Ring Direct Products .... Bland Problem 3(a), Problem Set 2.1

  • Thread starter Math Amateur
  • Start date
  • Tags
    Ring Set
And for the next line we have\begin{align*}\sum_{j=1}^n a_k e_k&= \sum_{k=1}^n \pi_k(c_k)e_k\quad\quad\textrm{since }\pi_k(c_k)=a_k\\&= \left(\sum_{k=1}^n \pi_k\right)\sum_{j=1}^n c_k e_k\quad\quad\textrm{since the functions in parentheses are a bijection}\\&= \sum_{j=1}^
  • #1
Math Amateur
Gold Member
MHB
3,998
48

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 someone to check my solution to Problem 3(a) of Problem Set 2.1 ...

Problem 3(a) of Problem Set 2.1 reads as follows:
Bland - Problem 3 - Problem Set 2.1 ... ....png

Homework Equations

The Attempt at a Solution



[/B]
My attempt at a solution follows:We claim that every right ideal of the ring ##R_1 \times R_2 \times \ ... \ ... \ \times R_n## is of the form ##A_1 \times A_2 \times \ ... \ ... \ \times A_n## ...Proof:

Suppose ##A## is a right ideal of ##R_1 \times R_2 \times \ ... \ ... \ \times R_n## ...

Let ##a \in A## and put ##A_1 = \pi_1 (A) , A_2 = \pi_2 (A) , \ ... \ ... \ , A_n = \pi_n (A)##
Now ##a \in A## ...##\Longrightarrow \pi_1(a) = a_1, \pi_2(a) = a_2, \ ... \ ... \ , \pi_n(a) = a_n##

for some ##a_1 \in A_1, a_2 \in A_2, \ ... \ ... \ , a_n \in A_n##Hence ...

##a = ( i_1 \pi_1 + i_2 \pi_2 + \ ... \ ... \ + i_n \pi_n ) (a)####= (a_1, 0, 0, \ ... \ ... \ , 0) + (0, a_2, 0, \ ... \ ... \ , 0) + \ ... \ ... \ + ( 0, 0, \ ... \ ... \ , a_n ) ####= ( a_1, a_2, \ ... \ ... \ , a_n)##Hence ... ##A \subseteq A_1 \times A_2 \times \ ... \ ... \ \times A_n## ... ... ... ... ... (1)
Conversely ...

Let ##a_1 \in A_1, a_2 \in A_2, \ ... \ ... \ , a_n \in A_n##Note that again ... ##A## is a right ideal of ##R_1 \times R_2 \times \ ... \ ... \ \times R_n## ...

... and ##a \in A## and put ##A_1 = \pi_1 (A) , A_2 = \pi_2 (A) , \ ... \ ... \ , A_n = \pi_n (A)##Then there are ##b_1, b_2, \ ... \ ... \ , b_n \in A## such that ...

##\pi_1 (b_1) = a_1, \pi_2 (b_2) = a_2, \ ... \ ... \ , \pi_n (b_n) = a_n## ... Hence ...

##b_1 ( 1,0,0, \ ... \ ... \ , 0 ) + b_2 ( 0, 1,0, \ ... \ ... \ , 0 ) + \ ... \ ... \ + b_n ( 0, 0,0, \ ... \ ... \ , 1 )####= ( i_1 \pi_1 + i_2 \pi_2 + \ ... \ ... \ + i_n \pi_n ) ( b_1 ( 1,0,0, \ ... \ ... \ , 0 ) + b_2 ( 0, 1,0, \ ... \ ... \ , 0 ) + \ ... \ ... \ + b_n ( 0, 0,0, \ ... \ ... \ , 1 ) )####= ( a_1, a_2, \ ... \ ... \ , a_n)##So ... ##A_1 \times A_2 \times \ ... \ ... \ \times A_n \subseteq A## ... ... ... ... ... (2)Now ... ##(1), (2) \Longrightarrow A = A_1 \times A_2 \times \ ... \ ... \ \times A_n##
Can someone please critique my proof ... and either confirm it is correct or point out the errors and shortcomings ... ...

