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.
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Hint
$$(2\sqrt{2}+1)(2\sqrt{2}-1)=7 \\ 1=7-2 \cdot 3$$
N. S.
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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 – user315567 Jan 12 '17 at 22:30
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@user315567 Yes. Rationalization is very helpful to get "nice" numbers in this type of problems. – N. S. Jan 12 '17 at 22:32
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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 for more. – Bill Dubuque Jan 12 '17 at 23:33
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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.
Bill Dubuque
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