Direct Products of Modules - Bland - Rings and Their Modules

[Math Amateur]

I am reading Paul E. Bland's book, Rings and Their Modules.

In Section 2.1: Direct Products and Direct Sums, Bland defines the direct product of a family of modules. He then, in Proposition 2.1.1 shows that there is a unique module homomorphism (or R-Linear mapping) from any particular R-module N to the direct product. He then re-defines the direct product.

My question is - what is going on? Why is Bland doing this?

Can anyone help explain what is going on here? What is the point of demonstrating that there is a unique homomorphism from any module to a given direct product? Further, why would one need to or want to demonstrate that every R-Module maps uniquely onto a direct product? Is this some way to show that the formally defined direct product actually exists? But even then how does the unique homomorphism assure this? What is the motivation for the proposition and the re-definition of the direct product?

Details of Bland's definition, proposition and re-definition follow.

The (first) definition of a direct product of modules is as follows:

https://www.physicsforums.com/attachments/2424

Proposition 2.1.1, preceded by an important definition, is as follow:

View attachment 2425

Finally, the following is Bland's re-definition of direct product in the light of Proposition 2.1.1.

View attachment 2426

Again, my question is - what is going on here? What is the motivation for Proposition 2.1.1 and what is achieved by the Proposition and the subsequent redefinition?

Hope someone can throw some light on what Bland is doing.

Peter

 

[Deveno]

Normally the direct product is defined like so:

We take two $R$-modules $M,N$ and define the underlying set of $M \times N$ to be the cartesian product of the underlying sets of the underlying sets of $M$ and $N$, and define the module addition and scalar multiplication "pointwise":

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

$r\cdot(m,n) = (r\cdot m,r \cdot n)$

Let's show that this is equivalent to the "re-definition".

First of all, we have the canonical projection $R$-module homomorphisms:

$p_1:M \times N \to M$

$p_2:M \times N \to N$

given by $p_1(m,n) = m$ and $p_2(m,n) = n$

Now suppose we have ANY pair of $R$-module homomorphisms $f:L \to M$, and $g: L \to N$.

I claim we can create an $R$-module homomorphism $h: L \to M \times N$ by defining:

$h(l) = (f(l),g(l))$

It should be clear that $h$ is uniquely determined by $f$ and $g$. This shows that the "normal definition" has the property named by the "re-definition".

On the other hand, suppose that we are given an $R$-module $P$ together with two maps:

$p_1:P \to M$
$p_2:P \to N$

such that for any $R$-module $L$, with any pairs of maps: $f: L \to M, g: L \to N$ we have a unique map:

$h: L \to P$ with:

$p_1h = f$
$p_2h = g$

I will show that $P \cong M \times N$ and that the $R$-module isomorphism $\theta:P \to M \times N$ is unique.

We define $\theta:P \to M \times N$ by $\theta(p) = (p_1(p),p_2(p))$. This is clearly an $R$-module homomorphism since $p_1,p_2$ are.

Consider the map $\psi: M \times N \to P$ given by $\psi (m,n) = h(m) + h'(n)$, where:

$h:M \to P$ (in other words $L = M$)
$h':N \to P$ (and here, $L = N$)

$p_1h = 1_M$ (here, we are taking $f = 1_M, g = 0$)
$p_2h = 0$

$p_1h' = 0$ (here we are taking $f = 0, g = 1_N$)
$p_2h' = 1_N$

(these maps are guaranteed to exist by our definition of $P$).

Then $\theta\psi(m,n) = \theta(\psi(m,n)) = \theta(h(m) + h'(n)) = \theta(h(m)) + \theta(h'(n))$

$(p_1(h(m)),p_2(h(m))) + (p_1(h'(n)),p_2(h'(n))) = (p_1h(m),p_1h(m)) + (p_1h'(n),p_2h'(n))$

$=(m,0) + (0,n) = (m,n)$.

This shows that $\theta$ has a right-inverse and is thus surjective.

To show $\theta$ is bijective, it suffices to show this right-inverse is unique.

To do this, let $L = M \times N$, with $f:M \times N \to M$ given by $f(m,n) = m$ and $g:M \times N \to N$ given by $g(m,n) = n$.

Then $p_1\psi(m,n) = p_1(h(m)) + p_1(h'(n)) = p_1h(m) + p_1h'(n) = m + 0 = m = f(m,n)$ and:

$p_2\psi(m,n) = p_2(h(m)) + p_2(h'(n)) = p_2h(n) + p_2h'(n) = 0 + n = n = g(m,n)$.

By the conditions of the "re-definition", then, the map $\psi$ is UNIQUE among ALL maps that have this property.

Now, suppose that $\phi:M \times N \to P$ with $\theta\phi = 1_{M \times N}$.

Then $\theta\phi(m,n) = (p_1\phi(m,n),p_2\phi(m,n)) = (m,n)$

so it follows by the definition of equality in $M \times N$, that:

$p_1\phi(m,n) = m = f(m,n)$
$p_2\phi(m,n) = n = g(m,n)$

that is:

$p_1\phi = f$
$p_2\phi = g$.

Since $\psi$ is the UNIQUE mapping having this property, it follows that $\phi = \psi$, so $\theta$ indeed has a unique right-inverse, and is thus bijective, and is thus an isomorphism.

It should (hopefully) be clear that the isomorphism $\theta$ is completely determined by $p_1,p_2$, and is thus unique.

*************

This shows that the "usual" definition of the product has the universal property, and that any $R$-module that has the universal property is (isomorphic to) the product. Now, to the larger question: why would we do this, why isn't the "standard" definition "good enough"?

Well, to define the "usual" way, we make use of "point-wise" addition and scalar multiplication. So to proceed, we need to know the "nitty gritty details" about how we add and act by scalars in the individual $R$-modules $M$ and $N$. Using the universal property definition, we only need to know about pairs of morphisms between the product and its "factors". So we've shifted the focus of our investigation from "elements" to "mappings". Often, we might deduce things about modules from just the mappings alone, without having to "look under the hood".

*************

I want to add in closing here that nothing changes if we deal with a direct product of 3 modules, or of arbitrarily finitely many, or even infinitely many. It is usually easier (notationally) to deal with an infinite number of factors by using an "index set" $I$. And it turns out that for a FINITE indexed set, the construction determined by the family of (indexed) projections, is the same as the construction determined by the family of (indexed) inclusions.

However, in the case that $I$ is an INFINITE set, this is NOT so: the reason being the "inclusion" construction (this is called the direct SUM) is "minimal" and is properly contained in the "projection" construction. I give a typical example:

Let $x$ be an element transcendental over a field $F$. We can regard the powers of $x$ as a set indexed by the natural numbers. It is clear that we have the inclusions:

$i_k: Fx^k \to F[x]$ given by: $\alpha x^k \mapsto 0+0x+\cdots+\alpha x^k$.

(Note that $Fx^k \cong F$ as an $F$-module).

It is clear that these inclusions satisfy the usual requirements:

if $f_k:Fx^k \to V$ is any family of (indexed) linear transformations then there exists a (unique) map:

$f:F[x] \to V$ given by:

$\displaystyle f\left(\sum_{k = 0}^n \alpha_kx^k\right) = \sum_{k=0}^n f_k(\alpha_kx^k)$

with $fi_k = f_k$.

One might be tempted, under these circumstances, to conclude that:

$\displaystyle \prod_{k = 0}^{\infty} Fx^k \cong F[x]$, but this would be untrue.

