TITLE: Why is $\mathbb{Z} [\sqrt{24}] \ne \mathbb{Z} [\sqrt{6}]$?
QUESTION [6 upvotes]: Why is $\mathbb{Z} [\sqrt{24}] \ne \mathbb{Z} [\sqrt{6}]$, while $\mathbb{Q} (\sqrt{24}) = \mathbb{Q} (\sqrt{6})$ ?
(Just guessing, is there some implicit division operation taking $2 = \sqrt{4}$ out from under the $\sqrt{}$ which you can't do in the ring?)
Thanks. (I feel like I should apologize for such a simple question.)

REPLY [6 votes]: We have
\begin{align*}
\mathbb{Z}[\sqrt{24}] &= \{a + b\sqrt{24} | a, b \in \mathbb{Z} \} \\
&= \{a + 2b\sqrt{6} | a, b \in \mathbb{Z} \} \\
&= \{a + b'\sqrt{6} | a, b' \in \mathbb{Z} \text{ with } b' \text { even}\}. 
\end{align*}
which is clearly a proper subring of $\mathbb{Z}[\sqrt{6}]$. On the other hand, 
\begin{align*}
\mathbb{Q}[\sqrt{24}] &= \{a + b\sqrt{24} | a, b \in \mathbb{Q} \} \\
&= \{a + 2b\sqrt{6} | a, b \in \mathbb{Q} \} \\
&= \{a + b'\sqrt{6} | a, b' \in \mathbb{Q}\} \\
&= \mathbb{Q}[\sqrt{6}].
\end{align*}
The point is that you can divide anything in $\mathbb{Q}$ by two, but not anything in $\mathbb{Z}$.