1. Finding all homomorphisms

My goal was to find all homomorphisms $\displaystyle \mathbb{Z} \oplus \mathbb{Z} \to \mathbb{Z} \oplus \mathbb{Z}$, and $\displaystyle \mathbb{Z} \oplus \mathbb{Z} \to \mathbb{Z}$, and $\displaystyle \mathbb{Q} \to \mathbb{Q}$. Here's what I have...

For $\displaystyle \mathbb{Z} \oplus \mathbb{Z} \to \mathbb{Z} \oplus \mathbb{Z}$ I believe they are $\displaystyle (a,b) \mapsto 0$, $\displaystyle (a,b) \mapsto (a,b)$, $\displaystyle (a,b) \mapsto (b,a)$, $\displaystyle (a,b) \mapsto (a,0)$, and $\displaystyle (a,b) \mapsto (0,b)$.

For $\displaystyle \mathbb{Z} \oplus \mathbb{Z} \to \mathbb{Z}$ I have the identity, the 0-map, $\displaystyle (a,b) \mapsto a$, and $\displaystyle (a,b) \mapsto b$.

And for $\displaystyle \mathbb{Q} \to \mathbb{Q}$ I just have the 0-map and the identity.

Did I get them all? And when there are more 'creative' ones like $\displaystyle (a,b) \mapsto (b,a)$, how do we know when we've found them all?

2. what about (a,b) ---> (ma,nb) and (a,b)--->(nb,ma)?

and for Z⊕Z -->Z, what about (a,b) --> ma+nb?

3. Neither of the first two work unless n, m= 0 or 1 because upon finding $\displaystyle \phi$(a,b)$\phi$(c,b)$$, there are $\displaystyle n^2$ and $\displaystyle m^2$ terms that do not appear with $\displaystyle \phi$(a,b)(c,d)$$, which just has n and m.

The third one doesn't work since $\displaystyle \phi$(a,b)(c,d)$=mac+nbd$ and $\displaystyle \phi$(a,b)$\phi$(c,b)$=m^2ac+mand+nbmc+n^2bd$

The second does give $\displaystyle (a,b) \mapsto (b,0)$ and $\displaystyle (a,b) \mapsto (0,a)$.

4. perhaps you should indicate what kind of structure the homomorphisms are to preserve?

5. Agreed. Ring homomorphisms?

6. Yes, ring homomorphisms. I suppose that would be good to know.

7. Originally Posted by AlexP
My goal was to find all homomorphisms $\displaystyle \mathbb{Z} \oplus \mathbb{Z} \to \mathbb{Z} \oplus \mathbb{Z}$, and $\displaystyle \mathbb{Z} \oplus \mathbb{Z} \to \mathbb{Z}$, and $\displaystyle \mathbb{Q} \to \mathbb{Q}$. Here's what I have...

For $\displaystyle \mathbb{Z} \oplus \mathbb{Z} \to \mathbb{Z} \oplus \mathbb{Z}$ I believe they are $\displaystyle (a,b) \mapsto 0$, $\displaystyle (a,b) \mapsto (a,b)$, $\displaystyle (a,b) \mapsto (b,a)$, $\displaystyle (a,b) \mapsto (a,0)$, and $\displaystyle (a,b) \mapsto (0,b)$. (a,b) → (a,a) ? (a,b) → (b,b) ?

For $\displaystyle \mathbb{Z} \oplus \mathbb{Z} \to \mathbb{Z}$ I have the identity, the 0-map, $\displaystyle (a,b) \mapsto a$, and $\displaystyle (a,b) \mapsto b$.

And for $\displaystyle \mathbb{Q} \to \mathbb{Q}$ I just have the 0-map and the identity.

Did I get them all? And when there are more 'creative' ones like $\displaystyle (a,b) \mapsto (b,a)$, how do we know when we've found them all?
A ring homomorphism has to take idempotents to idempotents. In the ring $\displaystyle \mathbb{Z} \oplus \mathbb{Z}$, the only idempotents are (0,0), (1,0), (0,1) and (1,1). Also, every element of $\displaystyle \mathbb{Z} \oplus \mathbb{Z}$ 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.

8. Thank you, that does clear things up. Ring homomorphisms are more rigid than I was thinking they are...I saw something like $\displaystyle (a,b) \mapsto (b,a)$ 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.