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

[Math Amateur]

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:

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

 

[andrewkirk]

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.

 

[member 587159]

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##.

 

[member 587159]

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!

 

[Math Amateur]

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 ......

Will be back in touch shortly ...

Peter

 

[Math Amateur]

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