Consider the $F$-module $F[[x]]$ of formal power series in $x$, together with the linear transformations:

$f_k:F[[x]] \to Fx^k$ given by:

$\displaystyle f_k\left(\sum_{j = 0}^{\infty} \alpha_jx^j\right) = \alpha_kx^k$

If $F[x]$ was the product of all the $Fx^k$ (along with the usual projection functions onto the $k$-th term), it would follow that there would be a linear transformation $T: F[[x]] \to F[x]$ with:

$p_kT = f_k$ for all $k \in \Bbb N$.

In particular, for the series, $s = 1 + x + x^2 +\cdots$, we would have to have:

$p_kT(s) = x^k$, for all $k$, but as $T(s) \in F[x]$, we must have some $n \geq 0$ with $p_kT(s) = 0$ for all $k > n$ (since all polynomials are of finite degree).

In other words:

$F \oplus Fx \oplus Fx^2 \oplus \cdots \cong F[x]$ but:

$F \times Fx \times Fx^2 \times \cdots \cong F[[x]]$.

 

[Math Amateur]

Normally the direct product is defined like so:

We take two $R$-modules $M,N$ and define the underlying set of $M \times N$ to be the cartesian product of the underlying sets of the underlying sets of $M$ and $N$, and define the module addition and scalar multiplication "pointwise":

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

$r\cdot(m,n) = (r\cdot m,r \cdot n)$

Let's show that this is equivalent to the "re-definition".

First of all, we have the canonical projection $R$-module homomorphisms:

$p_1:M \times N \to M$

$p_2:M \times N \to N$

given by $p_1(m,n) = m$ and $p_2(m,n) = n$

Now suppose we have ANY pair of $R$-module homomorphisms $f:L \to M$, and $g: N \to L$.

I claim we can create an $R$-module homomorphism $h: L \to M \times N$ by defining:

$h(l) = (f(l),g(l))$

It should be clear that $h$ is uniquely determined by $f$ and $g$. This shows that the "normal definition" has the property named by the "re-definition".

On the other hand, suppose that we are given an $R$-module $P$ together with two maps:

$p_1:P \to M$
$p_2:P \to N$

such that for any $R$-module $L$, with any pairs of maps: $f: L \to M, g: L \to N$ we have a unique map:

$h: L \to P$ with:

$p_1h = f$
$p_2h = g$

I will show that $P \cong M \times N$ and that the $R$-module isomorphism $\theta:P \to M \times N$ is unique.

We define $\theta:P \to M \times N$ by $\theta(p) = (p_1(p),p_2(p))$. This is clearly an $R$-module homomorphism since $p_1,p_2$ are.

Consider the map $\psi: M \times N \to P$ given by $\psi (m,n) = h(m) + h'(n)$, where:

$h:M \to P$ (in other words $L = M$)
$h':N \to P$ (and here, $L = N$)

$p_1h = 1_M$ (here, we are taking $f = 1_M, g = 0$)
$p_2h = 0$

$p_1h' = 0$ (here we are taking $f = 0, g = 1_N$)
$p_2h' = 1_N$

(these maps are guaranteed to exist by our definition of $P$).

Then $\theta\psi(m,n) = \theta(\psi(m,n)) = \theta(h(m) + h'(n)) = \theta(h(m)) + \theta(h'(n))$

$(p_1(h(m)),p_2(h(m))) + (p_1(h'(n)),p_2(h'(n))) = (p_1h(m),p_1h(m)) + (p_1h'(n),p_2h'(n))$

$=(m,0) + (0,n) = (m,n)$.

This shows that $\theta$ has a right-inverse and is thus surjective.

To show $\theta$ is bijective, it suffices to show this right-inverse is unique.

To do this, let $L = M \times N$, with $f:M \times N \to M$ given by $f(m,n) = m$ and $g:M \times N \to N$ given by $g(m,n) = n$.

Then $p_1\psi(m,n) = p_1(h(m)) + p_1(h'(n)) = p_1h(m) + p_1h'(n) = m + 0 = m = f(m,n)$ and:

$p_2\psi(m,n) = p_2(h(m)) + p_2(h'(n)) = p_2h(n) + p_2h'(n) = 0 + n = n = g(m,n)$.

By the conditions of the "re-definition", then, the map $\psi$ is UNIQUE among ALL maps that have this property.

Now, suppose that $\phi:M \times N \to P$ with $\theta\phi = 1_{M \times N}$.

Then $\theta\phi(m,n) = (p_1\phi(m,n),p_2\phi(m,n)) = (m,n)$

so it follows by the definition of equality in $M \times N$, that:

$p_1\phi(m,n) = m = f(m,n)$
$p_2\phi(m,n) = n = g(m,n)$

that is:

$p_1\phi = f$
$p_2\phi = g$.

Since $\psi$ is the UNIQUE mapping having this property, it follows that $\phi = \psi$, so $\theta$ indeed has a unique right-inverse, and is thus bijective, and is thus an isomorphism.

It should (hopefully) be clear that the isomorphism $\theta$ is completely determined by $p_1,p_2$, and is thus unique.

*************

This shows that the "usual" definition of the product has the universal property, and that any $R$-module that has the universal property is (isomorphic to) the product. Now, to the larger question: why would we do this, why isn't the "standard" definition "good enough"?

Well, to define the "usual" way, we make use of "point-wise" addition and scalar multiplication. So to proceed, we need to know the "nitty gritty details" about how we add and act by scalars in the individual $R$-modules $M$ and $N$. Using the universal property definition, we only need to know about pairs of morphisms between the product and its "factors". So we've shifted the focus of our investigation from "elements" to "mappings". Often, we might deduce things about modules from just the mappings alone, without having to "look under the hood".

*************

I want to add in closing here that nothing changes if we deal with a direct product of 3 modules, or of arbitrarily finitely many, or even infinitely many. It is usually easier (notationally) to deal with an infinite number of factors by using an "index set" $I$. And it turns out that for a FINITE indexed set, the construction determined by the family of (indexed) projections, is the same as the construction determined by the family of (indexed) inclusions.

However, in the case that $I$ is an INFINITE set, this is NOT so: the reason being the "inclusion" construction (this is called the direct SUM) is "minimal" and is properly contained in the "projection" construction. I give a typical example:

Let $x$ be an element transcendental over a field $F$. We can regard the powers of $x$ as a set indexed by the natural numbers. It is clear that we have the inclusions:

$i_k: Fx^k \to F[x]$ given by: $\alpha x^k \mapsto 0+0x+\cdots+\alpha x^k$.

(Note that $Fx^k \cong F$ as an $F$-module).

It is clear that these inclusions satisfy the usual requirements:

if $f_k:Fx^k \to V$ is any family of (indexed) linear transformations then there exists a (unique) map:

$f:F[x] \to V$ given by:

$\displaystyle f\left(\sum_{k = 0}^n \alpha_kx^k\right) = \sum_{k=0}^n f_k(\alpha_kx^k)$

with $fi_k = f_k$.

One might be tempted, under these circumstances, to conclude that:

$\displaystyle \prod_{k = 0}^{\infty} Fx^k \cong F[x]$, but this would be untrue.

Consider the $F$-module $F[[x]]$ of formal power series in $x$, together with the linear transformations:

$f_k:F[[x]] \to Fx^k$ given by:

$\displaystyle f_k\left(\sum_{j = 0}^{\infty} \alpha_jx^j\right) = \alpha_kx^k$

If $F[x]$ was the product of all the $Fx^k$ (along with the usual projection functions onto the $k$-th term), it would follow that there would be a linear transformation $T: F[[x]] \to F[x]$ with:

$p_kT = f_k$ for all $k \in \Bbb N$.