Problem/Issue

... there is part of the above proof I do not fully understand ... I will relate the issue to text solution for ##n = 2## ...In Bland's text on the problem we read the following:

"... ... Hence ##a(1,0) + b(0,1) = ( i_1 \pi_1 + i_2 \pi_2 ) ( a(1,0) + b(0,1) ) = (a_1, a_2)##, so ##A_1 \times A_2 \subseteq A.## ... ... "I have two questions regarding the above quote:(1) Exactly why/how is the equation ##a(1,0) + b(0,1) = ( i_1 \pi_1 + i_2 \pi_2 ) ( a(1,0) + b(0,1) ) = (a_1, a_2)## ... true?

Can someone please explain in detail why/how this is true ...(2) Exactly why/how does the above equation being true imply that ##A_1 \times A_2 \subseteq A## ... ?
Help with the above will be much appreciated ...

Peter
 

Attachments

  • Bland - Problem 3 - Problem Set 2.1 ... ....png
    Bland - Problem 3 - Problem Set 2.1 ... ....png
    53.3 KB · Views: 322
Physics news on Phys.org
  • #2
Math Amateur said:
In Bland's text on the problem we read the following:

"... ... Hence ##a(1,0) + b(0,1) = ( i_1 \pi_1 + i_2 \pi_2 ) ( a(1,0) + b(0,1) ) = (a_1, a_2)##, so ##A_1 \times A_2 \subseteq A.## ... ... "I have two questions regarding the above quote:(1) Exactly why/how is the equation ##a(1,0) + b(0,1) = ( i_1 \pi_1 + i_2 \pi_2 ) ( a(1,0) + b(0,1) ) = (a_1, a_2)## ... true?

Can someone please explain in detail why/how this is true ...
Presumably the functions ##i_j,\pi_j## are defined in the text outside the imaged portion. I imagine that ##\pi_j:\mathbf R\to R_j## is* the projection function that selects the ##j##th component of its input and ##i_j:R_j\to\mathbf R## is the injection function that gives a ##n##-tuple that is all zeros except the ##j##th component, which is equal to the input. Then for ##1\le j\le n##, the function ##i_j\pi_j:\mathbf R\to\mathbf R## zeros all components of the input expect the ##j##th, which it leaves unchanged. It follows that the function ##\sum_{j=1}^n i_j\pi_j## is the identity on ##\mathbf R##.

It is readily shown that ##i_j,\pi_j## are ring homomorphisms, although we only use the latter.

I find the author's use of ##a,b## as items in ##A## that will correspond to ##a_1,a_2## needlessly confusing, as the usual convention would be to have ##a=(a_1,a_2)##, which is not the case here.
So to remove the confusion let's use symbols ##c_1,c_2## in place of ##a,b##. Then we have ##c_1,c_2\in A## such that ##\pi_j(c_j)=a_j## for ##j\in\{1,2\}##. In fact, let's generalise this to the ##n##-dimensional case so that ##\pi_j(c_j)=a_j## for ##j\in\{1,2, ..., n\}##, where ##\forall j:c_j\in A##.

For ##1\leq k\leq n## let ##e_k## denote the element of ##\mathbf R\triangleq \prod_{k=1}^n R_k## that has all zero components except for a 1 in the ##k##th place.

Then we can write the line you were concerned about as follows:

\begin{align*}
\sum_{k=1}^n c_k e_k
&= \left(\sum_{j=1}^n i_j\pi_j\right)\sum_{j=1}^n c_k e_k
\quad\quad\textrm{since the function in parentheses is the identity}\\
&= \sum_{j,k=1}^n i_j\pi_j( c_k e_k)\\
&= \sum_{j,k=1}^n i_j(\pi(c_j)\cdot \pi_j( e_k))
\quad\quad\textrm{since $\pi_j$ is a ring homomorphism}\\
&= \sum_{j,k=1}^n i_j(a_j\cdot \pi_j( e_k))\\
&= \sum_{j=1}^n i_j(a_j\cdot 1)
\quad\quad\textrm{since $\pi_j(e_k)=1$ if $j= k$, otherwise 0}\\
&= \sum_{j=1}^n i_j(a_j)\\
&=(a_1,a_2,...,a_n)\\
\end{align*}

Recall that we started by choosing ##a_j\in A_j## for each ##j## in ##1,...,n##.

