# Existence of Tensor Products - Keith Conrad - Tensor Products I - Theorem 3.2

#### Peter

##### Well-known member
MHB Site Helper
I am reading and trying to follow the notes of Keith Conrad on Tensor products, specifically his notes: Tensor Products I (see attachment ... for the full set of notes see Expository papers by K. Conrad ).

I would appreciate some help with Theorem 3.2 which reads as follows: (see attachment page 7)

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Theorem 3.2. A tensor product of M and N exists.

(M and N are modules)

Proof: Consider $$\displaystyle M \times N$$ simply as a set. We form the free R-module on this set:

$$\displaystyle F_R (M \times N) = \bigoplus_{(m,n) \in M \times N} R \delta_{(m,n)}$$

(This is an enormous R-module. If $$\displaystyle R = M = N = \mathbb{R}$$ then $$\displaystyle F_{\mathbb{R}} ( \mathbb{R} \times \mathbb{R})$$ is a direct sum of $$\displaystyle \mathbb{R}^2$$-many copies of $$\displaystyle \mathbb{R}$$. The direct sum runs over all pairs of vectors, not just pairs coming from a basis, and for modules bases do not even usually exist)

Let D be the submodule of $$\displaystyle F_R (M \times N)$$ spanned by all the elements:

$$\displaystyle \delta_{(m + m', n)} - \delta_{(m,n)} - \delta_{(m',n)} , \ \ \delta_{(m, n + n')} - \delta_{(m,n)} - \delta_{(m,n')} , \ \$$

$$\displaystyle \delta_{(rm,n)} - \delta_{(m, rn)} , \ \ r \delta_{(m,n)} - \delta_{(rm,n)} , \ \ r\delta_{(m,n)} - \delta_{(m,rn)} , \ \$$

... ...

My problem with this is that Conrad does not define the symbols $$\displaystyle \delta_{(m,n)}$$ - can someone help me with their meaning, including the meaning of them in the direct sum above?

Such a clarification would be most helpful.

Peter

#### Deveno

##### Well-known member
MHB Math Scholar
It's just a notational symbol to avoid too much complexity.

Basically, $F_R(M \times N)$ takes every pair $(m,n) \in M \times N$ as a generating element.

Suppose we just look at a very simple example: $M = N = \Bbb R$. A "simple element" of $F_{\Bbb R}(\Bbb R \times \Bbb R)$ might look like this:

$\pi(1,2) - \sqrt{3}(2,\frac{1}{2}) + 8(e,7)$

and that is all the simplification we can achieve.

Of course we CAN do this:

$\pi(1,2) + 3(1,2) = (\pi + 3)(1,2)$

because we have the "common term" (1,2).

The "freeness" of $F_R(M \times N)$ means we can only collect terms when $(m,n) = (m',n')$ (in other words, we are not pre-supposing any algebraic relationship or "structure" on the set $M \times N$ even if one is possible).

So even if we can endow $M \times N$ with an $R$-module structure, in the FREE module it is NOT true that:

$(m,n) + (m',n) = (m+m',n)$

the left-hand side can not be simplified (we are regarding EVERY pair $(m,n)$ as "$R$-linearly independent").

By setting $D$ as the submodule of all possible $R$-linear combinations of elements like:

$(m+m',n) - (m,n) - (m',n)$

and so on, we are basically guaranteeing bilnearity of the tensor product.

I think Conrad is using the symbol $\delta_{(m,n)}$ instead of $(m,n)$ so that one is not tempted to think we can do anything with differently tagged deltas.

Extra credit: show $F_{\Bbb R}(\Bbb R \times \Bbb R)/D$ is spanned by $(1,1) + D$.

Conclude that the tensor product over the real numbers of the real numbers with itself is just $\Bbb R$ under ordinary real multiplication.

We have various ways of making bigger modules out of smaller modules. The two most important are:

$M \oplus N$

This is linear in the sense that:

$r(m+n) = rm + rn$

(linearity can be thought of as a form of distributivity), and

$M \otimes N$

This is bilinear in the sense that:

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

(bilinearity can be thought of as a form of "factoring").

These operations take place in the category $R$-Mod (this is too large an entity to be called a set).

The whole idea is that the operation $\otimes$ turns bilinear functions:

$B:M \times N \to A$

into linear functions:

$L: M \otimes N \to A$

just like the projection functions of the product turn pairs of functions into a product function (which is what all the diagrams are about).

Universal constructions like these often involve a statement such as "$\phi$ factors through $\psi$" and we can prove that any two such universals so constructed are isomorphic by "chasing arrows" (producing the isomorphism is usually a "no-brainer").

#### Peter

##### Well-known member
MHB Site Helper
It's just a notational symbol to avoid too much complexity.

Basically, $F_R(M \times N)$ takes every pair $(m,n) \in M \times N$ as a generating element.