In particular, for the series, $s = 1 + x + x^2 +\cdots$, we would have to have:

$p_kT(s) = x^k$, for all $k$, but as $T(s) \in F[x]$, we must have some $n \geq 0$ with $p_kT(s) = 0$ for all $k > n$ (since all polynomials are of finite degree).

In other words:

$F \oplus Fx \oplus Fx^2 \oplus \cdots \cong F[x]$ but:

$F \times Fx \times Fx^2 \times \cdots \cong F[[x]]$.

Thanks so much for this post Deveno ... it happens to address quite a number of issues that were puzzling me ... thank you ...

Just working carefully through your post ...

Just a trivial question regarding what may be a typo ... or a misunderstanding of mine ...

You write:

"Now suppose we have ANY pair of $R$-module homomorphisms $f:L \to M$, and $g: N \to L$."

Did you mean \(\displaystyle g: L \to N \)?

Peter

 

[Deveno]

Uhh...yep.

 

[Math Amateur]

Normally the direct product is defined like so:

We take two $R$-modules $M,N$ and define the underlying set of $M \times N$ to be the cartesian product of the underlying sets of the underlying sets of $M$ and $N$, and define the module addition and scalar multiplication "pointwise":

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

$r\cdot(m,n) = (r\cdot m,r \cdot n)$

Let's show that this is equivalent to the "re-definition".

First of all, we have the canonical projection $R$-module homomorphisms:

$p_1:M \times N \to M$

$p_2:M \times N \to N$

given by $p_1(m,n) = m$ and $p_2(m,n) = n$

Now suppose we have ANY pair of $R$-module homomorphisms $f:L \to M$, and $g: L \to N$.

I claim we can create an $R$-module homomorphism $h: L \to M \times N$ by defining:

$h(l) = (f(l),g(l))$

It should be clear that $h$ is uniquely determined by $f$ and $g$. This shows that the "normal definition" has the property named by the "re-definition".

On the other hand, suppose that we are given an $R$-module $P$ together with two maps:

$p_1:P \to M$
$p_2:P \to N$

such that for any $R$-module $L$, with any pairs of maps: $f: L \to M, g: L \to N$ we have a unique map:

$h: L \to P$ with:

$p_1h = f$
$p_2h = g$

I will show that $P \cong M \times N$ and that the $R$-module isomorphism $\theta:P \to M \times N$ is unique.

We define $\theta:P \to M \times N$ by $\theta(p) = (p_1(p),p_2(p))$. This is clearly an $R$-module homomorphism since $p_1,p_2$ are.

Consider the map $\psi: M \times N \to P$ given by $\psi (m,n) = h(m) + h'(n)$, where:

$h:M \to P$ (in other words $L = M$)
$h':N \to P$ (and here, $L = N$)

$p_1h = 1_M$ (here, we are taking $f = 1_M, g = 0$)
$p_2h = 0$

$p_1h' = 0$ (here we are taking $f = 0, g = 1_N$)
$p_2h' = 1_N$

(these maps are guaranteed to exist by our definition of $P$).

Then $\theta\psi(m,n) = \theta(\psi(m,n)) = \theta(h(m) + h'(n)) = \theta(h(m)) + \theta(h'(n))$

$(p_1(h(m)),p_2(h(m))) + (p_1(h'(n)),p_2(h'(n))) = (p_1h(m),p_1h(m)) + (p_1h'(n),p_2h'(n))$

$=(m,0) + (0,n) = (m,n)$.

This shows that $\theta$ has a right-inverse and is thus surjective.

To show $\theta$ is bijective, it suffices to show this right-inverse is unique.

To do this, let $L = M \times N$, with $f:M \times N \to M$ given by $f(m,n) = m$ and $g:M \times N \to N$ given by $g(m,n) = n$.

Then $p_1\psi(m,n) = p_1(h(m)) + p_1(h'(n)) = p_1h(m) + p_1h'(n) = m + 0 = m = f(m,n)$ and:

$p_2\psi(m,n) = p_2(h(m)) + p_2(h'(n)) = p_2h(n) + p_2h'(n) = 0 + n = n = g(m,n)$.

By the conditions of the "re-definition", then, the map $\psi$ is UNIQUE among ALL maps that have this property.

Now, suppose that $\phi:M \times N \to P$ with $\theta\phi = 1_{M \times N}$.

Then $\theta\phi(m,n) = (p_1\phi(m,n),p_2\phi(m,n)) = (m,n)$

so it follows by the definition of equality in $M \times N$, that:

$p_1\phi(m,n) = m = f(m,n)$
$p_2\phi(m,n) = n = g(m,n)$

that is:

$p_1\phi = f$
$p_2\phi = g$.

Since $\psi$ is the UNIQUE mapping having this property, it follows that $\phi = \psi$, so $\theta$ indeed has a unique right-inverse, and is thus bijective, and is thus an isomorphism.

It should (hopefully) be clear that the isomorphism $\theta$ is completely determined by $p_1,p_2$, and is thus unique.

*************

This shows that the "usual" definition of the product has the universal property, and that any $R$-module that has the universal property is (isomorphic to) the product. Now, to the larger question: why would we do this, why isn't the "standard" definition "good enough"?

Well, to define the "usual" way, we make use of "point-wise" addition and scalar multiplication. So to proceed, we need to know the "nitty gritty details" about how we add and act by scalars in the individual $R$-modules $M$ and $N$. Using the universal property definition, we only need to know about pairs of morphisms between the product and its "factors". So we've shifted the focus of our investigation from "elements" to "mappings". Often, we might deduce things about modules from just the mappings alone, without having to "look under the hood".

*************

I want to add in closing here that nothing changes if we deal with a direct product of 3 modules, or of arbitrarily finitely many, or even infinitely many. It is usually easier (notationally) to deal with an infinite number of factors by using an "index set" $I$. And it turns out that for a FINITE indexed set, the construction determined by the family of (indexed) projections, is the same as the construction determined by the family of (indexed) inclusions.

However, in the case that $I$ is an INFINITE set, this is NOT so: the reason being the "inclusion" construction (this is called the direct SUM) is "minimal" and is properly contained in the "projection" construction. I give a typical example:

Let $x$ be an element transcendental over a field $F$. We can regard the powers of $x$ as a set indexed by the natural numbers. It is clear that we have the inclusions:

$i_k: Fx^k \to F[x]$ given by: $\alpha x^k \mapsto 0+0x+\cdots+\alpha x^k$.

(Note that $Fx^k \cong F$ as an $F$-module).

It is clear that these inclusions satisfy the usual requirements:

if $f_k:Fx^k \to V$ is any family of (indexed) linear transformations then there exists a (unique) map:

$f:F[x] \to V$ given by:

$\displaystyle f\left(\sum_{k = 0}^n \alpha_kx^k\right) = \sum_{k=0}^n f_k(\alpha_kx^k)$

with $fi_k = f_k$.

One might be tempted, under these circumstances, to conclude that:

$\displaystyle \prod_{k = 0}^{\infty} Fx^k \cong F[x]$, but this would be untrue.

Consider the $F$-module $F[[x]]$ of formal power series in $x$, together with the linear transformations:

$f_k:F[[x]] \to Fx^k$ given by:

$\displaystyle f_k\left(\sum_{j = 0}^{\infty} \alpha_jx^j\right) = \alpha_kx^k$

If $F[x]$ was the product of all the $Fx^k$ (along with the usual projection functions onto the $k$-th term), it would follow that there would be a linear transformation $T: F[[x]] \to F[x]$ with:

$p_kT = f_k$ for all $k \in \Bbb N$.

