Describe all the ring homomorphisms of $\mathbb{Z}\times \mathbb{Z}$ into $\mathbb{Z}_7$.
My Attempt: $\mathbb{Z}_7$ is a field and having two idempotent element $0$ and $1$. By using this idempotent elements it will be solve. Can anyone help me with this?
On
Of course, there is (possibly, if you allowed it) the zero homomorphism. Let's set it aside for now.
Other than that, the only idempotent that $(1,1)$ can map to is $1\in\mathbb Z_7$. So the image of the map is an integral domain.
This means that the kernel of your map is a prime ideal of $\mathbb Z\times \mathbb Z$. Now, the prime ideals of this product of two rings are easy: you get them by finding $P\times \mathbb Z$ and $\mathbb Z\times P$ for each prime ideal $P\lhd \mathbb Z$.
So there are only four (types of) possibilities: $\{0\}\times \mathbb Z$,$(p)\times \mathbb Z, \mathbb Z\times \{0\}$, and $\mathbb Z\times (p)$ for primes $p$.
But the characteristic of $\mathbb Z_7$, and that cuts down on these possibilities dramatically. The quotient by the first or the third gives you a ring of characteristic $0$, and a quotient by the second or fourth gives a ring of characteristic $p$.
So in fact, the only two choices for kernels are $\mathbb Z\times (7)$ and $(7)\times \mathbb Z$. The first one corresponds to $(0,1)\mapsto 1$ and the second one corresponds to $(1,0)\mapsto 1$, and since $\mathbb Z_7$ is additively cyclic, the maps are completely determined.
$\Bbb Z \times \Bbb Z$ is generated by $a=(1,0)$ and $b=(0,1)$, so it is sufficient to note what $\varphi(a)$ and $\varphi(b)$ is, where $\varphi$ is the desired ring homomorphism.
It should be noted that $\varphi(a) \varphi(b) = \varphi(ab) = \varphi(0,0) = 0$. Since $\Bbb Z_7$ is a field, we have either $\varphi(a)=0$ or $\varphi(b)=0$.
It should also be noted that $\forall p,q:\varphi(1,1)\varphi(p,q)=\varphi(p,q)$ implies $\varphi(1,1)=1$ or $\forall p,q:\varphi(p,q)=0$, since $\Bbb Z_7$ is a field.
This leaves us with the homomorphism $\varphi(p,q)=0$, $\varphi(p,q) = p1$ or $\varphi(p,q) = q1$.