Suppose we just look at a very simple example: $M = N = \Bbb R$. A "simple element" of $F_{\Bbb R}(\Bbb R \times \Bbb R)$ might look like this:

$\pi(1,2) - \sqrt{3}(2,\frac{1}{2}) + 8(e,7)$

and that is all the simplification we can achieve.

Of course we CAN do this:

$\pi(1,2) + 3(1,2) = (\pi + 3)(1,2)$

because we have the "common term" (1,2).

The "freeness" of $F_R(M \times N)$ means we can only collect terms when $(m,n) = (m',n')$ (in other words, we are not pre-supposing any algebraic relationship or "structure" on the set $M \times N$ even if one is possible).

So even if we can endow $M \times N$ with an $R$-module structure, in the FREE module it is NOT true that:

$(m,n) + (m',n) = (m+m',n)$

the left-hand side can not be simplified (we are regarding EVERY pair $(m,n)$ as "$R$-linearly independent").

By setting $D$ as the submodule of all possible $R$-linear combinations of elements like:

$(m+m',n) - (m,n) - (m',n)$

and so on, we are basically guaranteeing bilnearity of the tensor product.

I think Conrad is using the symbol $\delta_{(m,n)}$ instead of $(m,n)$ so that one is not tempted to think we can do anything with differently tagged deltas.

Extra credit: show $F_{\Bbb R}(\Bbb R \times \Bbb R)/D$ is spanned by $(1,1) + D$.

Conclude that the tensor product over the real numbers of the real numbers with itself is just $\Bbb R$ under ordinary real multiplication.

We have various ways of making bigger modules out of smaller modules. The two most important are:

$M \oplus N$

This is linear in the sense that:

$r(m+n) = rm + rn$

(linearity can be thought of as a form of distributivity), and

$M \otimes N$

This is bilinear in the sense that:

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

(bilinearity can be thought of as a form of "factoring").

These operations take place in the category $R$-Mod (this is too large an entity to be called a set).

The whole idea is that the operation $\otimes$ turns bilinear functions:

$B:M \times N \to A$

into linear functions:

$L: M \otimes N \to A$

just like the projection functions of the product turn pairs of functions into a product function (which is what all the diagrams are about).

Universal constructions like these often involve a statement such as "$\phi$ factors through $\psi$" and we can prove that any two such universals so constructed are isomorphic by "chasing arrows" (producing the isomorphism is usually a "no-brainer").
Hi Deveno,

Thanks for a really helpful post ... still working through it ...

Just a point of clarification ...

In describing one way of making bigger modules out of smaller modules, you write:

" $M \oplus N$

This is linear in the sense that:

$r(m+n) = rm + rn$ "

In the above, I am assuming that you are referring to the direct product or external direct sum of M and N. Is that correct?

On page 353, Dummit and Foote define the direct product (external direct sum) as follows: (see attachment)

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Definition. Let $$\displaystyle M_1, M_2, ... \ ... ,M_k$$ be a collection of R-modules. The collection of k-tuples $$\displaystyle (m_1, m_2, ... \ ... ,m_k)$$ where $$\displaystyle m_i \in M_i$$ with addition and action of R defined componentwise is called the direct product of $$\displaystyle M_1, M_2, ... \ ... ,M_k$$, denoted $$\displaystyle M_1 \times M_2 \times ... \ ... \times M_k$$.

It is evident that the direct product of a collection or R-modules is again an R-module. The direct product of $$\displaystyle M_1, M_2, ... \ ... ,M_k$$ is also referred to as the (external) direct sum of $$\displaystyle M_1, M_2, ... \ ... ,M_k$$ and denoted $$\displaystyle M_1 \oplus M_2 \oplus ... \ ... \oplus M_k$$. The direct product and direct sum of an infinite product of modules (which are different in general) are defined in Exercise 20.

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So given the quote from D&F, the external direct sum for M and N would be

$$\displaystyle M \oplus N = \{(m,n) \ | \ m \in M, n \in N \}$$ ... ... (1)

where addition and the action of R on M and N is defined componentwise.

So then, following this definition and its notation, addition in $$\displaystyle M \oplus N$$ would take place as follows:

$$\displaystyle (m,n) +_{M \oplus N} (m',n') = (m +_{M} m', n +_{N} n')$$

where addition in M+N is denoted $$\displaystyle +_{M \oplus N}$$, and addition in M, N is denoted $$\displaystyle +_{M}$$ and $$\displaystyle +_{N}$$ respectively.

$$\displaystyle r \times_{M \oplus N} (m,n) = (r \times_{M} m, r \times_{N} n)$$

where the action of $$\displaystyle R$$ on $$\displaystyle M \oplus N$$ is denoted $$\displaystyle \times_{M \oplus N}$$ and the action of R on M is denoted $$\displaystyle \times_{M}$$, and similarly the action of R on N is denoted $$\displaystyle \times_{N}$$.