In particular, for the series, $s = 1 + x + x^2 +\cdots$, we would have to have:

$p_kT(s) = x^k$, for all $k$, but as $T(s) \in F[x]$, we must have some $n \geq 0$ with $p_kT(s) = 0$ for all $k > n$ (since all polynomials are of finite degree).

In other words:

$F \oplus Fx \oplus Fx^2 \oplus \cdots \cong F[x]$ but:

$F \times Fx \times Fx^2 \times \cdots \cong F[[x]]$.

Thanks again Deveno, your post goes to the heart of the matter for me in this situation ...

Just working through your post carefully ... just a point ... hope I am not being to pedantic ... but ... you write:

"We take two $R$-modules $M,N$ ... ... "

and then you write:

"Now suppose we have ANY pair of $R$-module homomorphisms $f:L \to M$, and $g: L \to N$. ... ... "

My question is as follows:

How do we know for a given L that such homomorphisms f, g exist ... indeed how do we know they exist for every module L ...

Are you able to clarify this point ...

Peter

 

[Deveno]

Well, for ANY $R$-module $L$, we always have the 0-maps. Usually, the availability of maps isn't that big a problem. Remember as well, that we are talking about ANY possible module $L$, so some $L$'s might have a lot of such maps, and some $L$'s might not have so many. It doesn't matter, what we are saying is that given such a pair of maps, we can make a map $L \to M \times N$ such that our pair of maps "factor through the projections": taking the $M$ part of the pair is the same as "projecting onto $M$ the map we say exists" and likewise for $N$.

The map we are asserting the existence of is often called "the product of $f$ and $g$", and written:

$(f \times g)(l) = (f(l),g(l))$.

In other words, we are really just talking about "pairs" of morphisms, and the product is the "universal way" (default canonical construction) to make this happen. In other words we split the pair of maps like so:

$L \stackrel{f}{\to} M$
$L \stackrel{g}{\to} N$

as:

$L \stackrel{h}{\to} M \times N \stackrel{p_1}{\to} M$
$L \stackrel{h}{\to} M \times N \stackrel{p_2}{\to} N$

As you can see, the product serves as "the middleman" in pairing $f$ and $g$. It doesn't matter which 2 maps we are pairing, the SAME $R$-module works as a "universal" middleman.

 

[Math Amateur]

Well, for ANY $R$-module $L$, we always have the 0-maps. Usually, the availability of maps isn't that big a problem. Remember as well, that we are talking about ANY possible module $L$, so some $L$'s might have a lot of such maps, and some $L$'s might not have so many. It doesn't matter, what we are saying is that given such a pair of maps, we can make a map $L \to M \times N$ such that our pair of maps "factor through the projections": taking the $M$ part of the pair is the same as "projecting onto $M$ the map we say exists" and likewise for $N$.

The map we are asserting the existence of is often called "the product of $f$ and $g$", and written:

$(f \times g)(l) = (f(l),g(l))$.

In other words, we are really just talking about "pairs" of morphisms, and the product is the "universal way" (default canonical construction) to make this happen. In other words we split the pair of maps like so:

$L \stackrel{f}{\to} M$
$L \stackrel{g}{\to} M$

as:

$L \stackrel{h}{\to} M \times N \stackrel{p_1}{\to} M$
$L \stackrel{h}{\to} M \times N \stackrel{p_2}{\to} M$

As you can see, the product serves as "the middleman" in pairing $f$ and $g$. It doesn't matter which 2 maps we are pairing, the SAME $R$-module works as a "universal" middleman.

Thanks Deveno ... anyone who wants to fully understand the construction of direct products should read your last two posts ... better than the texts I have been reading ...

But just checking a possible typo ...

You write:

"In other words we split the pair of maps like so:

$L \stackrel{f}{\to} M$
$L \stackrel{g}{\to} M$ ... ..."

Should this be as follows:

"In other words we split the pair of maps like so:

$L \stackrel{f}{\to} M$
$L \stackrel{g}{\to} N$ ..."

A similar typo occurs in the following:

"as:

$L \stackrel{h}{\to} M \times N \stackrel{p_1}{\to} M$
$L \stackrel{h}{\to} M \times N \stackrel{p_2}{\to} M$ ... "

which I think should read:

"as:

$L \stackrel{h}{\to} M \times N \stackrel{p_1}{\to} M$
$L \stackrel{h}{\to} M \times N \stackrel{p_2}{\to} N$"

Mind you the two typos (I think they are typos anyway - please let me know if they are not ...) are very minor in what is actually a very valuable post to me ...

Peter

PS Probably getting late in Texas ... while it is only mid-afternoon (the next day) here in Tasmania ... :)

 

[Deveno]

Indeed, I saw those typos just after I posted. Derp.

 

[Math Amateur]

Normally the direct product is defined like so:

We take two $R$-modules $M,N$ and define the underlying set of $M \times N$ to be the cartesian product of the underlying sets of the underlying sets of $M$ and $N$, and define the module addition and scalar multiplication "pointwise":

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

$r\cdot(m,n) = (r\cdot m,r \cdot n)$

Let's show that this is equivalent to the "re-definition".

First of all, we have the canonical projection $R$-module homomorphisms:

$p_1:M \times N \to M$

$p_2:M \times N \to N$

given by $p_1(m,n) = m$ and $p_2(m,n) = n$

Now suppose we have ANY pair of $R$-module homomorphisms $f:L \to M$, and $g: L \to N$.

I claim we can create an $R$-module homomorphism $h: L \to M \times N$ by defining:

$h(l) = (f(l),g(l))$

It should be clear that $h$ is uniquely determined by $f$ and $g$. This shows that the "normal definition" has the property named by the "re-definition".

On the other hand, suppose that we are given an $R$-module $P$ together with two maps:

$p_1:P \to M$
$p_2:P \to N$

such that for any $R$-module $L$, with any pairs of maps: $f: L \to M, g: L \to N$ we have a unique map:

$h: L \to P$ with:

$p_1h = f$
$p_2h = g$

I will show that $P \cong M \times N$ and that the $R$-module isomorphism $\theta:P \to M \times N$ is unique.

We define $\theta:P \to M \times N$ by $\theta(p) = (p_1(p),p_2(p))$. This is clearly an $R$-module homomorphism since $p_1,p_2$ are.

Consider the map $\psi: M \times N \to P$ given by $\psi (m,n) = h(m) + h'(n)$, where:

$h:M \to P$ (in other words $L = M$)
$h':N \to P$ (and here, $L = N$)

$p_1h = 1_M$ (here, we are taking $f = 1_M, g = 0$)
$p_2h = 0$

$p_1h' = 0$ (here we are taking $f = 0, g = 1_N$)
$p_2h' = 1_N$

(these maps are guaranteed to exist by our definition of $P$).

Then $\theta\psi(m,n) = \theta(\psi(m,n)) = \theta(h(m) + h'(n)) = \theta(h(m)) + \theta(h'(n))$

$(p_1(h(m)),p_2(h(m))) + (p_1(h'(n)),p_2(h'(n))) = (p_1h(m),p_1h(m)) + (p_1h'(n),p_2h'(n))$

$=(m,0) + (0,n) = (m,n)$.

This shows that $\theta$ has a right-inverse and is thus surjective.

To show $\theta$ is bijective, it suffices to show this right-inverse is unique.

To do this, let $L = M \times N$, with $f:M \times N \to M$ given by $f(m,n) = m$ and $g:M \times N \to N$ given by $g(m,n) = n$.

Then $p_1\psi(m,n) = p_1(h(m)) + p_1(h'(n)) = p_1h(m) + p_1h'(n) = m + 0 = m = f(m,n)$ and:

$p_2\psi(m,n) = p_2(h(m)) + p_2(h'(n)) = p_2h(n) + p_2h'(n) = 0 + n = n = g(m,n)$.

