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

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    $\begingroup$ What is $Z*Z$ ? $\endgroup$ Nov 15, 2010 at 7:57
  • $\begingroup$ the product of Z and Z $\endgroup$
    – KEVEIN
    Nov 15, 2010 at 7:58
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    $\begingroup$ For a map $\varphi:\mathbb{Z}\times \mathbb{Z}\rightarrow \mathbb{Z}$, look at $\varphi(1,0), \;\varphi(0,1)$. what are the conditions on them so that $\varphi$ would be a ring homomorphism? $\endgroup$
    – Ofir
    Nov 15, 2010 at 8:11
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    $\begingroup$ The direct product of rings $R$ and $S$ is denoted by $R\times S$, not by $R*S$. For your question, note that under a ring homomorphism, idempotents go to idempotents. $\endgroup$ Nov 15, 2010 at 8:58

2 Answers 2

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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 - $f_1([m,n])=m,\; \; \;f_2([m,n])= n,\; \; \;f_3([m,n])=0$

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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.

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