External Direct Sums & Direct Products .... Bland Ex. 1b, 2.1

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In summary: Sure. For any set ##A##, the notation ##\prod_{a\in A} X_a## is a short way to write the set of all functions with domain ##A## and values in the sets ##X_a##. So in this case, the elements of ##\prod_{\alpha\in\Delta} M_\alpha## are functions with domain ##\Delta##, meaning that they are a set of functions. In particular, for any function ##y## in this set, ##y(\alpha)## is the value that the function ##y## takes at the point ##\alpha## in the domain.
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Homework Statement



I am reading Paul E. Bland's book: Rings and Their Modules and am currently focused on Section 2.1 Direct Products and Direct Sums ... ...

I need help with Problem 1(b) of Problem Set 2.1 ...

Problem 1(b) of Problem Set 2.1 reads as follows:Bland Problem 1, Section 2.1 reads as follows:

Bland - Problem 1 ... Problem Set 2.1 ... .png


I need some help with 1(b) ...

Homework Equations



The definition of External Direct Sum is as follows:
Bland - Defn of External Direct Sums ... page 43 ... .png


The Attempt at a Solution



I have had difficulty in formulating a rigorous and convincing proof of the statement in Problem 1(b) ... can someone please

(1) critique my attempt at a proof (see below)

(2) provide an alternate rigorous and convincing proofMy attempt at a proof is as follows:

We need to demonstrate that ##\prod_\Delta M_\alpha = \bigoplus_\Delta M_\alpha## if and only if ##\Delta## is a finite set ...Assume ##\prod_\Delta M_\alpha = \bigoplus_\Delta M_\alpha##

The above equality would require all of the terms ##(x_\alpha)## of ##\prod_\Delta M_\alpha## to have a finite number of components or elements in each ##(x_\alpha)## ... thus ##\Delta## is a finite set ...
Assume ##\Delta## is a finite set

... then ##\prod_\Delta M_\alpha## has terms of the form ##(x_\alpha) = ( x_1, x_2, \ ... \ ... \ , x_n )## for some ##n \in \mathbb{Z}## ... ...

and

... ## \bigoplus_\Delta M_\alpha## has the same terms given that each of the above terms ##(x_\alpha)## has a finite number of components ...
Hope someone can indicate how to formulate a better proof ...

Peter
 

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  • #2
Let ##y\in \prod_{\alpha\in\Delta} M_\alpha##. Then ##y## is a function with domain ##\Delta## and range ##\bigcup_{\alpha\in\Delta} M_\alpha## such that ##y(\alpha)\in M_\alpha##. Define ##N(y)=\{\alpha\in\Delta\ :\ y(\alpha)\neq 0_{M_\alpha}\}##, which is the set of 'nonzero' components of ##y##. Then by definition ##y## is also in the direct sum ##\sum_{\alpha\in\Delta} M_\alpha## iff ##|N(y)|<\infty##, that is, if ##y## has only a finite number of nonzero components.

If ##\Delta## is finite then every ##y## has a finite number of components, hence a finite number of nonzero components, hence is in the direct sum.

To prove the reverse direction we want to show that if ##\Delta## is infinite, there exists a ##y## in the direct product that is not in the direct sum, ie which has an infinite number of nonzero components. In fact that will only be true if an infinite number of modules in ##\Delta## are nontrivial. If that is the case then, to construct our counterexample element ##y##, for every nontrivial module ##M_\alpha##, we choose a nonzero element in the ##\alpha## position. Then ##y## will have an infinite number of nonzero components and hence is not in the direct sum.

Note that that para uses the axiom of choice, which is always a little sad. We can avoid it if we are given some more info about each module, such as a basis. Say that we are given a generating-set function ##B:\Delta\times\mathbb N\to \bigcup_{\alpha\in\Delta} M_\alpha## such that ##B(\alpha,k)## is the ##k##th element of the specified generating-set for ##M_\alpha##, and ##0_{M_\alpha}## if ##k## is larger than the size of the generating set of ##\alpha##. Then we can construct our counterexample ##y## as ##y(\alpha)=B(\alpha,1)## without using the axiom of choice.

So, to state the theorem part (b) properly, we would say that the direct product of a set ##\Delta## of nontrivial modules is equal to the direct sum thereof if the set is finite and that, assuming the axiom of choice, if the direct product is equal to the direct sum, the set ##\Delta## must be finite.

It is possible we can avoid using the axiom of choice, but it is late and my brain is tired and I can't see how right now.
 
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  • #3
andrewkirk said:
Let ##y\in \prod_{\alpha\in\Delta} M_\alpha##. Then ##y## is a function with domain ##\Delta## and range ##\bigcup_{\alpha\in\Delta} M_\alpha## such that ##y(\alpha)\in M_\alpha##. Define ##N(y)=\{\alpha\in\Delta\ :\ y(\alpha)\neq 0_{M_\alpha}\}##, which is the set of 'nonzero' components of ##y##. Then by definition ##y## is also in the direct sum ##\sum_{\alpha\in\Delta} M_\alpha## iff ##|N(y)|<\infty##, that is, if ##y## has only a finite number of nonzero components.

If ##\Delta## is finite then every ##y## has a finite number of components, hence a finite number of nonzero components, hence is in the direct sum.

To prove the reverse direction we want to show that if ##\Delta## is infinite, there exists a ##y## in the direct product that is not in the direct sum, ie which has an infinite number of nonzero components. In fact that will only be true if an infinite number of modules in ##\Delta## are nontrivial. If that is the case then, to construct our counterexample element ##y##, for every nontrivial module ##M_\alpha##, we choose a nonzero element in the ##\alpha## position. Then ##y## will have an infinite number of nonzero components and hence is not in the direct sum.