By the conditions of the "re-definition", then, the map $\psi$ is UNIQUE among ALL maps that have this property.

Now, suppose that $\phi:M \times N \to P$ with $\theta\phi = 1_{M \times N}$.

Then $\theta\phi(m,n) = (p_1\phi(m,n),p_2\phi(m,n)) = (m,n)$

so it follows by the definition of equality in $M \times N$, that:

$p_1\phi(m,n) = m = f(m,n)$
$p_2\phi(m,n) = n = g(m,n)$

that is:

$p_1\phi = f$
$p_2\phi = g$.

Since $\psi$ is the UNIQUE mapping having this property, it follows that $\phi = \psi$, so $\theta$ indeed has a unique right-inverse, and is thus bijective, and is thus an isomorphism.

It should (hopefully) be clear that the isomorphism $\theta$ is completely determined by $p_1,p_2$, and is thus unique.

*************

This shows that the "usual" definition of the product has the universal property, and that any $R$-module that has the universal property is (isomorphic to) the product. Now, to the larger question: why would we do this, why isn't the "standard" definition "good enough"?

Well, to define the "usual" way, we make use of "point-wise" addition and scalar multiplication. So to proceed, we need to know the "nitty gritty details" about how we add and act by scalars in the individual $R$-modules $M$ and $N$. Using the universal property definition, we only need to know about pairs of morphisms between the product and its "factors". So we've shifted the focus of our investigation from "elements" to "mappings". Often, we might deduce things about modules from just the mappings alone, without having to "look under the hood".

*************

I want to add in closing here that nothing changes if we deal with a direct product of 3 modules, or of arbitrarily finitely many, or even infinitely many. It is usually easier (notationally) to deal with an infinite number of factors by using an "index set" $I$. And it turns out that for a FINITE indexed set, the construction determined by the family of (indexed) projections, is the same as the construction determined by the family of (indexed) inclusions.

However, in the case that $I$ is an INFINITE set, this is NOT so: the reason being the "inclusion" construction (this is called the direct SUM) is "minimal" and is properly contained in the "projection" construction. I give a typical example:

Let $x$ be an element transcendental over a field $F$. We can regard the powers of $x$ as a set indexed by the natural numbers. It is clear that we have the inclusions:

$i_k: Fx^k \to F[x]$ given by: $\alpha x^k \mapsto 0+0x+\cdots+\alpha x^k$.

(Note that $Fx^k \cong F$ as an $F$-module).

It is clear that these inclusions satisfy the usual requirements:

if $f_k:Fx^k \to V$ is any family of (indexed) linear transformations then there exists a (unique) map:

$f:F[x] \to V$ given by:

$\displaystyle f\left(\sum_{k = 0}^n \alpha_kx^k\right) = \sum_{k=0}^n f_k(\alpha_kx^k)$

with $fi_k = f_k$.

One might be tempted, under these circumstances, to conclude that:

$\displaystyle \prod_{k = 0}^{\infty} Fx^k \cong F[x]$, but this would be untrue.

Consider the $F$-module $F[[x]]$ of formal power series in $x$, together with the linear transformations:

$f_k:F[[x]] \to Fx^k$ given by:

$\displaystyle f_k\left(\sum_{j = 0}^{\infty} \alpha_jx^j\right) = \alpha_kx^k$

If $F[x]$ was the product of all the $Fx^k$ (along with the usual projection functions onto the $k$-th term), it would follow that there would be a linear transformation $T: F[[x]] \to F[x]$ with:

$p_kT = f_k$ for all $k \in \Bbb N$.

In particular, for the series, $s = 1 + x + x^2 +\cdots$, we would have to have:

$p_kT(s) = x^k$, for all $k$, but as $T(s) \in F[x]$, we must have some $n \geq 0$ with $p_kT(s) = 0$ for all $k > n$ (since all polynomials are of finite degree).

In other words:

$F \oplus Fx \oplus Fx^2 \oplus \cdots \cong F[x]$ but:

$F \times Fx \times Fx^2 \times \cdots \cong F[[x]]$.

Thanks again Deveno ... still working through this very helpful post ... trying to ensure that I fully understand the post, so I hope you will forgive me trying to clarify some issues ...

In the post you write:

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

"To show $\theta$ is bijective, it suffices to show this right-inverse is unique.

To do this, let $L = M \times N$, with $f:M \times N \to M$ given by $f(m,n) = m$ and $g:M \times N \to N$ given by $g(m,n) = n$.

Then $p_1\psi(m,n) = p_1(h(m)) + p_1(h'(n)) = p_1h(m) + p_1h'(n) = m + 0 = m = f(m,n)$ and:

$p_2\psi(m,n) = p_2(h(m)) + p_2(h'(n)) = p_2h(n) + p_2h'(n) = 0 + n = n = g(m,n)$.

By the conditions of the "re-definition", then, the map $\psi$ is UNIQUE among ALL maps that have this property."

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

So you are showing that \(\displaystyle \psi \) is unique ... and essentially you show that \(\displaystyle p_1 \psi = f \) and \(\displaystyle p_2 \psi = g \) ... and then you say that \(\displaystyle \psi \) is unique by the conditions of the "re-definition" (i.e. Bland's Definition 2.1.2) ... BUT ... what exactly are the conditions to which you refer?

You also state that the map \(\displaystyle \psi \) is unique among all maps sharing this property ... but what is the property to which you refer?

Can you help clarify these issues?

Peter

 

[Deveno]

The property in question is THIS one:

If:

$L$ is any $R$-module, and $f,g$ are any pair of mappings (i.e., $R$-module homomorphisms):

$f:L \to M$
$g:L \to N$

then:

there is a UNIQUE $R$-module homomorphism $h: L \to P$ with:

$p_1h = f$
$p_2h = g$.

In the matter at hand, we are taking $L = M \times N$ and showing any right inverse of the specific mapping:

$\theta:P \to M \times N$ given by: $\theta(p) = (p_1(p),p_2(p))$

has this property ($f$ and $g$ are just such a right-inverse "factored through the projections").

If you can wade through (what is in my opinion) the tedious abstraction, what we are really saying is:

for the particular $L = M \times N$, the mappings $f,g$ and $p_1,p_2$ are "really the same, just the names have changed".

The "universal property" of the "structure maps" (i.e., the canonical projections from the product to the factors) and the associated "structure (i.e., the direct product) says that the map we make from $L$ (which is typically called $f \times g$ when $P = M \times N$) is unique.

*********************

Let's look at a simpler example of this same thing: in the realm of sets, say we have two sets: $A$ and $B$.

For any element $(a,b)$, we can do two things without any "extra knowledge" of which particular sets we have: we can select the first coordinate, or select the second coordinate (well, we could also do some other things, that aren't quite as illuminating: select both elements (this is just the identity function), or reverse the order (which gives us a rather dull bijection), or send every pair to some constant value (there are, of course, LOTS of "constant functions" possible, but they are all rather distressingly alike)).

So with the cartesian product, $A \times B$, we get two functions "for free":

$p_1:A \times B \to A$ given by $p_1((a,b)) = a$
$p_2:A \times B \to B$ given by $p_2((a,b)) = b$.

Now, what does a function $f:C \to A \times B$ look like? It looks like this:

$f(c) = (a_c,b_c)$.

We get "two functions for free" out of this as well:

$f_1:C \to A$ which sends $c \mapsto a_c$
$f_2:C \to B$ which sends $c \mapsto b_c$.

