what about (a,b) ---> (ma,nb) and (a,b)--->(nb,ma)?
and for Z⊕Z -->Z, what about (a,b) --> ma+nb?
My goal was to find all homomorphisms , and , and . Here's what I have...
For I believe they are , , , , and .
For I have the identity, the 0-map, , and .
And for I just have the 0-map and the identity.
Did I get them all? And when there are more 'creative' ones like , how do we know when we've found them all?
A ring homomorphism has to take idempotents to idempotents. In the ring , the only idempotents are (0,0), (1,0), (0,1) and (1,1). Also, every element of is a linear combination of (1,0) and (0,1), so a homomorphism is specified by what it does to those two elements. That severely restricts the possible choices for a homomorphism.
Thank you, that does clear things up. Ring homomorphisms are more rigid than I was thinking they are...I saw something like as being somewhat random and didn't know how it fit in with the properties of homomorphisms, but now I see what the basis for it is. Thanks to everyone else too.