# Factor ring with zero-divisors

I am asked if the ideal $I = \left \{ \left ( x,7y \right ): x,y \in \mathbb{Z} \right \}$ of a ring R= $\mathbb{Z}\oplus \mathbb{Z}$ is a prime ideal.

Looking at the factor ring, $\mathbb{Z}\oplus \mathbb{Z} / I$, there exists zero-divisors;

in particular $\left ( x_{1},0 \right ) and \left ( 0,y_{2} \right ) \in \mathbb{Z}\bigoplus \mathbb{Z} \forall$ $x_{1}, y_{2}$ non-zero.

Hence, there exists a zero-divisor and hence the factor ring is not an integral domain and so the ideal $I$ is not prime.

However, my solution says the factor ring is an integral domain.

Am I wrong or is my solution wrong?

Note that the zero element of $\mathbb{Z}\oplus \mathbb{Z} / I$ is $I=0+I.$

Let $$((x_1,y_1)+I)((x_2,y_2)+I)=0+I$$ Then $$(x_1x_2,y_1y_2)\in I$$ Thus $$7\mid y_1y_2$$ As $7$ is prime, therefore $7 \mid y_1$ or $7 \mid y_2$. Hence, $(x_1,y_1) \in I$ or $(x_2,y_2)\in I$.

So, $(x_1,y_1)+I = 0 + I$ or $(x_2,y_2)+I=0+I$. So, $\mathbb{Z}\oplus \mathbb{Z} / I$ is an integral domain.

$(x_1, 0)\in I$, and therefore does not represent a zero divisor in the quotient ring; it represents zero.

As for proving that the quotient ring is an integral domain, you could start by showing that each element in the quotient ring has a representative of the form $(0, y)$, where $0\leq y \leq 6$, and then study how these elements multiply with one another.

It would be a good exercise to prove the following results.

Result-1: In $\Bbb Z \times \Bbb Z$ the prime ideals are of the form, $\Bbb Z/p\Bbb Z\times \Bbb Z$ or $\Bbb Z\times\Bbb Z/p\Bbb Z$ where $p$ is prime.

In fact you can generalize the above result as follows.

Result-2: Let $R_i$ be a comutative ring with unity for each $i=1,...,n$. The prime ideals of $R_1\times ...\times R_n$ are of the form $P_1\times...\times P_n$ where for some $j$, $P_j$ is a prime ideal of $R_j$ and $P_i=R_i$ for $i\not=j$.

In your case, $I=\Bbb Z \times \Bbb Z/7\Bbb Z$. Hence it follows that $I$ is a prime ideal.