Consider the ideal $I = \langle(3,4) \rangle$ of the ring $\mathbb{Z} ×\mathbb{Z}$. Prove that $(\mathbb{Z}×\mathbb{Z})/I$ is not a domain.

Question

Having some trouble with this. Need to show that there exist some Zero Divisor, but not really sure what that would look like in $(\mathbb{Z}×\mathbb{Z})/I$

Thanks in advance

Answer 1

Hint: $(3,1) \cdot (1,4) = (3,4) \in I$.

Answer 2

Things in $(\mathbb{Z} \times \mathbb{Z})/I$ look like elements of $\mathbb{Z} \times \mathbb{Z}$ — that is, pairs of integers. The only difference between the two rings is that the former has an additional equivalence relation imposed upon it.

---

Sometimes, a change of representation of the elements can help. I often find it useful to apply the isomorphism $$ \mathbb{Z}[x] / \langle x(1-x) \rangle \cong \mathbb{Z} \times \mathbb{Z} $$ The isomorphism is determined by the map $f \mapsto (f(0), f(1))$. (check that this sends $x(1-x)$ to zero!) To compute the inverse, note, $a + (b-a)x \mapsto (a, b)$.

If you're comfortable manipulating quotients of polynomial rings, you can translate the given problem to this form and solve.