Note that we can re-state this fact as:

$f_1 = p_1f$
$f_2 = p_2f$ (*)

So all we are really saying is "given the coordinate functions" $f_1,f_2$, we can "reconstitute $f$" as the unique function for which (*) holds.

And ALL we are doing with the direct product of $R$-modules is replacing "function" with "$R$-module homomorphism" and "set" with "$R$-module".

In other words we are replacing an INTERNAL description of how the direct product acts based on its "components", with an EXTERNAL characterization based on certain homomorphisms going to and from it.

This can be compared to the two different characterizations of ideals in rings:

Internal description: An ideal is an additive subgroup of a ring with certain multiplicative properties.

External description: An ideal is a kernel of a surjective homomorphism.

With ideals, to prove the first condition holds, you have to pick "typical elements". For the second condition, you just look at maps.

 

[Math Amateur]

The property in question is THIS one:

If:

$L$ is any $R$-module, and $f,g$ are any pair of mappings (i.e., $R$-module homomorphisms):

$f:L \to M$
$g:L \to N$

then:

there is a UNIQUE $R$-module homomorphism $h: L \to P$ with:

$p_1h = f$
$p_2h = g$.

In the matter at hand, we are taking $L = M \times N$ and showing any right inverse of the specific mapping:

$\theta:P \to M \times N$ given by: $\theta(p) = (p_1(p),p_2(p))$

has this property ($f$ and $g$ are just such a right-inverse "factored through the projections").

If you can wade through (what is in my opinion) the tedious abstraction, what we are really saying is:

for the particular $L = M \times N$, the mappings $f,g$ and $p_1,p_2$ are "really the same, just the names have changed".

The "universal property" of the "structure maps" (i.e., the canonical projections from the product to the factors) and the associated "structure (i.e., the direct product) says that the map we make from $L$ (which is typically called $f \times g$ when $P = M \times N$) is unique.

*********************

Let's look at a simpler example of this same thing: in the realm of sets, say we have two sets: $A$ and $B$.

For any element $(a,b)$, we can do two things without any "extra knowledge" of which particular sets we have: we can select the first coordinate, or select the second coordinate (well, we could also do some other things, that aren't quite as illuminating: select both elements (this is just the identity function), or reverse the order (which gives us a rather dull bijection), or send every pair to some constant value (there are, of course, LOTS of "constant functions" possible, but they are all rather distressingly alike)).

So with the cartesian product, $A \times B$, we get two functions "for free":

$p_1:A \times B \to A$ given by $p_1((a,b)) = a$
$p_2:A \times B \to B$ given by $p_2((a,b)) = b$.

Now, what does a function $f:C \to A \times B$ look like? It looks like this:

$f(c) = (a_c,b_c)$.

We get "two functions for free" out of this as well:

$f_1:C \to A$ which sends $c \mapsto a_c$
$f_2:C \to B$ which sends $c \mapsto b_c$.

Note that we can re-state this fact as:

$f_1 = p_1f$
$f_2 = p_2f$ (*)

So all we are really saying is "given the coordinate functions" $f_1,f_2$, we can "reconstitute $f$" as the unique function for which (*) holds.

And ALL we are doing with the direct product of $R$-modules is replacing "function" with "$R$-module homomorphism" and "set" with "$R$-module".

In other words we are replacing an INTERNAL description of how the direct product acts based on its "components", with an EXTERNAL characterization based on certain homomorphisms going to and from it.

This can be compared to the two different characterizations of ideals in rings:

Internal description: An ideal is an additive subgroup of a ring with certain multiplicative properties.

External description: An ideal is a kernel of a surjective homomorphism.

With ideals, to prove the first condition holds, you have to pick "typical elements". For the second condition, you just look at maps.

Thanks Deveno, that was an extremely helpful post ... especially your statement (and the examples following it):

"In other words we are replacing an INTERNAL description of how the direct product acts based on its "components", with an EXTERNAL characterization based on certain homomorphisms going to and from it."

That is the first statement that indicates a point to all this ... which you seem to indicate is a point that, to an extent anyway, you wonder about as well, given your comment:

"If you can wade through (what is in my opinion) the tedious abstraction, ... ... ... "

I must say that getting the point as to why Paul Bland was doing what he was doing was as big a puzzle to me as following the mechanics of the constructions.

Mind you an external description based on certain homomorphisms is well in line with category theory's imperative (obsession?) or tendency to describe everything in terms of the 'arrows' (morphisms) rather than depending to heavily on the nature of the objects ... at least I think that is a tendency or characteristic of category theory ...

Just a note in passing that I found the example of the sets very helpful as well.

If you ever write a book on rings and modules ... let me know ... ...

Thanks again!

Peter

 

[Math Amateur]

Normally the direct product is defined like so:

We take two $R$-modules $M,N$ and define the underlying set of $M \times N$ to be the cartesian product of the underlying sets of the underlying sets of $M$ and $N$, and define the module addition and scalar multiplication "pointwise":

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

$r\cdot(m,n) = (r\cdot m,r \cdot n)$

Let's show that this is equivalent to the "re-definition".

First of all, we have the canonical projection $R$-module homomorphisms:

$p_1:M \times N \to M$

$p_2:M \times N \to N$

given by $p_1(m,n) = m$ and $p_2(m,n) = n$

Now suppose we have ANY pair of $R$-module homomorphisms $f:L \to M$, and $g: L \to N$.

I claim we can create an $R$-module homomorphism $h: L \to M \times N$ by defining:

$h(l) = (f(l),g(l))$

It should be clear that $h$ is uniquely determined by $f$ and $g$. This shows that the "normal definition" has the property named by the "re-definition".

On the other hand, suppose that we are given an $R$-module $P$ together with two maps:

$p_1:P \to M$
$p_2:P \to N$

such that for any $R$-module $L$, with any pairs of maps: $f: L \to M, g: L \to N$ we have a unique map:

$h: L \to P$ with:

$p_1h = f$
$p_2h = g$

I will show that $P \cong M \times N$ and that the $R$-module isomorphism $\theta:P \to M \times N$ is unique.

We define $\theta:P \to M \times N$ by $\theta(p) = (p_1(p),p_2(p))$. This is clearly an $R$-module homomorphism since $p_1,p_2$ are.

Consider the map $\psi: M \times N \to P$ given by $\psi (m,n) = h(m) + h'(n)$, where:

$h:M \to P$ (in other words $L = M$)
$h':N \to P$ (and here, $L = N$)

$p_1h = 1_M$ (here, we are taking $f = 1_M, g = 0$)
$p_2h = 0$

$p_1h' = 0$ (here we are taking $f = 0, g = 1_N$)
$p_2h' = 1_N$

(these maps are guaranteed to exist by our definition of $P$).

Then $\theta\psi(m,n) = \theta(\psi(m,n)) = \theta(h(m) + h'(n)) = \theta(h(m)) + \theta(h'(n))$

$(p_1(h(m)),p_2(h(m))) + (p_1(h'(n)),p_2(h'(n))) = (p_1h(m),p_1h(m)) + (p_1h'(n),p_2h'(n))$

$=(m,0) + (0,n) = (m,n)$.

This shows that $\theta$ has a right-inverse and is thus surjective.

To show $\theta$ is bijective, it suffices to show this right-inverse is unique.

To do this, let $L = M \times N$, with $f:M \times N \to M$ given by $f(m,n) = m$ and $g:M \times N \to N$ given by $g(m,n) = n$.

Then $p_1\psi(m,n) = p_1(h(m)) + p_1(h'(n)) = p_1h(m) + p_1h'(n) = m + 0 = m = f(m,n)$ and:

$p_2\psi(m,n) = p_2(h(m)) + p_2(h'(n)) = p_2h(n) + p_2h'(n) = 0 + n = n = g(m,n)$.

