Ring Theory

[Steven G]

I am trying to help my cousin and have a few question in ring theory and some answers verified.

Let R = Z + Z with addition and multiplication defined component-wise (this I understand)

Let A = < (2,3)>, the principal ideal of R generated by the element (2,3). (I do know what an ideal is as well as what a pid is)

I need to answer the following about A and the factor ring R/A

a)Describe in words or by giving a pattern the elements of A.
A = {(2m,3m) | m is in Z} I think i am happy with this result.

b) List or give a pattern for all elements of R/A
I thought that this would be easy as we just mod out (2,3) from R, ie (2,3) behaves like (0,0).
Clearly {(x,y) | 0<=x < 2 and 0<= y <3} is in R/A (??) But what about elements like (108, 1) and (1, 324)? Are they in R/A? I think so. If I am right, then how would one describe the elts. in R/A? For the record I know that for example (7,9) in Z + Z would be (1,0) + A in R/A

c) Is the element (1,2) + A a unit in the factor ring R/A? I know that a nonzero element, x, of a commutative ring with unity does not have to have a multiplicative inverse but when it does we say x is a unit of the ring. That is, x is a unit if x^(-1) exist. I would say that (1,2) + A is not a unit as all units are in the form (2m,3m) + A where m is in Z

d) Is the factor ring R/A an integral domain? Before I answer this I would 1st like to know the elements in R/A (part b above).

I truly would appreciate any help.

Thanks,
Jomo

 

[Steven G]

Would the answer to part b be {(x,y) | 0<= x <2 or 0<= y <3 }
But what about when x or y is negative. I guess I still do not see this answer. Ah, we add enough (2,3) to get the element of the form above. OK, so I just talked myself into the above answer. Am I correct?

 

[HallsofIvy]

I am trying to help my cousin and have a few question in ring theory and some answers verified.

Let R = Z + Z with addition and multiplication defined component-wise (this I understand)

Don't you mean \(\displaystyle R= Z\times Z\)the set of ordered pairs of integers?

Let A = < (2,3)>, the principal ideal of R generated by the element (2,3). (I do know what an ideal is as well as what a pid is)

I need to answer the following about A and the factor ring R/A

a)Describe in words or by giving a pattern the elements of A.
A = {(2m,3m) | m is in Z} I think i am happy with this result.

Yes, that looks good.

b) List or give a pattern for all elements of R/A
I thought that this would be easy as we just mod out (2,3) from R, ie (2,3) behaves like (0,0).
Clearly {(x,y) | 0<=x < 2 and 0<= y <3} is in R/A (??) But what about elements like (108, 1) and (1, 324)? Are they in R/A? I think so. If I am right, then how would one describe the elts. in R/A? For the record I know that for example (7,9) in Z + Z would be (1,0) + A in R/A

There are a couple of different ways of looking at "mod n". Some people and texts, think of it as "numbers from 0 to n-1" others define it as equivalence classes so that an element of "Z mod 3" would be the sets {0, 3, -3, 6, -6, ...}, {1, -2, 4, -5, 7 ...}, and {2, -1, 5, -4, 8, ...}. Depending on which view point you take (the latter is a more "sophisticated" way of looking at it) you can say that the elements are of F/A are (0, 0), (0, 1), (0, 2), (1, 0), (1, 1), (1, 2) or that they are the six equivalence classes that contain those. A variation of latter would be to say that (108, 1) is "the same as" (0, 1) since "108= 0 (mod 2)" and that (1, 324) is "the same as" (1, 0) since 324= 0 (mod 3). Yes, because 7= 2(3)+1 and 9= 3(3)+ 0, 7= 1 (mod 2) and 9= 0 (mod 3) so (7, 9) is the same (or in the same equivalence class) as (1, 0).

c) Is the element (1,2) + A a unit in the factor ring R/A? I know that a nonzero element, x, of a commutative ring with unity does not have to have a multiplicative inverse but when it does we say x is a unit of the ring. That is, x is a unit if x^(-1) exist. I would say that (1,2) + A is not a unit as all units are in the form (2m,3m) + A where m is in Z

"(1, 2)+ A" is the equivalence class containing the pair (1, 2). That will be a unit, as you say, of R/A if there exist a pair (a, b) such that (1, 2)(a, b)= (a, 2b) is in the same equivalence class as the multiplicative identity for R, (1, 1). That means that we must have a= 1 (mod 2) and 2b= 1 (mod 3). 2b= 1 (mod 3) as long as 2b= 1+ 3n for some integer n. In particular, taking n= 1, we have 2b= 1+ 3= 4 so b= 2. We can take as a "representative" of the equivalence class satisfying this (1, 2). (1, 2)(1, 2)= (1, 4) which is equivalent to (1, 1) (mod (2, 3)) so, yes, (1, 2) is a "unit" and, in fact, its multiplicative inverse is itself!
(While "Z mod n" is not, in general, a "field" (not all elements have multiplicative inverses), if n is a prime number, it is a field.)

d) Is the factor ring R/A an integral domain? Before I answer this I would 1st like to know the elements in R/A (part b above).

