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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?

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2 Answers 2

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$\Bbb Z \times \Bbb Z$ 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 $\varphi(p,q) = q1$.

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  • $\begingroup$ Depending on the definition of "ring homomorphism", the requirement $\varphi(1,1) = 1$ implies $k=1$. $\endgroup$
    – pisco
    Sep 9, 2017 at 8:50
  • $\begingroup$ @pisco125 or $k=0$, as I have edited. $\endgroup$
    – Kenny Lau
    Sep 9, 2017 at 8:55
  • $\begingroup$ how to we prove p and q belongs to Z_7 ? $\endgroup$
    – user293581
    Sep 9, 2017 at 9:13
  • $\begingroup$ @rajendra sorry, they don't. I've edited. $\endgroup$
    – Kenny Lau
    Sep 9, 2017 at 9:16
  • $\begingroup$ sorry i donot understand what is p1 and q1 $\endgroup$
    – user293581
    Sep 9, 2017 at 9:23
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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.

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