<s>[INST] Prove that in the ring $Z[\sqrt2]$: $\langle3+8\sqrt2, 7\rangle$ = $\langle3+\sqrt2\rangle$ So, 
I know that $\langle a\rangle$ = $aR$ and $\langle a,b\rangle$ = $aR + bR$ for any ring R.
I then multiplied each side by $\sqrt2$ and computed them.
These are the steps that I took:
$\langle3+8\sqrt2, 7\rangle$ = $\langle3+\sqrt2\rangle$
$(3+8\sqrt2)\sqrt2 + 7\sqrt2 = (3+\sqrt2)\sqrt2$
$3\sqrt2 + 16 + 7\sqrt2 = 3\sqrt2 + 2$
$16 + 10\sqrt2 = 2+3\sqrt2$
I'm not really sure where to go from here.
Any help would be much appreciated. 
 [/INST] Note that since $1,-\sqrt{2} \in\mathbb{Z}[\sqrt{2}],$ 
$(3+8\sqrt{2})\cdot1-7\cdot\sqrt{2}=3+\sqrt{2}\in\langle 3+8\sqrt{2},7\rangle.$ 
Since $\forall r\in\mathbb{Z},1\cdot r\in\mathbb{Z}$ and $-\sqrt{2}\cdot r\in\mathbb{Z},$ we have that $\forall r\in\mathbb{Z},(3+\sqrt{2})r\in\langle 3+8\sqrt{2},7\rangle\\
\Rightarrow \langle 3+\sqrt{2}\rangle\in\langle3+8\sqrt{2},7\rangle.$ 
Divide $3+8\sqrt{2}$ by $3+\sqrt{2}$ to get a quotient of $-1+3\sqrt{2}\in\mathbb{Z}[\sqrt{2}].$ 
Similarly, the quotient of $\dfrac{7}{3+\sqrt{2}}$ is $3-\sqrt{2}\in\mathbb{Z}[\sqrt{2}].$ So we have that $\forall r\in \langle3+8\sqrt{2},7\rangle,r\in\langle3+\sqrt{2}\rangle\\
\Rightarrow \langle 3+8\sqrt{2},7\rangle\subseteq \langle 3+\sqrt{2}\rangle.$
This is because $3+8\sqrt{2}$ and $7$ are both multiples of $3+\sqrt{2}$ in $\mathbb{Z}[\sqrt{2}].$ 
Thus, since $\langle 3+\sqrt{2}\rangle\subseteq\langle 3+8\sqrt{2},7\rangle$ and $\langle 3+8\sqrt{2},7\rangle\subseteq \langle 3+\sqrt{2}\rangle,$ we have that $\langle 3+8\sqrt{2},7\rangle = \langle 3+\sqrt{2}\rangle.$
Both conditions must be satisfied, or else they may not be equal. For instance, the ideal
$\langle 6+2\sqrt{2}\rangle\subseteq\langle3+\sqrt{2}\rangle$ but
$\langle3+\sqrt{2}\rangle \not\subseteq\langle6+2\sqrt{2}\rangle,$ so the ideals are unequal.
As well, $\langle 3+\sqrt{2}\rangle \subseteq \langle 1+\sqrt{2}\rangle$ but $\langle 1+\sqrt{2}\rangle\not\subseteq \langle3+\sqrt{2}\rangle.$ 
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