Yes, this is an integral domain. If (a, b) and (c, d) are non- zero (a and c are odd and b and d are not multiples of 3) then the same is true of (a, b)(c, d)= (ab, cd). Again, this depends on the fact that 2 and 3 are prime. If, instead of "mod 3" we were working with "mod 6" neither "2" nor "3" is 0 but there product is 6= 0 (mod 6). In fact, as I point out above it is not just an "integral domain" but is, in fact, a field.

I truly would appreciate any help.

Thanks,
Jomo

 

[Steven G]

There are a couple of different ways of looking at "mod n". Some people and texts, think of it as "numbers from 0 to n-1" others define it as equivalence classes so that an element of "Z mod 3" would be the sets {0, 3, -3, 6, -6, ...}, {1, -2, 4, -5, 7 ...}, and {2, -1, 5, -4, 8, ...}. Depending on which view point you take (the latter is a more "sophisticated" way of looking at it) you can say that the elements are of F/A are (0, 0), (0, 1), (0, 2), (1, 0), (1, 1), (1, 2) or that they are the six equivalence classes that contain those. A variation of latter would be to say that (108, 1) is "the same as" (0, 1) since "108= 0 (mod 2)" and that (1, 324) is "the same as" (1, 0) since 324= 0 (mod 3). Yes, because 7= 2(3)+1 and 9= 3(3)+ 0, 7= 1 (mod 2) and 9= 0 (mod 3) so (7, 9) is the same (or in the same equivalence class) as (1, 0).
I am a bit confused here. I thought that the elements of A are in the form (2m, 3m)-actually you agreed with this. I am sure that you are correct about what you are saying but I am still confused about things. I thought that you can only add any multiple of (2,3) to anything in ZxZ. That is why I do not see that (108,1) is in the same class as (0,1). It seems that you are saying that you can add any multiple of 2 to the 1st component and any multiple of 3 to the 2nd component. This is what I am not seeing. Can you please justify that?

Please see my comment above

 

[Ishuda]

I am a bit confused here. I thought that the elements of A are in the form (2m, 3m)-actually you agreed with this. I am sure that you are correct about what you are saying but I am still confused about things. I thought that you can only add any multiple of (2,3) to anything in ZxZ. That is why I do not see that (108,1) is in the same class as (0,1). It seems that you are saying that you can add any multiple of 2 to the 1st component and any multiple of 3 to the 2nd component. This is what I am not seeing. Can you please justify that?
Please see my comment above

The elements of A are of the form, (2m, 3n), i.e.
A = {(2m, 3n); m, n \(\displaystyle \epsilon\, Z\)}. SEE BELOW, THIS IS AN IDEAL BUT NOT THE PRINCIPAL IDEAL WE ARE TALKING ABOUT

However the factor ring/quotient ring/difference ring/residue class ring is, in maybe the broadest sense, a set of equivalence classes which one can say is the equivalence classes of module arithmetic. That is, isomorphic to the set of sets (set of equivalence classes)
{{(j, k) + A}; j = 0, 1; k = 0, 1, 2} = {{(j + 2m, k+ 2n)}; j = 0, 1; k = 0, 1, 2; ; m, n \(\displaystyle \epsilon\, Z\)}

EDIT: BTW: Yes, I know it the same thing said by HallsofIvy above, just in different words.

 

[Steven G]

The elements of A are of the form, (2m, 3n), i.e.
A = {(2m, 3n); m, n \(\displaystyle \epsilon\, Z\)}.

However the factor ring/quotient ring/difference ring/residue class ring is, in maybe the broadest sense, a set of equivalence classes which one can say is the equivalence classes of module arithmetic. That is, isomorphic to the set of sets (set of equivalence classes)
{{(j, k) + A}; j = 0, 1; k = 0, 1, 2} = {{(j + 2m, k+ 2n)}; j = 0, 1; k = 0, 1, 2; ; m, n \(\displaystyle \epsilon\, Z\)}

EDIT: BTW: Yes, I know it the same thing said by HallsofIvy above, just in different words.

So my definition of A was wrong?

 

[HallsofIvy]

So my definition of A was wrong?

Yes. In your first post you said
"a)Describe in words or by giving a pattern the elements of A.
A = {(2m,3m) | m is in Z}
and I agreed with that because I misread it! I thought you had written "A= {(2m, 3n)|m, n in Z}. That is what Ishuda told you. (And what "I shouda" said.)
The two components are independent of one another so require different multipliers.

 

[Ishuda]

So my definition of A was wrong?

In this case, a Principal Ring with generator a would be defined as the set of element {m a; m \(\displaystyle \epsilon\, Z\)} so guess who's wrong So we have
A = {...., (-12, -18), (-10, -15), (-8, -12), ..., (0, 0), ..., (8, 12), (10, 15), (12, 18), ...}
The equivalence classes for the factor ring R/A are defined by
x ~ y iff x-y is in A
so, I'm going to have to think about this.

EDIT: This sounds right at the moment. First note that x and y don't have to belong to A, just their difference. Let x= (x1, y1) and y=(x2,y2) then
x - y = (x1-x2, y1-y2)
Thus x1 - x2 is even and y1 - y2 is a multiple of 3, i.e.
x1 = x2 mod(2)
y1 = y1 mod(3)
Thus the base for the equivalent classes for x is {0, 1} and the base for the equivalent classes for y is {0, 1, 2}. So we have the same we had before (if we stop soon enough) and that is
R/A = {{(j, k) + A}; j = 0, 1; k = 0, 1, 2}