So, looking at the sequence of equalities in reverse order, an arbitrary element ##(a_1,...,a_n)## of ##\prod_{j=1}^n A_j## is shown by the above to be equal to##\sum_{k=1}^n c_k e_k##, which is the sum of terms, each of which is an element ##c_j## of ##A## right-multiplied by an element ##e_j## of ##\mathbf R##, and that product is in ##A## since ##A## is a right ideal. The sum is in ##A## because ideals are closed under addition.

I note by the way that the author's proof omits the essential sub-proof that each ##A_j\triangleq \pi_j(A)## is an ideal of ##R_j##. I suggest you try to construct the missing proof of that.

* EDIT: I forgot to specify that I am using ##\mathbf R## to denote the product ring ##\prod_{j=1}^n R_j##.
Oh wait, I did define it, only further down, which is too late. That's what happens when I move things around I suppose.
 
Last edited:
  • Like
Likes Math Amateur and member 587159
  • #3
I don't fully understand the notations used.

Therefore, I will offer an alternative (?) proof.

I will offer the proof for ##n=2##. It's easy to generalise.

Let ##A \times B## be a right ideal of ##R \times S##. Our goal is to show that ##A## is a right ideal in ##R##.

So, let ##a \in A, r \in R##. Our goal is to show that ##ar \in A##. Other axioms can be done in the same way:

We know that ##(a,0)r = (ar,0) \in A \times B##, because ##A \times B## is a right ideal. Hence ##ar\in A##.
 
  • Like
Likes Math Amateur
  • #4
andrewkirk said:
I note by the way that the author's proof omits the essential sub-proof that each ##A_j\triangleq \pi_j(A)## is an ideal of ##R_j##. I suggest you try to construct the missing proof of that.

It is probably already proven that the image of an ideal under a ring morphism is again an ideal. But very good post!
 
  • #5
Thanks to Andrew and Math_QED for their posts ...

Sorry to be slow in responding but one of my two large standard poodles chewed up my glasses ...:frown:...

Will be back in touch shortly ...

Peter
 
  • #6
Math_QED said:
It is probably already proven that the image of an ideal under a ring morphism is again an ideal. But very good post!
Thanks to Andrew and Math_QED ...

Have now worked through your posts ... ... most helpful!

Thanks again,

Peter
 

FAQ: Ring Direct Products .... Bland Problem 3(a), Problem Set 2.1

1. What are "Ring Direct Products"?

"Ring Direct Products" refer to a specific type of chemical compounds that contain a ring structure in their molecular formula. These compounds are commonly used in organic synthesis and drug development.

2. How are "Ring Direct Products" different from other chemical compounds?

Unlike linear compounds, "Ring Direct Products" have a closed-loop structure, which can affect their physical and chemical properties. They also have a higher degree of stability and are often more resistant to reactions with other compounds.

3. What is "Bland Problem 3(a)"?

"Bland Problem 3(a)" refers to a specific problem in a set of chemistry problems, specifically in Problem Set 2.1. It is named after the author or creator of the problem, and is used as a way to differentiate it from other problems in the set.

4. How can "Bland Problem 3(a)" be solved?

The solution to "Bland Problem 3(a)" will depend on the specific problem and its context. In general, it involves applying relevant concepts and principles from chemistry to come up with a logical and accurate answer or solution.

5. What is the purpose of "Problem Set 2.1"?

"Problem Set 2.1" is a collection of chemistry problems designed to test and reinforce understanding of concepts and principles covered in a specific course or textbook chapter. It is used as a tool for learning and practice in the field of chemistry.

Similar threads

Back
Top