Note that that para uses the axiom of choice, which is always a little sad. We can avoid it if we are given some more info about each module, such as a basis. Say that we are given a generating-set function ##B:\Delta\times\mathbb N\to \bigcup_{\alpha\in\Delta} M_\alpha## such that ##B(\alpha,k)## is the ##k##th element of the specified generating-set for ##M_\alpha##, and ##0_{M_\alpha}## if ##k## is larger than the size of the generating set of ##\alpha##. Then we can construct our counterexample ##y## as ##y(\alpha)=B(\alpha,1)## without using the axiom of choice.

So, to state the theorem part (b) properly, we would say that the direct product of a set ##\Delta## of nontrivial modules is equal to the direct sum thereof if the set is finite and that, assuming the axiom of choice, if the direct product is equal to the direct sum, the set ##\Delta## must be finite.

It is possible we can avoid using the axiom of choice, but it is late and my brain is tired and I can't see how right now.
Hi Andrew ... thanks so much for the help ...

BUT ... I need further help to understand what you have said ...

You write:

" ... ... Let ##y\in \prod_{\alpha\in\Delta} M_\alpha##. Then ##y## is a function with domain ##\Delta## and range ##\bigcup_{\alpha\in\Delta} M_\alpha## such that ##y(\alpha)\in M_\alpha##. ... ... "Can you please explain (slowly ... :smile:... ) how/why ##y## is a function with domain ##\Delta## and range ##\bigcup_{\alpha\in\Delta} M_\alpha## ...

Sorry to be slow to catch on ...

Peter
 
  • #4
Math Amateur said:
Hi Andrew ... thanks so much for the help ...

BUT ... I need further help to understand what you have said ...

You write:

" ... ... Let ##y\in \prod_{\alpha\in\Delta} M_\alpha##. Then ##y## is a function with domain ##\Delta## and range ##\bigcup_{\alpha\in\Delta} M_\alpha## such that ##y(\alpha)\in M_\alpha##. ... ... "Can you please explain (slowly ... :smile:... ) how/why ##y## is a function with domain ##\Delta## and range ##\bigcup_{\alpha\in\Delta} M_\alpha## ...

Sorry to be slow to catch on ...

Peter
It is an equivalent way of viewing direct products. In the book, the elements are ##(y_\alpha)_{\alpha \in \Delta}##, but what does that mean? It means we have for each ##\alpha## an element ##y_\alpha \in M_\alpha##. But this is as well a function, because it maps ##\alpha \longmapsto y_\alpha##, i.e. a function from ##\Delta \longrightarrow \Pi_{\alpha\in\Delta}M_\alpha##. So the elements of direct products are often called a function of the index set: each element corresponds one-to-one to a function.
 
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  • #5
fresh_42 said:
It is an equivalent way of viewing direct products. In the book, the elements are ##(y_\alpha)_{\alpha \in \Delta}##, but what does that mean? It means we have for each ##\alpha## an element ##y_\alpha \in M_\alpha##. But this is as well a function, because it maps ##\alpha \longmapsto y_\alpha##, i.e. a function from ##\Delta \longrightarrow \Pi_{\alpha\in\Delta}M_\alpha##. So the elements of direct products are often called a function of the index set: each element corresponds one-to-one to a function.
Thanks fresh_42 ... helpful and interesting ...

Do you know a text that treats direct products in this way ... or at least treats indexed sets in this way ...

Peter
 
  • #6
Math Amateur said:
Do you know a text that treats direct products in this way ... or at least treats indexed sets in this way ...
Yes, but in the wrong language. I assume that any book about homological algebra does it this way. Haven't you started to read one? Have a look then. A direct product is defined as the solution of a universal mapping problem. It is basically not different from the algebraic version in your book above, but simply in the more general language of categories and functors. The direct product is defined with projections ##\Pi_{\alpha \in \Delta}M_\alpha \twoheadrightarrow M_\alpha##, the direct sum with injections ##M_\alpha \hookrightarrow \oplus_{\alpha \in \Delta} M_\alpha##. The direct sum is the co-product.

Here's a book I can recommend and which is in English:
https://www.amazon.com/dp/0387948236/?tag=pfamazon01-20
 
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  • #7
Thanks fresh_42 ... appreciate your help ...

Will pursue this ...

Peter
 
  • #8
Math Amateur said:
Thanks fresh_42 ... appreciate your help ...

Will pursue this ...

Peter
The book I've quoted has the direct product and sum on page 20 f., so I wouldn't buy a book just to cover these single cases.
 

Related to External Direct Sums & Direct Products .... Bland Ex. 1b, 2.1

What is an external direct sum?

An external direct sum is a mathematical concept where two or more groups or vector spaces can be combined to form a larger group or vector space. It is denoted by ⊕ and is defined as the set of all possible combinations of elements from the individual groups or vector spaces.

What is the difference between an external direct sum and a direct product?

The main difference between an external direct sum and a direct product is the operation used to combine the groups or vector spaces. In an external direct sum, the groups or vector spaces are combined using addition, while in a direct product, they are combined using multiplication.

How do you determine the size of an external direct sum?

The size of an external direct sum is equal to the product of the sizes of the individual groups or vector spaces. For example, if A and B are groups of sizes m and n respectively, then the size of the external direct sum A ⊕ B is mn.

Can an external direct sum be infinite?

Yes, an external direct sum can be infinite if at least one of the individual groups or vector spaces is infinite. In this case, the external direct sum will also be infinite.

What are some real-life applications of external direct sums?

External direct sums are commonly used in mathematics and physics to describe systems that can be broken down into smaller, independent components. They are also used in coding theory and cryptography to analyze and manipulate information. Additionally, external direct sums are used in economics to model complex systems that have multiple independent variables.

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