Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
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 2

up vote 4 down vote accepted

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.

share|improve this answer
    
+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.

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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.