Can $\mathbb{Z}\times\mathbb{Z}$ be realized as a quotient ring of some integral domain?

Question

Question. Does there exist an integral domain $R$ and an ideal $I\subset R$ such that the quotient ring $R/I$ is isomorphic to $\mathbb{Z}\times \mathbb{Z}$?

Context. There is a rather standard exercise in abstract algebra which asks the following: Give an example to show that a quotient ring of an integral domain may have zero divisors. The simplest answer is to let $R=\mathbb{Z}$ and $I=4\mathbb{Z}$, and observe that $R/I$ has some zero divisors. It is natural to ask whether we can solve this exercise with other rings. In particular, if we can cook up an integral domain $R$ and some ideal $I$ such that $R/I \cong \mathbb{Z}\times\mathbb{Z}$, that would solve the problem too.

Bonus Question. Is there a characterization of commutative rings that can be realized as a quotient ring of some integral domain? Maybe every ring can be realized in this fashion?

Attempt. For $\mathbb{Q}\times \mathbb{Q}$, I can do this as follows. Let $R=\mathbb{Q}[x]$ and $I=(x^2-1)$. Then $R/I$ is isomorphic to $\mathbb{Q}[x]/(x-1) \times \mathbb{Q}[x]/(x+1)\cong \mathbb{Q}\times \mathbb{Q}$. Here, I used the fact that $(x-1)$ and $(x+1)$ are co-maximal, that is, $(x-1)+(x+1) = \mathbb{Q}[x]$, so I can apply the Chinese Remainder Theorem for rings: $R/(I\cap J) \cong R/I \times R/J$ where $I+J=R$.

Thank you for your time!

Answer 1

Here is another example: Consider the ring

$$ Q = \mathbb{Z}[x]/\langle x^2 - x \rangle, $$

and let $y = 1 - x$. Then it is easy to check the following:

1. $Q = \{ mx + ny : m, n \in \mathbb{Z} \}$. In fact, the quotient map $\varphi : \mathbb{Z}[x] \to Q$ is given by $$ f(x) \mapsto f(1)x + f(0)y. $$

2. $x^2 = x$, $y^2 = y$, and $xy = 0$.

Consequently, $Q = x\mathbb{Z} \oplus y\mathbb{Z} \simeq \mathbb{Z} \times \mathbb{Z} $.

Answer 2

Every ring is the quotient of an integral domain, for the silly reason that "free" rings are always integral domains.

Let's look at your example as a special case. We'll look at the integral domain

$$ \mathbb{Z}[x_{(m,n)} \mid (m,n) \in \mathbb{Z} \times \mathbb{Z}] $$

which is a polynomial ring in countably many variables. Now there's an obvious surjection (given by the universal property of polynomial rings)

$$ \varphi : \mathbb{Z}[x_{(m,n)}] \twoheadrightarrow \mathbb{Z} \times \mathbb{Z} \\ x_{(m,n)} \mapsto (m,n) $$

so the first isomorphism theorem tells us that $\mathbb{Z} \times \mathbb{Z}$ is isomorphic to the quotient $\mathbb{Z}[x_{(m,n)}] \big / \text{Ker}(\varphi)$.

Of course, there's nothing special here about $\mathbb{Z} \times \mathbb{Z}$, and we could play the same game for any ring $R$. It's always a quotient of the integral domain $\mathbb{Z}[x_r \mid r \in R]$.

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This turns out to have some pretty useful consequences! You should think of the "free ring" $\mathbb{Z}[x_r \mid r \in R]$ as providing a kind of "syntax" for $R$, with the ring $R$ itself provding the semantics.

It turns out that this syntactic ring has nice properties (like being an integral domain) that $R$ itself might lack, and so we can prove things at the level of syntax (where these "niceness" assumptions hold) and then project down to $R$ itself! You can see an old blog post of mine for examples of this idea in action, as well as some old mse questions like this one.

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I hope this helps ^_^