# Must certain rings be isomorphic to $\mathbb{Z}[\sqrt{a}]$ for some $a$

Consider the group $(\mathbb{Z}\times\mathbb{Z},+)$, where $(a,b)+(c,d)=(a+c,b+d)$. Let $\times$ be any binary operation on $\mathbb{Z}\times\mathbb{Z}$ such that $(\mathbb{Z}\times\mathbb{Z},+,\times)$ is a ring. Must there exist a non-square integer "$a$" such that $$(\mathbb{Z}\times\mathbb{Z},+,\times)\cong\mathbb{Z}[\sqrt{a}]?$$

Thank you.

Edit: Chris Eagle noted that setting $x\times y=0$ for all $x,y\in\mathbb{Z}\times\mathbb{Z}$ would provide a counterexample. I would like to see other ecounterexamples though.

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Do you require your rings to have a unit? – Miha Habič Jul 11 '13 at 9:15
@MihaHabic No. not necessarily. – Amr Jul 11 '13 at 9:15
Then the answer is obviously no: you could declare every product to be $0$, for exampele. – Chris Eagle Jul 11 '13 at 9:16
@ChrisEagle You are right. I would also be glad to see another counterexample. – Amr Jul 11 '13 at 9:17
How would you express $\mathbb{Z}\times\mathbb{Z}$ with componentwise multiplication as a ring of your form? – Miha Habič Jul 11 '13 at 9:20

## 2 Answers

Probably the most natural counterexample is the following:

If the operation $\times$ is defined such that the resulting ring is simply product of two copies of the usual ring $(\mathbb{Z},+,\times)$ (that is, if we set $(a,b)\times(c,d)=(ac,bd)$), then, again, no isomorphism exists, since the resulting ring $\mathbb{Z}\times \mathbb{Z}$ is not an integral domain and $\mathbb{Z}[\sqrt{a}]$ is.

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+1 I actually asked a bad question. – Amr Jul 11 '13 at 9:36

No. For example, the zero ring on the group $\Bbb{Z} \times \Bbb{Z}$ is not of this form.

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+1 Thank you. It would be great if there are other examples. – Amr Jul 11 '13 at 9:21