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Is there a ring homomorphism between $\mathbb{Z}$ into $\mathbb{Z} \times \mathbb{Z}$?

Also for any rings $R_{1}$ and $R_{2}$ does there exist a ring homomorphism $\phi$ : $R_{1} \rightarrow$ $R_{1} \times R_{2}$?

Note: I am allowing 1 in both rings. So any homomorphism must map the identity on one ring to the identity on the other ring.

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    $\begingroup$ Well there's the zero map, but I'm assuming you're talking about unital maps? $\endgroup$ – stochasm Apr 5 '16 at 22:48
  • $\begingroup$ Since $\Bbb Z$ is cyclic, any isomorphism $\phi: \Bbb Z \to R$ is determined by $\phi(1)$. $\endgroup$ – Travis Apr 5 '16 at 22:49
  • $\begingroup$ If by unital you mean those that send 1 to 1 then yes. $\endgroup$ – user1058860 Apr 5 '16 at 22:50
  • $\begingroup$ with $\mathbb{Z}^2$ you meant $\mathbb{Z}[i]$ ? $\endgroup$ – reuns Apr 5 '16 at 22:56
  • $\begingroup$ @Travis does that necessarily imply existence of $\phi$? Could you give an example? $\endgroup$ – user1058860 Apr 5 '16 at 22:59
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Suppose $R$, $S$ and $T$ are rings. Giving a ring homomorphism $f\colon R\to S\times T$ is the same as giving homomorphisms $g\colon R\to S$ and $h\colon R\to T$.

Let's see why. First, the projection maps $p\colon S\times T\to S$ and $q\colon S\times T\to T$ are ring homomorphisms, so if we are given $f\colon R\to S\times T$, we get $p\circ f\colon R\to S$ and $q\circ f\colon R\to T$.

Suppose instead we are given $g\colon R\to S$ and $h\colon R\to T$. Define $$ f(x)=(g(x),h(x)) $$ It's easy to see that $f$ is a ring homomorphism and that $g=p\circ f$, $h=q\circ f$.

In general terms this is the statement that $S\times T$ is the product in the category of rings.

This answers your second question: in order to have a ring homomorphism $R\to R\times T$ you need a ring homomorphism $R\to T$ (and use, for instance, the identity as the homomorphism $R\to R$). So, for a counterexample find rings $R$ and $T$ such that there is no ring homomorphism $R\to T$; take $R$ to have characteristic $2$ and $T$ to have characteristic $3$, for instance.

In the case of $\mathbb{Z}$, it's a standard result that, for any ring $R$, there is a unique ring homomorphism $\mathbb{Z}\to R$ (when unital rings are concerned and homomorphisms are required to preserve the identity element).

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For any ring $R$ there is a unique homomorphism $\phi:\mathbb{Z}\to R$ defined $\phi(1)=1_R$. This is unique because then $\phi(n)=n\cdot 1_R$ for all $n\in\mathbb{Z}$. The category theoretic term for this is that $\mathbb{Z}$ is the initial object in the category of rings.

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  • $\begingroup$ Is there a homomorphism defined in the same way for $\mathbb{Z} \times \mathbb{Z} \rightarrow X$ for a ring $X$ $\endgroup$ – user1058860 Apr 5 '16 at 23:21
  • $\begingroup$ Not quite, because a homomorphism $\phi:\mathbb{Z}\times \mathbb{Z}\to X$ is no longer determined by $\phi(1)$. $\endgroup$ – carmichael561 Apr 6 '16 at 0:04
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As for the second question:

There is no homomorphism $\mathbb Z_5\rightarrow \mathbb Z_5\times \mathbb Z$.

This is because the image must be isomorphic to a quotient of $\mathbb Z_5$, and is therefore finite. On the other hand it contains $n(1,1)$ for every $n$, this gives us an infinite number of elements.

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