (left) $R$-module = (left) $S$-module.

The commutativity of $S$ is essential, here: it is what allows the usual left action of $R$ to be considered a right action.

In general, the tensor product does this:

$-\otimes_R -: \mathbf{Mod-}R \times R\mathbf{-Mod} \to \mathbf{AbGrp}$

If the first "factor" is an ($R,R$)-bimodule, the tensor product has a natural (left) $R$ action.

As you might suspect, it makes sense to study tensor products of abelian groups before jumping on to more complicated rings than the integers.

#### Peter

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MHB Site Helper
As I have remarked before, the tensor product needs "a ring in the middle" (and this is why we do algebra over commutative rings FIRST).

What is actually happening with extension of scalars is this:

($S,R$)-bimodule tensor (left) $R$-module = (left) $S$-module.

The commutativity of $S$ is essential, here: it is what allows the usual left action of $R$ to be considered a right action.

In general, the tensor product does this:

$-\otimes_R -: \mathbf{Mod-}R \times R\mathbf{-Mod} \to \mathbf{AbGrp}$

If the first "factor" is an ($R,R$)-bimodule, the tensor product has a natural (left) $R$ action.

As you might suspect, it makes sense to study tensor products of abelian groups before jumping on to more complicated rings than the integers.

Thanks Deveno, ThePerfectHacker ... appreciate your help ...

Deveno, do you have a reference (that you think is good) that starts tensor products with tensor products of abelian groups?

I am still uncertain about some of D&Fs reasoning in Example 3 (Section 10.4, page 369)

In particular, in Example 3 D&F state the following:

"The induced map $$\displaystyle \Phi \ : \ \mathbb{Z} / mZ \otimes_\mathbb{Z} \mathbb{Z} / nZ \to \mathbb{Z} / dZ$$ from Corollary 12 (see D&F, page 368) maps $$\displaystyle 1 \otimes 1$$ to the element $$\displaystyle 1 \in \mathbb{Z} / dZ$$, which is an element of order d."

I am uncertain of why exactly $$\displaystyle \Phi$$ maps $$\displaystyle 1 \otimes 1$$.

My thinking is as follows:

On page 368, D&F write the following regarding $$\displaystyle \Phi$$ :

"Moreover, on simple tensors

$$\displaystyle \Phi ( (rm) \otimes n ) = \phi (rm, n) = r \phi (m,n) = r \Phi ( m \otimes n )$$"

Mind you, I am not entirely sure why this is true, but anyway, in particular we have:

$$\displaystyle \Phi ( (rm) \otimes n ) = r \phi (m,n)$$

and taking r = m = n = 1 in this relation we have

$$\displaystyle \Phi ( 1 \otimes_\mathbb{Z} 1) = 1 \cdot \phi (1,1)$$

and then since

$$\displaystyle \phi (a \ mod m, b \ mod n = ab mod d$$ we have

$$\displaystyle \Phi ( 1 \otimes_\mathbb{Z} 1) = 1 \cdot \phi (1,1) = \phi (1,1) = 1.1 = 1$$

Can someone please confirm that the above reasoning is OK?

Can someone also explain why the element $$\displaystyle 1 \in \mathbb{Z} / dZ$$ is an element of order d?

My thinking is that it is because $$\displaystyle d (1 \otimes_\mathbb{Z} 1) = 0$$ in $$\displaystyle \mathbb{Z} / mZ \otimes_\mathbb{Z} \mathbb{Z} / nZ$$ and this order is preserved through the operation of $$\displaystyle \Phi$$ - so 1 is of order d in $$\displaystyle \mathbb{Z} / dZ$$.

Can someone please confirm if this reasoning is correct?

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

"In particular $$\displaystyle \mathbb{Z} / mZ \otimes_\mathbb{Z} \mathbb{Z} / nZ$$ has order at least d"

Can someone explain why this is the case?

Finally, D&F write the following conclusion to the Example:

"Hence $$\displaystyle 1 \otimes 1$$ is an element of order d and $$\displaystyle \Phi$$ gives an isomorphism $$\displaystyle \mathbb{Z} / mZ \otimes_\mathbb{Z} \mathbb{Z} / nZ \cong \mathbb{Z} / dZ$$."

Can someone please explain how the two points in this text follow?

Would appreciate some help.

Peter

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#### Deveno

##### Well-known member
MHB Math Scholar
From the universal property of the tensor product (a lot of proofs will start this way, so get used to it) we have:

$\Phi\circ\otimes(a,b) = ab\text{ (mod }d)$

that is:

$\Phi(a\otimes b) = ab\text{ (mod }d)$, since the map:

$([a]_m,_n) \mapsto [ab]_d$ is $\Bbb Z$-bilinear.

Now we have:

$a\otimes b = a(1 \otimes b) = (ab)(1 \otimes 1)$ from the $\Bbb Z$-bilinearity of $\otimes$, which shows that $1 \otimes 1$ spans the tensor product (any sum of elementary tensors is a sum of integer multiples of the elementary tensor $1 \otimes 1$)

Remember that "order" in ($\Bbb Z_d,+$) is the least positive integer (for any element $[a]_d$) $k$ such that $k[a]_d = [ka]_d = [0]_d$.

Thus:

$d[1]_d = [d]_d = [0]_d$

and no other integer $1 \leq k < d$ has this property, since:

$[0]_d,[1]_d,[2]_d,\dots,[d-1]_d$ are all distinct.

It should be clear that $\Phi$ is onto, since the generator of $\Bbb Z_d$ (namely: $[1]_d$) is in the image of $\Phi$.