By the conditions of the "re-definition", then, the map $\psi$ is UNIQUE among ALL maps that have this property.

Now, suppose that $\phi:M \times N \to P$ with $\theta\phi = 1_{M \times N}$.

Then $\theta\phi(m,n) = (p_1\phi(m,n),p_2\phi(m,n)) = (m,n)$

so it follows by the definition of equality in $M \times N$, that:

$p_1\phi(m,n) = m = f(m,n)$
$p_2\phi(m,n) = n = g(m,n)$

that is:

$p_1\phi = f$
$p_2\phi = g$.

Since $\psi$ is the UNIQUE mapping having this property, it follows that $\phi = \psi$, so $\theta$ indeed has a unique right-inverse, and is thus bijective, and is thus an isomorphism.

It should (hopefully) be clear that the isomorphism $\theta$ is completely determined by $p_1,p_2$, and is thus unique.

*************

This shows that the "usual" definition of the product has the universal property, and that any $R$-module that has the universal property is (isomorphic to) the product. Now, to the larger question: why would we do this, why isn't the "standard" definition "good enough"?

Well, to define the "usual" way, we make use of "point-wise" addition and scalar multiplication. So to proceed, we need to know the "nitty gritty details" about how we add and act by scalars in the individual $R$-modules $M$ and $N$. Using the universal property definition, we only need to know about pairs of morphisms between the product and its "factors". So we've shifted the focus of our investigation from "elements" to "mappings". Often, we might deduce things about modules from just the mappings alone, without having to "look under the hood".

*************

I want to add in closing here that nothing changes if we deal with a direct product of 3 modules, or of arbitrarily finitely many, or even infinitely many. It is usually easier (notationally) to deal with an infinite number of factors by using an "index set" $I$. And it turns out that for a FINITE indexed set, the construction determined by the family of (indexed) projections, is the same as the construction determined by the family of (indexed) inclusions.

However, in the case that $I$ is an INFINITE set, this is NOT so: the reason being the "inclusion" construction (this is called the direct SUM) is "minimal" and is properly contained in the "projection" construction. I give a typical example:

Let $x$ be an element transcendental over a field $F$. We can regard the powers of $x$ as a set indexed by the natural numbers. It is clear that we have the inclusions:

$i_k: Fx^k \to F[x]$ given by: $\alpha x^k \mapsto 0+0x+\cdots+\alpha x^k$.

(Note that $Fx^k \cong F$ as an $F$-module).

It is clear that these inclusions satisfy the usual requirements:

if $f_k:Fx^k \to V$ is any family of (indexed) linear transformations then there exists a (unique) map:

$f:F[x] \to V$ given by:

$\displaystyle f\left(\sum_{k = 0}^n \alpha_kx^k\right) = \sum_{k=0}^n f_k(\alpha_kx^k)$

with $fi_k = f_k$.

One might be tempted, under these circumstances, to conclude that:

$\displaystyle \prod_{k = 0}^{\infty} Fx^k \cong F[x]$, but this would be untrue.

Consider the $F$-module $F[[x]]$ of formal power series in $x$, together with the linear transformations:

$f_k:F[[x]] \to Fx^k$ given by:

$\displaystyle f_k\left(\sum_{j = 0}^{\infty} \alpha_jx^j\right) = \alpha_kx^k$

If $F[x]$ was the product of all the $Fx^k$ (along with the usual projection functions onto the $k$-th term), it would follow that there would be a linear transformation $T: F[[x]] \to F[x]$ with:

$p_kT = f_k$ for all $k \in \Bbb N$.

In particular, for the series, $s = 1 + x + x^2 +\cdots$, we would have to have:

$p_kT(s) = x^k$, for all $k$, but as $T(s) \in F[x]$, we must have some $n \geq 0$ with $p_kT(s) = 0$ for all $k > n$ (since all polynomials are of finite degree).

In other words:

$F \oplus Fx \oplus Fx^2 \oplus \cdots \cong F[x]$ but:

$F \times Fx \times Fx^2 \times \cdots \cong F[[x]]$.

Hi Deveno ... as I have mentioned before, your post was most helpful ... and to me the central point of your post was to show that the internal definition/construction of the direct product in terms of the definition of module addition and scalar multiplication is isomorphic to the external definition in terms of homomorphisms. ... ... indeed and further showing that this isomorphism is unique (although I do not think I yet understand the significance of the uniqueness ... ?) ... that is, your post demonstrates the following"

" ... ... $P \cong M \times N$ and that the $R$-module isomorphism $\theta:P \to M \times N$ is unique"

I felt that establishing this isomorphism was essential to seeing that the two definitions were the same ... SO ... I looked for Bland's proof of the same ... BUT ... did not find such a proof ... ...

I did, however, find a statement of the existence of the isomorphism on page 42 after Proposition 2.1.3 ... ...

What Bland does is as follows:

... ... after Proposition 2.1.1 (see my post above) and the 're-definition' (i.e. Definition 2.1.1 - see my post above), Bland proves Proposition 2.1.3. The statement of Proposition 2.1.3 is as follows:

View attachment 2457

Following Proposition 2.1.3, Bland makes the following statement:

View attachment 2458

Now it seems that Bland is claiming that Proposition 2.1.3 has established the unique isomorphism \(\displaystyle \theta \) ... BUT how does Proposition 2.1.3 actually establish that

$P \cong M \times N$

Can you help?

As a side issue, could you please comment of the significance/meaning of \(\displaystyle \theta \) being unique?

Peter

EDIT: In case MHB members reading the post wish to get the complete context of Bland's treatment of the direct product construction I have attached the relevant pages of Paul E Bland's book, "Rings and Their Modules" - see attached.

 

[Deveno]

What he is saying is not only that for any $P$ that satisfies the defining universal property of the product , if $P'$ is another such module, $P \cong P'$ but that there is only ONE possible isomorphism between them, that makes the diagram commute.

For example, if the "projections" associated with $P$ are $\pi_1,\pi_2$ and the projections associated with $P'$ are $\pi'_1,\pi'_2$, then the isomorphism $\phi:P' \to P$ has to map $\pi'_j(p')$ to $\pi_j(p)$

Let's look at a concrete example, to see how this actually works.

Consider the 2 $\Bbb R$-modules $\Bbb C$ and $\Bbb R^2$. The isomorphism between them, is, of course:

$a+bi \mapsto (a,b)$.

Now, pretend we don't know that, for just a moment.

Let $\pi_1: \Bbb C \to \Bbb R$ be the map $\pi_1(z) = \mathfrak{Re}(z)$

Similarly, let: $\pi_2: \Bbb C \to \Bbb R$ be the map $\pi_2(z) = \mathfrak{Im}(z)$

Define the maps: $p_1,p_2: \Bbb R^2 \to \Bbb R$ by:

$p_1(x,y) = x$
$p_2(x,y) = y$.Suppose we want to know how many possible isomorphisms $\theta:\Bbb C \to \Bbb R^2$ there are such that:

$p_1(\theta(z)) = \pi_1(z)$
$p_2(\theta(z)) = \pi_2(z)$.(**)

We will write $z \in \Bbb C$ as $z = a+bi$.

Now, $\theta(z) = \theta(a+bi) = \theta(a+0i) + \theta(0+bi)$, using the additive property of module homomorphisms.

Hence, by (**), we have:

$p_1(\theta(a+0i)) = \pi_1(a+0i) = a$
$p_2(\theta(0+bi)) = \pi_2(0+bi) = b$.

It follows that $\theta(z) = (\mathfrak{Re}(z),\mathfrak{Im}(z))$, for any complex $z$, which is just the map given above.

