Why is $\mathbb{Z} [\sqrt{24}] \ne \mathbb{Z} [\sqrt{6}]$? 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.) 
 A: No need to apologize; and your instinct is correct. Note that $\sqrt{6}\notin\mathbb{Z}[\sqrt{24}]$; indeed,
$$\mathbb{Z}[\sqrt{24}]=\{a+b\sqrt{24}\mid a,b\in\mathbb{Z}\}=\{a+2c\sqrt{6}\mid a,c\in \mathbb{Z}\}.$$
Thus, $\mathbb{Z}[\sqrt{24}]$ consists of the elements of $\mathbb{Z}[\sqrt{6}]$ for which the number of times $\sqrt{6}$ occurs is even. However, when thinking about $\mathbb{Q}$, we now have $\frac{1}{2}$ available to us, and
$$\mathbb{Q}(\sqrt{24})=\{a+b\sqrt{24}\mid a,b\in\mathbb{Q}\}=\{a+2c\sqrt{6}\mid a,c\in \mathbb{Q}\}=\{a+d\sqrt{6}\mid a,d\in\mathbb{Q}\}=\mathbb{Q}(\sqrt{6})$$
because we can take $c=\frac{d}{2}$.
A: 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 $\mathbb{Z}$.
A: Since $\sqrt{6}\not\in\mathbb{Z} [\sqrt{24}]$.