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Now I am trying to "tie up" my (excessively formal) notation with your notation.

We could write (I think) the external direct sum for M and N

$$\displaystyle M \oplus N = \{(m,n) \ | \ m \in M, n \in N \}$$

as

$$\displaystyle M \oplus N = \{ m \oplus n \ | \ m \in M, n \in N \}$$ ... ... (2)

Then we could write

$$\displaystyle r \times_{M \oplus N} (m,n)$$ as $$\displaystyle r(m \oplus n)$$

and

$$\displaystyle (r \times_{M} m, r \times_{N} n )$$ as $$\displaystyle rm \oplus rn$$

so we get

r(m+n) = rm + rn as you specify (if we drop the formalism of using $$\displaystyle \oplus$$ for addition in $$\displaystyle M \oplus N$$ and just use +.

Can you please confirm that the above reasoning is correct and that the notation is coherent and meaningful.

Peter

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#### Deveno

##### Well-known member
MHB Math Scholar
That looks right except it is more usual to write:

$r\cdot_Mm$

$r\times_Mm$

for the scalar product if you wish to emphasize which module the scalar product is defined for.

Usually this is suppressed though because we identify the scalar product:

$\cdot:R\times M \to M$

with the family of functions:

$r\cdot(\_):M \to M$ for each $r \in R$.

This is in keeping with thinking of scalar multiplication as an "action" of $R$ upon $M$ which means this:

the mapping $r \mapsto r\cdot(\_)$ is a ring homomorphism

$R \to \text{End}(M)$ (considering $M$ as just an abelian group)

(this is a more compact way of expressing the module axioms).

#### Peter

##### Well-known member
MHB Site Helper
That looks right except it is more usual to write:

$r\cdot_Mm$

$r\times_Mm$

for the scalar product if you wish to emphasize which module the scalar product is defined for.

Usually this is suppressed though because we identify the scalar product:

$\cdot:R\times M \to M$

with the family of functions:

$r\cdot(\_):M \to M$ for each $r \in R$.

This is in keeping with thinking of scalar multiplication as an "action" of $R$ upon $M$ which means this:

the mapping $r \mapsto r\cdot(\_)$ is a ring homomorphism

$R \to \text{End}(M)$ (considering $M$ as just an abelian group)

(this is a more compact way of expressing the module axioms).
Thanks again Deveno ... still reflecting n many things you have said regarding tensor products ...

Just another small issue ...

In the expression $$\displaystyle F_R (M \times N) = \bigoplus_{(m,n) \in M \times N} R \delta_{(m,n)}$$

would the external direct sum

$$\displaystyle \bigoplus_{(m,n) \in M \times N} R \delta_{(m,n)}$$

be equal to RA where

$$\displaystyle A = \{ (m,n) \ | \ m \in M , n \in N \}$$?

Is this correct?

If it is it may be a better way to express it ... but I am essentially just checking my understanding of the nature of $$\displaystyle \bigoplus_{(m,n) \in M \times N} R \delta_{(m,n)}$$

Peter

#### Deveno

##### Well-known member
MHB Math Scholar
No.

The set $RA$ just consists of elements $ra$ with $r \in R$, and $a \in A$.

We need to include elements like $r_1a_1 + r_2a_2$, which is why $R$ is "inside" the direct summation.

#### Peter

##### Well-known member
MHB Site Helper
No.

The set $RA$ just consists of elements $ra$ with $r \in R$, and $a \in A$.

We need to include elements like $r_1a_1 + r_2a_2$, which is why $R$ is "inside" the direct summation.

Thanks Deveno ... but I am finding the notation RA a bit confusing ...

In Dummit and Foote, Section 1.3 Generation of Modules, Direct Sums and Free Modules on page 351 we read the following:

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Definition. Let M be an R-module and let $$\displaystyle N_1, N_2, ... \ ... ,N_n$$ be submodules of M.

(1) ... ...

(2) For any subset A of M let

$$\displaystyle RA = \{ r_1a_1 + r_2a_2 \ + ... \ ... + \ r_ma_m \ | \ r_1, r_2, ... \ ... ,r_m \in R, \ a_1, a_2, ... \ ... ,a_m \in A, \ m \in \mathbb{Z}^{+} \}$$

where by convention $$\displaystyle RA = \{ 0 \}$$ if $$\displaystyle a = \emptyset$$

If A is the finite set $$\displaystyle \{ a_1, a_2, ... \ ... , a_n \}$$ we shall write $$\displaystyle Ra_1 + Ra_2 + ... \ ... + Ra_n$$ for $$\displaystyle RA$$.

