$Z[\sqrt3$] = $\{a + b\sqrt3 \mid a,b \in Z \}$
To find out if the statement is true or not i tried the following. I created a surjective morphism that has $<x^2-27>$ as a kernel. And this way i've proved that the $<Z[x]/x^2-27>$ is isomorphic to $Z[3\sqrt3]$.
Now i want to prove that $Z[3\sqrt3]$ is not isomorphic with $Z[\sqrt3]$ ( or maybe is?).
Well i said that $Z[3\sqrt3]$ $= \{ a + 3*b\sqrt3 \}$. Well now i change the notation of $3*b$ with a number $c \in Z$. So i have $a + c\sqrt3$. But i can see that the elements $1\sqrt3 $ and $ 2\sqrt3$ are lost. I've been thinking that this might lack of elements and as such the isomorphism may not be true. But it is true also that i can still define a bijection between $Z[\sqrt3] - \{1\sqrt3,2\sqrt3,-1\sqrt3,-2\sqrt3\}$ and $Z[\sqrt3]$ which makes me wonder what do i have to do next.
Can someone help me out? A help with this approach would me wonderful, but of course any other approach is perfect for me!
Thank you in advance!
if $(a+b\sqrt{27})^2 = 3$, then $(a^2+27b^2)+2ab\sqrt {27} = 3$, and so $a^2+27b^2 = 3$ and $ab=0$.
If $a=0$ you get $9b^2 = 1$, and if $b=0$ you get $a^2 = 3$, both of which are impossible.
So there is no element in $\Bbb Z[\sqrt{27}]$ that squares to $3$