(explicitly, we are using the injections defined by the projections to "piece together the coordinates of $\theta(z)$").

As an aside, it's perfectly reasonable to view a product as (perhaps) infinite sequences of coordinates (using the identification of $P$ with $\prod_{\alpha} M_{\alpha}$). The uniqueness stipulations say there is only one way to do this once we decide "which coordinate is which". Of course, this is unwieldy if the "indexing set" is "large" (say we have a separate module for each real number), but it helps clarify how it all goes together.

 

[Math Amateur]

What he is saying is not only that for any $P$ that satisfies the defining universal property of the product , if $P'$ is another such module, $P \cong P'$ but that there is only ONE possible isomorphism between them, that makes the diagram commute.

For example, if the "projections" associated with $P$ are $\pi_1,\pi_2$ and the projections associated with $P'$ are $\pi'_1,\pi'_2$, then the isomorphism $\phi:P' \to P$ has to map $\pi'_j(p')$ to $\pi_j(p)$

Let's look at a concrete example, to see how this actually works.

Consider the 2 $\Bbb R$-modules $\Bbb C$ and $\Bbb R^2$. The isomorphism between them, is, of course:

$a+bi \mapsto (a,b)$.

Now, pretend we don't know that, for just a moment.

Let $\pi_1: \Bbb C \to \Bbb R$ be the map $\pi_1(z) = \mathfrak{Re}(z)$

Similarly, let: $\pi_2: \Bbb C \to \Bbb R$ be the map $\pi_2(z) = \mathfrak{Im}(z)$

Define the maps: $p_1,p_2: \Bbb R^2 \to \Bbb R$ by:

$p_1(x,y) = x$
$p_2(x,y) = y$.Suppose we want to know how many possible isomorphisms $\theta:\Bbb C \to \Bbb R^2$ there are such that:

$p_1(\theta(z)) = \pi_1(z)$
$p_2(\theta(z)) = \pi_2(z)$.(**)

We will write $z \in \Bbb C$ as $z = a+bi$.

Now, $\theta(z) = \theta(a+bi) = \theta(a+0i) + \theta(0+bi)$, using the additive property of module homomorphisms.

Hence, by (**), we have:

$p_1(\theta(a+0i)) = \pi_1(a+0i) = a$
$p_2(\theta(0+bi)) = \pi_2(0+bi) = b$.

It follows that $\theta(z) = (\mathfrak{Re}(z),\mathfrak{Im}(z))$, for any complex $z$, which is just the map given above.

(explicitly, we are using the injections defined by the projections to "piece together the coordinates of $\theta(z)$").

As an aside, it's perfectly reasonable to view a product as (perhaps) infinite sequences of coordinates (using the identification of $P$ with $\prod_{\alpha} M_{\alpha}$). The uniqueness stipulations say there is only one way to do this once we decide "which coordinate is which". Of course, this is unwieldy if the "indexing set" is "large" (say we have a separate module for each real number), but it helps clarify how it all goes together.

Thanks Deveno ... again, that post helped, especially in helping me to understand Proposition 2.1.3 ... but I would be less than honest if I did not say I am still puzzled about an issue ... apologies if, in some way, you have actually answered the following question/issue ...

So ... you have explained that Proposition 2.1.3 shows/demonstrates that where P is a module that satisfies the defining universal of the product, and P' is another such module, that \(\displaystyle P \cong P' \).

But Paul Bland seems to be asserting that somehow (?) Proposition 2.1.3 assures us (demonstrates that ...) \(\displaystyle P \cong M \times N \) when he writes the following on page 42, just after the proof of Proposition 2.1.3:

"If \(\displaystyle ( P, p_\alpha )_\Delta \) is (also) a direct product of \(\displaystyle \{ M_\alpha \}_\Delta \), then there is a unique isomorphism \(\displaystyle \phi \ : \ P \to \prod_\Delta M_\alpha \) such that \(\displaystyle \pi_\alpha \phi = p_\alpha \) for each \(\displaystyle \alpha \in \Delta \)"

Now it seems to me that Bland has NOT shown that \(\displaystyle P \cong \prod_\Delta M_\alpha \) (or \(\displaystyle P \cong M \times N \) in your exposition of this matter) ... never mind the uniqueness of the isomorphism ... the isomorphism itself does not seem to have been established/demonstrated!

Can you help with why Paul Bland's Proposition 2.1.3 actually demonstrates \(\displaystyle P \cong M \times N \) ? ... ... ... ... or am I just misinterpreting the whole thing in some way and not asking a valid question?

... ... ...

I also wonder if you could clarify another issue pertaining to Paul Bland's remarks on page 42 that follow the proof of Proposition 2.1.3. Bland makes the following rather mystical/mysterious statement:

"In view of Proposition 2.1.3 we see that \(\displaystyle ( \prod_\Delta M_\alpha , \pi_\alpha ) \) is actually a model for every direct product of a family \(\displaystyle \{ M_\alpha \}_\Delta \) of R-modules. ... ... "

What does Bland mean by this statement? In particular, what exactly does he mean by a 'model"? Further puzzlement arises for me as I thought that since the definition of \(\displaystyle ( \prod_\Delta M_\alpha , \pi_\alpha ) \) was quite general and abstract, a statement assuring us that \(\displaystyle ( \prod_\Delta M_\alpha , \pi_\alpha ) \) is a model for every direct product of a family of modules was unnecessary! Paul Bland is talking as if he gave a specific example and the example is standing as a 'model' for the general case ?

Can you clarify this for me ... since I feel I must be missing something here?

Hoping you can help.

Peter

 

[Deveno]

The uniqueness and isomorphism he is asserting occur "by definition" (this is what is so strange about "universal property proofs").

Of course, defining an entity by the properties we wish it to have is all very well and good, IF: we can produce at least "one thing" that actually HAS that property (from just the definition alone, it could conceivably be that NO $R$-module has the universal mapping property of the product).

This is what he means by "model", an actual construction that possesses the universal property.

It should be clear then, that:

$P = \displaystyle \prod_{\alpha \in \Delta} M_{\alpha}$

together with the maps:

$\pi_{\alpha}:P \to M_{\alpha}$ given by: $\pi_{\alpha}(p) = m_{\alpha}$

(where $p = (m_{\alpha})_{\Delta}$...this notation is cumbersome for an infinite index set $\Delta$, and is a lot easier to write for a finite index set),

is an actual construction that possesses the desired property.

(In other words, an "actual" direct product of a family of $R$-modules has the CARTESIAN product of the underlying sets of the individual factor modules as ITS underlying set, and the addition and scalar multiplication is done "component-wise":

$(m_{\alpha})_{\Delta} + (m'_{\alpha})_{\Delta} = (m_{\alpha} + m'_{\alpha})_{\Delta}$

$r \cdot (m_{\alpha})_{\Delta} = (r\cdot m_{\alpha})_{\Delta}$

When $\Delta = \{1,2\}$ this has the more usual familiar form:

$(m_1,m_2) + (m'_1,m'_2) = (m_1+m'_1,m_2+m'_2)$

$r\cdot(m_1,m_2) = (r\cdot m_1,r \cdot m_2)$.

In my example above, BOTH the complex plane, and the Euclidean plane are "models" of a direct product of the real numbers (as a real module over itself) with itself. They are not "the same thing", for example the Euclidean plane does not have a multiplicative structure on it (in fact, we could give it one totally different than the complex numbers if we wanted to, by setting:

$(x,y)\ast(x',y') = (xx',yy')$

But AS $\Bbb R$-MODULES, they have few discernible differences.