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
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.
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
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... .