So it remains to be shown that $\text{ker }\Phi = \{0\}$.

Since we know $\Bbb Z_m \otimes \Bbb Z_n$ is cyclic (with generator $1 \otimes 1$), it clearly suffices to show $1 \otimes 1$ has order $d$. To do this, write:

$d = ra + db$, and note that this is the least positive $\Bbb Z$-linear combination of $a$ and $b$, by definition of the gcd.

#### Peter

##### Well-known member
MHB Site Helper
From the universal property of the tensor product (a lot of proofs will start this way, so get used to it) we have:

$\Phi\circ\otimes(a,b) = ab\text{ (mod }d)$

that is:

$\Phi(a\otimes b) = ab\text{ (mod }d)$, since the map:

$([a]_m,_n) \mapsto [ab]_d$ is $\Bbb Z$-bilinear.

Now we have:

$a\otimes b = a(1 \otimes b) = (ab)(1 \otimes 1)$ from the $\Bbb Z$-bilinearity of $\otimes$, which shows that $1 \otimes 1$ spans the tensor product (any sum of elementary tensors is a sum of integer multiples of the elementary tensor $1 \otimes 1$)

Remember that "order" in ($\Bbb Z_d,+$) is the least positive integer (for any element $[a]_d$) $k$ such that $k[a]_d = [ka]_d = [0]_d$.

Thus:

$d[1]_d = [d]_d = [0]_d$

and no other integer $1 \leq k < d$ has this property, since:

$[0]_d,[1]_d,[2]_d,\dots,[d-1]_d$ are all distinct.

It should be clear that $\Phi$ is onto, since the generator of $\Bbb Z_d$ (namely: $[1]_d$) is in the image of $\Phi$.

So it remains to be shown that $\text{ker }\Phi = \{0\}$.

Since we know $\Bbb Z_m \otimes \Bbb Z_n$ is cyclic (with generator $1 \otimes 1$), it clearly suffices to show $1 \otimes 1$ has order $d$. To do this, write:

$d = ra + db$, and note that this is the least positive $\Bbb Z$-linear combination of $a$ and $b$, by definition of the gcd.

Thanks Deveno, really appreciate your help.

Still working through and reflecting on your post ... but just one small issue ... related to the start of your post ...

You write:

"From the universal property of the tensor product (a lot of proofs will start this way, so get used to it) we have:

$\Phi\circ\otimes(a,b) = ab\text{ (mod }d)$

that is:

$\Phi(a\otimes b) = ab\text{ (mod }d)$, since the map:

$([a]_m,_n) \mapsto [ab]_d$ is $\Bbb Z$-bilinear."

My issue is understanding why $\Phi(a\otimes b) = ab\text{ (mod }d)$ depends on the fact that $([a]_m,_n) \mapsto [ab]_d$ is $\Bbb Z$-bilinear, that is on $$\displaystyle \phi$$ being $\Bbb Z$-bilinear ... ...

My understanding of why it is the case follows ...

We have $$\displaystyle \phi = \Phi \cdot \otimes$$ ... ... ... ... (1)

where $$\displaystyle \phi \ : \ \mathbb{Z} / mZ \times \mathbb{Z} / nZ \to \mathbb{Z} / dZ$$

is defined by

$$\displaystyle \phi ( a \text{ mod } m , b \text{ mod } n ) = ab \text{ mod } d$$ ... ... ... ... (2)

and $$\displaystyle \otimes \ : \ \mathbb{Z} / mZ \times \mathbb{Z} / nZ \to \mathbb{Z} / mZ \otimes_\mathbb{Z} \mathbb{Z} / nZ$$

is defined by

$$\displaystyle \otimes (a,b) = (a \otimes b)$$ ... ... ... ... ... (3)

Now, given the above, we have, by simple composition of functions (see (1) above) - indeed, by substituting (3) into (1) we have:

$$\displaystyle \phi ( [a]_m, _n ) = \Phi \cdot \otimes ( [a]_m, _n )$$ $$\displaystyle = \Phi ( [a]_m \otimes _n )$$ ... ... ... ... (4)

and we also have by (2) that

$$\displaystyle \phi ( [a]_m, _n ) = ab \text{ mod } d$$

So, then, (1) and (2) $$\displaystyle \Longrightarrow$$

$$\displaystyle \Phi (a \otimes b) = ab \text{ mod } d$$ ... ... (5)

My question is again ... that (5) seems to flow from a simple use of composition of functions ... so in what way does our conclusion depend on the $$\displaystyle \mathbb{Z}$$-bilinearity of $$\displaystyle \phi$$?

Can you help?

Peter

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#### Deveno

##### Well-known member
MHB Math Scholar
When you invoke the universal property of the tensor product, you do it to "factor" a bilinear map.

So, first, you establish that the map:

$\phi([a]_m,_n) = [ab]_d$ is a bilinear map. So first, we check "linearity in $a$":

$\phi([a]_m + [a']_m,_n) = \phi([a+a']_m,_n)$

$= [(a+a')b]_d = [ab + a'b]_d = [ab]_d + [a'b]_d$

$= \phi([a]_m,_n) + \phi([a']_m,_n)$

(I am writing the subscripts so it is clear which integers we are modding by).

Now take any $k \in \Bbb Z$:

$\phi(k[a]_m,_n) = \phi([ka]_m,_n)$

$= [(ka)b]_d = [k(ab)]_d = k[ab]_d = k\phi([a]_m,_n)$.

"Linearity in $b$" is done in much the same way, and I urge you to work through it one time.

Once we have bilinearity established, we can THEN write the composition you have in (1), which does indeed lead directly to (5).