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Recall that $\mathbb{Z}[i]=\{a+bi:a,b \in \mathbb{Z}\}$, i.e., the Gaussian integers, and $\mathbb{Z}[\sqrt{2}]=\{a+b\sqrt{2}:a,b \in \mathbb{Z}\}$.

I want to show that $\mathbb{Z}[i] \not\cong \mathbb{Z}[\sqrt{2}]$.

Suppose, to the contrary, that they are isomorphic. Then there exists a bijective ring homomorphism $\phi: \mathbb{Z}[i] \rightarrow \mathbb{Z}[\sqrt{2}]$. Since $\phi$ is a homomorphism, it preserves additive and multiplicative identities, so $\phi(0)=0$ and $\phi(1)=1$. It also preserves sums, so for any $n \in \mathbb{Z}$, $\phi(n)=n$.

We know that $\phi(a+bi)=a'+b'\sqrt{2}$. I am trying to find an element in $\mathbb{Z}[i]$ that will map to an element in $\mathbb{Z}[\sqrt{2}]$ that will give me an equation in $\mathbb{Z}[\sqrt{2}]$ that has no solution. Some help?

Thank you.

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  • $\begingroup$ Have you tried finding an equation that involves the element $i$? $\endgroup$ – Erick Wong Apr 5 '17 at 16:40
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    $\begingroup$ The more conceptual definitions $\mathbb{Z}[i] = \mathbb{Z}[X]/(X^2+1)$ and $\mathbb{Z}[\sqrt{2}] = \mathbb{Z}[X]/(X^2-2)$, including the universal properties of such rings, suggest immediately what has to be done. $\endgroup$ – HeinrichD Apr 5 '17 at 16:43
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You are on the right track. Note that $\phi(i)=x+y\sqrt{2}$ and $$ -1=\phi(-1)=\phi(i^2)=\phi(i)^2=(x+y\sqrt{2})^2=(x^2+2y^2)+2xy\sqrt{2} $$ Can you solve this equation?

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$Z[\sqrt2]\subset \bf R$ contains no element of square root $-1$.

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  • $\begingroup$ Does this prove that the two rings are not isomorphic ? I do not think so. $\endgroup$ – Peter Apr 5 '17 at 17:46
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    $\begingroup$ If they are isomorphic the number of solutions of the equation $x^2+1=0$ is the same. $\endgroup$ – Thomas Apr 5 '17 at 19:38

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