Prove that in the ring $Z[\sqrt2]$: $\langle3+8\sqrt2, 7\rangle$ = $\langle3+\sqrt2\rangle$

Question

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.

Answer 1

$\langle3+8\sqrt2, 7\rangle \supseteq \langle3+\sqrt2\rangle$

Indeed, $3+\sqrt2= (1)\cdot(3+8\sqrt2)+(-\sqrt2)\cdot(7)$

$\langle3+8\sqrt2, 7\rangle \subseteq \langle3+\sqrt2\rangle$

Indeed, $$ \frac{3+8\sqrt2}{3+\sqrt2} = \frac{(3+8\sqrt2)(3-\sqrt2)}{(3+\sqrt2)(3-\sqrt2)} = \frac{-7+21\sqrt2}{7} = -1+3\sqrt2 $$ $$ \frac{7}{3+\sqrt2} = \frac{(7)(3-\sqrt2)}{(3+\sqrt2)(3-\sqrt2)} = \frac{(7)(3-\sqrt2)}{7} = 3-\sqrt2 $$

Answer 2

Hint: To show $\ \langle 3+8\sqrt{2}, 7\rangle=\langle 3+\sqrt{2}\rangle\ $, you need to show:

• $\ 3+8\sqrt{2} = r_1\left(3+\sqrt{2}\right)\ $ and $\ 7 = r_2\left(3+\sqrt{2}\right)\ $ for some $\ r_1\ $ and $\ r_2\ $ in $\ \mathbb{Z}\left[\sqrt{2}\right]\ $; and
• $\ 3+\sqrt{2}=s_1\left(3+8\sqrt{2}\right) + s_2 7\ $ for some $\ s_1\ $ and $\ s_2\ $ in $\ \mathbb{Z}\left[\sqrt{2}\right]\ $.

The first part is very straightforward: simply calculate $\ r_1=\frac{3+8\sqrt{2}}{3+\sqrt{2}}\ $ and $\ r_1=\frac{7}{3+\sqrt{2}}\ $ in $\ \mathbb{Q}\left[\sqrt{2}\right]\ $, and check to see whether the results both lie in $\ \mathbb{Z}\left[\sqrt{2}\right]\ $. The second is not much more difficult. A solution for $\ s_1\ $ a small integer, and $\ s_2\ $ a small multiple of $\ \sqrt{2}\ $, is fairly easy to find by trial and error.

Answer 3

Simply multiplying the generators of the ideals by $\,\sqrt{2}\,$ will not work.

To be more precise, $\,\langle a\rangle = \{\alpha a \mid \alpha \in R\}\,$ and $\,\langle b,c\rangle = \{\beta b + \gamma c\mid \beta, \gamma \in R\}.\,$

To show that two ideals are equal, it is sufficient to show that the each generator of one is an $\,R$-linear combination of the generators of the other, and vice versa. In your specific case, you need to show that:

(i)$\quad\,7=r_1(3+\sqrt2),\,$ for some $\,r_2 \in R\,;\,$ and

(ii) $\quad\,3+8\sqrt2=r_2(3+\sqrt2),\,$ for some $\,r_1 \in R\,;$

(iii)$\quad3+\sqrt2=r_3(3+8\sqrt2)+r_4(7),\,$ for some $\,r_3, r_4 \in R\,.$

For part (i), you might note by inspection that $\,r_1 = 3-\sqrt2,\,$ as $\,(3-\sqrt2)(3+\sqrt2)=9-2=7;\,$ if that's not immediately apparent, you can just do $\,r_1=\frac{7}{3+\sqrt2}\,$ etc.

Then for part (ii), note that $\,3+8\sqrt2=(3+\sqrt2)+7\sqrt{2},\,$ and then you can just substitute the $\,7\,$ with the expression found in part (i) and solve for $\,r_2\,$.

Finally for part (iii), note that $\,3+\sqrt2=(3+8\sqrt2)-(\sqrt{2})(7),\,$ so it should be immediately clear what $\,r_3\,$ and $\,r_4\,$ should be.

The complete solutions are below in case any of the steps are proving elusive.

(i)

By inspection, $\,r_1=3-\sqrt{2};\, \text{or } \,r_1=\dfrac{7}{3+\sqrt2}=\dfrac{7(3-\sqrt2)}{(3+\sqrt2)(3-\sqrt2)}=\dfrac{7(3-\sqrt2)}{9-2}=3-\sqrt2\,$

(ii)

$\,3+8\sqrt2 \, = (3+\sqrt2)+7\sqrt{2} \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad = (3+\sqrt2) + (3+\sqrt2)(3-\sqrt2)\sqrt{2} \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad = (3+\sqrt2) \bigl[\,1+(3-\sqrt2)\sqrt{2}\,\bigr] \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad= (3+\sqrt2) \bigl[\,1+3\sqrt2-2\,\bigr] \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad = (3+\sqrt2) (3\sqrt2-1)\, \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \text{so} \,r_2 = (3\sqrt2-1).$

(iii)

$\quad3+\sqrt2=(3+8\sqrt2)-(\sqrt{2})(7),\,$ so $\,r_3=1\,$ and $\,r_4=-\sqrt{2}.$

Answer 4

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 $\langle 1+\sqrt{2}\rangle\not\subseteq \langle3+\sqrt{2}\rangle.$

Answer 5

A la Euclid: $\ (\color{#c00}7,\,3\!+\!8\sqrt2) = (7,\!\!\!\!\!\!\!\!\!\overbrace{3\!+\!\sqrt 2}^{\textstyle 3\!+\!8\sqrt2\!-\!\color{#c00}7\sqrt 2)}\!\!\!\!\!\!\!\!\!\!) = (3\!+\!\sqrt 2)\,(3\!-\!\sqrt2,\,1) = (3\!+\!\sqrt 2)$