Can anyone help me with an effective approach to this problem

Find all possible ring homomorphisms betweeen the indicated rings

(a) :

(b) Q( 2) Q( 3)

Would be most grateful of help.

Peter

Printable View

- Sep 5th 2011, 03:25 AMBernhardRing Homomorphisms
Can anyone help me with an effective approach to this problem

Find all possible ring homomorphisms betweeen the indicated rings

(a) :

(b) Q( 2) Q( 3)

Would be most grateful of help.

Peter - Sep 5th 2011, 03:59 AMSwlabrRe: Ring Homomorphisms
If then note that there are no elements of order 3 in . So, if is to map anywhere non-trivially it must map to a copy of . So, do this and see where it takes you.

There are no non-trivial ones between and . This is a classic question - basically, show that 0 must map to 0, and 1 to 1, but 1+1=2, which has a square root in the pre-image, and 1+1+1=3 which has a square root in the image. This is a contradiction (can you see why?). I will leave it to you to formalise it all. - Sep 5th 2011, 12:16 PMDrexel28Re: Ring Homomorphisms
- Sep 6th 2011, 02:39 AMSwlabrRe: Ring Homomorphisms
- Sep 6th 2011, 01:47 PMDrexel28Re: Ring Homomorphisms
- Sep 7th 2011, 03:05 AMSwlabrRe: Ring Homomorphisms
- Sep 7th 2011, 03:33 AMBernhardRe: Ring Homomorphisms
Thanks to those posting help ... but I am still struggling to determine an answer ...

In answer to the question:

"Find all possible homomorphisms between the indicated groups: : "

Swlabr writes: (thanks for help by the way!)

"If : , then note that there are no elements of order 3 in . So if is to map anywhere nontrivially it must map to a copy of . So do this and see where it takes you."

However as far as I got was as follows:

I can see that the homomorphism is completely determined by ( ) =

and, since has order 6 then we have 0( ) | 6

Thus 0( ) is 6, 3,2 or 1 (Correct?)

... but ... what exactly are the next few steps?

Would be grateful for some help

Peter - Sep 7th 2011, 04:50 AMBernhardRe: Ring Homomorphisms
Just a note:

I can see that if you check all elements of that you find that only the element has an order that divides 6.

But what is a convenient way to do this if enumeration is difficult?

Peter - Sep 7th 2011, 04:52 AMOpalgRe: Ring Homomorphisms
The first essential is that you need to be clear about whether you are looking for

*group*homomorphisms or*ring*homomorphisms. In a previous posting, you asked about group homomorphisms from to As you correctly say in the above quote, if is a group homomorphism with then [m] must have order 6, 3, 2 or 1 in the additive group of integers mod 10. But that group has no elements of order 3 or 6. Therefore [m] must have order 1 or 2.

But there is an extra complication if you are looking for a ring homomorphism, because ring homomorphisms have to preserve multiplication (in this case, multiplication mod 10) as well as addition. In fact, a ring homomorphism between unital rings is either identically zero, or it has to take the identity element to an idempotent element. In this case, that means that [m] must be [0], [1] or [5]. But it can't be [1] because that has order 10... .