All homomorphisms from the ring $\mathbb Z \times \mathbb Z \to \mathbb Z$

Question

How to describe all ring homomorphism from $\mathbb Z \times \mathbb Z \to \mathbb Z$ ?

Answer 1

To sum up all the comments\answers so far:

As Chandru wrote, if $f:\mathbb{Z}\times \mathbb{Z} \rightarrow \mathbb{Z}$ is a ring homomorphism, then it is also an abelian group homomorphism (for the addition operation), so $f([m,n]) = m \cdot f([1,0]) + n \cdot (f([0,1])$.

Since you also want the function to preserve multiplication, then you have other conditions. For example, as Robin commented f preserves idempotent -$f([0,1])=f([0,1][0,1])=f([0,1])f([0,1])$ so $x=f([0,1])$ is an element of $\mathbb{Z}$ that satisfies $x^2=x$ - therefore it must be 1 or 0.

f also preserves zero divisors so $0=f([0,0])=f([0,1][1,0])=f([0,1])f([1,0])$ which means that at least one of them is zero.

you are now left with three possibilities - $f_1([m,n])=m,\; \; \;f_2([m,n])= n,\; \; \;f_3([m,n])=0$

Answer 2

HINT $\ $ hom's preserve orthogonal idempotents, so $\ \{\:(0,1), (1,0)\:\}\to \{0,1\}\ $ if nontrivial

Answer 3

In general, the product of two rings $R \times S$ has a universal property concerning maps into it. But actually you can also describe maps on it: A homomorphism $R \times S \to T$ corresponds to an idempotent $e \in T$ together with homomorphisms $R \to eT, S \to (1-e)T$. Here, $eT$ is considered as a ring with unit $e$. A more natural definition of this is probably the localization $T_e$.

Thus if $T$ has no nontrivial idempotents (for example if $T$ is an integral domain), then you get either a homomorphism $R \to T$ (and on $S$ it's $0$) or a homomorphism $S \to T$.

In your example, remark that $\mathbb{Z}$ is the initial ring.