# $\mathbb{Z}[\sqrt{d_1}]$ and $\mathbb{Z}[\sqrt{d_2}]$ are not isomorphic

If $d_1, d_2$, and $d_1/d_2$ are not squares in $\mathbb{Z}\backslash\left\{ 0\right\}$, show that $\mathbb{Z}[\sqrt{d_1}]$ and $\mathbb{Z}[\sqrt{d_2}]$ are not isomorphic

-

Well, suppose we have an isomorphism $i$, that means in $\mathbb{Z}[\sqrt{d_2}]$ we have an element $a + b\sqrt{d_2}$ which squares to $d_1$ (namely $i(\sqrt{d_1})$).
We read off: $$a^2 + 2ab\sqrt{d_2} + d_2 b^2 = d_1$$For the first term to lie in $\mathbb{Z}$, the middle term has to be zero (since $d_2$ wasn't a square), hence $a$ or $b$ is zero.
If $a$ is zero, we have $d_2 b^2 = d_1$, which would mean $d_1/d_2$ isn't square free, if $b$ is zero, we have $a^2 = d_1$, meaning $d_1$ isn't square free.
you mean $i(d_1)=d_1$, i don't think this is true –  Firmino Oct 29 '12 at 10:13
@Firmino $i(d_1)$ HAS to equal $d_1$ for $i$ to be an isomorphism (remember, $d_1$ is a non-zero integer). And again, since $i$ is an isomorphism, $i(\sqrt{d_1}^2) = i(\sqrt{d_1})^2$. Setting $i(\sqrt{d_1}) = a + b\sqrt{d_2}$, you get the above argument. –  Arthur Oct 29 '12 at 12:18
@Firmino to see why $i(d_1) = d_1$, we have $1 \mapsto 1$ always (to be a ring map), so $1 + 1 \mapsto 1 + 1$, $3 \mapsto 3$, and (inductively) $d_1 \mapsto d_1$. –  uncookedfalcon Oct 29 '12 at 20:28
Yes, I see, $i(1)=1$ –  Firmino Nov 4 '12 at 5:43