Determine the homomorphisms from $R\times R\to R$, where $R=\mathbb{Z}[i]$. 
Consider the ring of Gaussian integers $R=\mathbb{Z}[i]$. Determine all ring homomorphisms $R\times R\to R$ that map the identity of $R\times R$ to the identity of $R$.

I know that a homomorphism must map the generators of $R\times R$ to $R$, and in general looking at generators seems to be the strategy for these types of questions. However, $R\times R$ and $R$ are not generated by any single element. I know $R\times R$ is generated by the set $\{(1,0),(i,0), (0,1),(0,i)\}$. Is the homomorphism determined by where it send these elements? 
Apart from the identity homomorphism I know the following projection homomorphisms:
$$(a+ib, c + id) \mapsto (a+ib)$$ 
$$(a+ib, c+id) \mapsto (c+id)$$
and 
$$(a+ib, c + id) \mapsto (a-ib)$$ 
$$(a+ib, c+id) \mapsto (c-id)$$
How can I determine all the homomorphisms?
 A: I have an equivalent answer presented a slightly different way.
A homomorphic image of $R$ in $\mathbb Z[i]$ carrying $1$ to $1$ would have to have characteristic $0$, and it would be a domain. The kernel, then, is a prime ideal of $\mathbb Z[i]$.
It is useful to know that $\mathbb Z[i]\cong \mathbb Z[x]/(x^2+1)$ because the prime ideals are all known. There is, namely, the zero ideal $(x^2+1)$, and also $(p, x^2+1)$ for every prime $p$ such that $x^2+1$ is irreducible over $F_p$. Once you know the prime ideals of $R$, the prime ideals of $R\times R$ are just the ideals of the form $P\times R$ and $R\times P$ where $P$ is a prime ideal of $R$. 
Now, you don't have to worry about any of the cases involving a prime $p$, because once you introduce such a $p$, the characteristic becomes positive, and we have ruled that out above. So the only two possible kernels are $0\times R$ and $R\times 0$. Modulo the kernel, the map becomes an isomorphsim of $R\to R$. So now the question is, what are the isomorphisms of $R$ to $R$?
This is where the solution begins to converge with the others. If $\phi$ is such an isomorphism, you have, necessarily, that $\phi$ permutes the two roots of $x^2+1$. So either $i\mapsto i$ or $i\mapsto -i$. After having determined how $\phi$ acts on $\mathbb Z$ and $i$, you know how $\phi$ acts (linearly) on the rest of the elements of $R$, and that produces an isomorphism.
Lifting back these two cases for each of the two possible kernels, you recover exactly the maps you described, and now you can be confident there aren't any others. (We ruled them out early on.)
A: If you know where $(1,0)$ goes then you know where $(2,0)$ and $(3,0)$ and so on go because, for example, $f(2,0) = f(1,0) + f(1,0)$. So you can see that it is enough to know where $(1,0), (0,1), (i, 0)$, and $(0, i)$ go.


*

*You must have $f(1,0) + f(0,1) = 1$ since $f(1,1) = f(1_{R \times R}) = 1$.

*You must have $f(i,i)^2 = f(-1,-1) = -1$.

*You must have $f(1,0)^2 = f(1,0)$ and $f(0,1)^2 = f(0,1)$. So either $f(1,0) = 0$ or $f(1,0) = 1$ and similarly, $f(0,1) = f(1,1) - f(1,0) = 1 - f(1,0)$ is either $0$ or $1$.
etc. With enough properties like this you should eventually narrow down the ring homomorphisms. If I'm not mistaken, it is exactly those 4 maps you wrote down.
A: Yes, $R\times R$ is a free group generated by $a=(1,0), b=(0,1), c=(i,0), $ and $d=(0,i)$. Now consider an arbitrary homomorphism $\varphi$ such that $\varphi(1,1)= 1$. We must have that $\varphi(a) + \varphi(b)  = \varphi(a+b)= \varphi(1,1) = 1$. This produces two cases. 
Case 1: $\varphi(a) = 1$ and $\varphi(b) = 0$. Now note that $\varphi(c)^2 = \varphi(c^2) = \varphi(-a) = -\varphi(a) = -1$. Thus $\varphi(c) = \pm i$. Note that $\varphi(d)^2 = \varphi(d^2) = \varphi(-b) = -\varphi(b) = 0$. Thus $\varphi(d) = 0$. 
Case 2: $\varphi(a) = 0$ and $\varphi(b) = 1$. Now note that $\varphi(d)^2 = \varphi(d^2) = \varphi(-b) = -\varphi(b) = -1$. Thus $\varphi(d) = \pm i$. Note that $\varphi(c)^2 = \varphi(c^2) = \varphi(-c) = -\varphi(c) = 0$. Thus $\varphi(c) = 0$.  
We have found how a homomorphism must behaves on the generating set. Since homomorphisms are determined by where they send the generators we have the following possibilities. From Case 1 we get the following two homomorphisms
$$(x+iy, v+wi) \mapsto x+iy $$
$$(x+iy, v+wi) \mapsto x-iy $$
From Case 2 we have the following two homomorphisms:
$$(x+iy, v+wi) \mapsto v+wi $$
$$(x+iy, v+wi) \mapsto v-wi $$ 
Thus these are all the possible homomorphisms that map $(1,1)$ to $1$.
