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How can I show, that the Ideal $(3, 1+2 \cdot \sqrt{2}) \subset \mathbb{Z}[\sqrt{2}]$ contains 1?

I tried to write 1 as a combination of $3$ and $1+2 \cdot \sqrt{2}$ but could not find any combination.

2 Answers 2

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Hint

$$(2\sqrt{2}+1)(2\sqrt{2}-1)=7 \\ 1=7-2 \cdot 3$$

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    ok that means, 7 is in the Ideal and 1 can be written by a linear combination of 7 and 3. thanks – 2017-01-12
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    @user315567 Yes. Rationalization is very helpful to get "nice" numbers in this type of problems. – 2017-01-12
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    @user315567 It is worth emphasizing that this "rationalization" is by way of the **norm** map, and this map often proves very useful to relate the multiplicative properties of $\,\Bbb Z\,$ and the ambient number ring, e.g. in some nice cases the number ring has unique factorization iff its monoid of norms does - [see here](http://math.stackexchange.com/a/95015/242) for more. – 2017-01-12
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${\rm mod}\,\ (\color{#c00}{\bf 3},\,1+2\sqrt{2})\!:\ \left[\, 1\equiv {\color{#c00}{\bf -2}\sqrt{2}\equiv \color{#c00}{\bf 1}\sqrt2}\:\right]^{\large 2}\! $ $\Rightarrow$ $\, 1\equiv 2\,\Rightarrow\, 0\equiv 1$

Remark $\ $ The common way is that since the ideal contains $\,w = 1+2\sqrt2,\,$ it also contains its norm $\,Nw = ww' = -7,\,$ so it contains $\,(3,7)=1.\,$ But the above modular approach is often easier in many cases, so it is worth knowing.