Is there a ring homomorphism between $\mathbb{Z}$ into $\mathbb{Z} \times \mathbb{Z}$? 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.
 A: 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).
A: 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.
A: 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.
