Isomorphic Gaussian Integers

Question

Can we show that the ring of Gaussian integers $$\mathbb{Z}[\sqrt{17}]:=\{a+b\sqrt{17}:a,b\in\mathbb{Z}\}$$ $$\mathbb{Z}[\sqrt{11}]:=\{a+b\sqrt{11}:a,b\in\mathbb{Z}\}$$

equipped with standard addition and multiplication are not isomorphic?

If they were isomorphic, then wouldn't they both have the same number of solutions of the equation $x^2=11$? — Old John, Oct 27 '13 at 18:16

score 3 · Answer 1 · answered Oct 27 '13 at 18:17

3

Hint: Both rings can be equipped with multiplicative norms. Use this to show that $11$ is the product of nontrivial divisors in one ring, but not the other.

answered Oct 27 '13 at 18:17

user7530

49,280

score 3 · Accepted Answer · answered Oct 27 '13 at 18:17

3

An isomorphism of rings would map $1$ to $1$. Hence it would fix all integers. In $\mathbb Z[\sqrt{17}]$ we have $(\sqrt{17})^2 = 17$. It is easy to check that no element $x$ of $\mathbb Z[\sqrt{11}]$ satisfies $x^2 = 17$.

answered Oct 27 '13 at 18:17

Ayman Hourieh

39,603

Isomorphic Gaussian Integers

2 Answers2