Call RA the submodule of M generated by A.

If N is a submodule of M (possibly N = M) and N = RA, for some subset A of M, we call A a set of generators or generating set for N, and we say N is generated by A.

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So it seems that by D&F's notation that for finite sets A, we can use either $$\displaystyle RA$$ or $$\displaystyle Ra_1 + Ra_2 + ... \ ... + Ra_n$$ for a set/submodule containing all elements of the form $$\displaystyle r_1a_1 + r_2a_2 \ + ... \ ... + \ r_ma_m$$ but for infinite sets A, we use RA.

You write:

"The set $RA$ just consists of elements $ra$ with $r \in R$, and $a \in A$."

So I am uncertain about the notation ... can you clarify ....

Peter

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#### Deveno

##### Well-known member
MHB Math Scholar
That is because in a subset $A$ of a module $M$ we already have some $m$ we can replace $a_1 + a_2$ (by virtue of closure in $M$) with.

We don't want to do this with free objects, we're building up, not down. For any set $X$, the free module $F_R(X)$ is usually going to be lots, lots bigger than $X$, whereas in $M$ by $RA$ we mean some sub-module of $M$.

If we consider $A$ as just a set, then $RA \subset M$ will be some QUOTIENT of $F_R(A)$...there will be a UNIQUE ring homomoprhism $\phi:F_R(A) \to RA$ with $\phi(a) = a$

If, in $M$, we have $a^n = 1$, for example, then $a^n - 1$ will be in the kernel of $\phi$. If it were the case that $a_1 + a_2 = a_4$ in $M$, then $a_1 + a_2 - a_4$ would be in the kernel of $\phi$. But in $F_R(A)$ we DO NOT HAVE $a_1 + a_2 = a_4$.

Free objects tend to be "pretty big". The free group generated by even two elements is HUGE, but there exists groups generated by two elements as small as order 4. With free things, we "add everything we have to until we finally get closure", with objects WITHIN a given structure, we have closure at the outset.

You can think of the free module on a set $X$ this way:

For each $x \in X$, we create a module $Rx$ by taking formal sums of $x$ over $R$. So we have:

$1x = x$
$2x = (1 + 1)x = x + x$, etc.
.....
$(r+s)x = rx + sx$

this is isomorphic to $R$ of course, but it's "tagged by $x$".

If $X$ is finite, say $|X| = n$, the resulting free module is just isomorphic to $R^n$:

$r_1x_1 + r_2x_2 + \cdots +r_nx_n \mapsto (r_1,r_2,\dots,r_n)$

In other words, $X$ becomes a BASIS. We can't "combine" basis elements, they each live in their own "copy" of $R$.

In a module $M$, we may have several relations in $M$ between the generators in $A$. For example, $A$ may not be minimal, it could be that the set $A$ is $R$-linearly dependent.

Another example:

If $M$ is the $\Bbb Z$-module $\Bbb Z_3$ and $N$ is the $\Bbb Z$-module $\Bbb Z_4$, then our basis for $F_{\Bbb Z}(M \times N)$ has 12 elements, and the resulting free $\Bbb Z$-module is infinite, we can't say:

$3(1,2) = (0,2)$

what we get instead is isomorphic to 12 copies of $\Bbb Z$. We may as well call:

$\{(0,0),(1,0),(2,0),(0,1),(1,1),(2,1),(0,2),(1,2),(2,2),(0,3),(1,3),(2,3)\}$

the set:

$\{x_1,x_2,x_3,x_4,x_5,x_6,x_7,x_8,x_9,x_{10},x_{11},x_{12}\}$

we "forget" all the structure $\Bbb Z_3 \times \Bbb Z_4$ has as any other structure. It's just a SET, and the only salient algebraic feature of a set is its cardinality.

To create the tensor product, we're going to "force bilinearity" by "modding out" certain elements (setting them equal to the 0 of the free module). This is parallel to the process by where we "force finiteness" in $\Bbb Z_n$, by setting $n = 0$ in the free $\Bbb Z$-module $\Bbb Z$ (and we accomplish this by taking the QUOTIENT $\Bbb Z/(n)$, or in the terminology of free modules:

$F_{\Bbb Z}(\{1\})/D$

where $D = \langle 1 + 1 + \cdots + 1\rangle$, where we have $n$ summands.

Note "1" is just a formal symbol here, and any other formal symbol would do as well, (like "A" or "elephant").

Free objects are usually pretty awkward to deal with, they have a lot of stuff in them, and expressions can be tedious to write out long-hand. They usually express "the most general (insert structure type here) possible" and as such there's usually few short-cuts available for them (except the ones afforded by the structure axioms themselves, for example in the free group we can always replace $AA^{-1}$ with $e$).