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I am starting to learn about rings and ideals in abstract algebra. I came across a textbook problem that I am having a lot of trouble solving:

Prove that for any positive integer $n$ ending in $7$, the ideal generated by $n$ and $1+\sqrt{11}$ in $\mathbb{Z}[\sqrt{11}]$ is trivial. In other words, that it is the same as the whole ring.

I would appreciate any help on this!

user26857
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John
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  • I have gone through the ring $Z$ and I think I have a good understanding of it and its ideals. I just saw this question, and I was very curious. I do not know yet about Norm-Euclidean rings. Also related to the question above, how do you represent elements in the ideal $\mathbb{Z} [\sqrt{11}$? Are they of the form $(n, 1+\sqrt{11}) = {nx + y ( 1+ \sqrt{11}) | x, y ; \in \mathbb{Z}[\sqrt{11}] }$? – John Oct 26 '20 at 20:08
  • Norm Euclidean implies that we have a gcd. This is used in the answers. See for example this post, where there is no gcd of tow elements. – Dietrich Burde Oct 26 '20 at 20:10
  • Is it possible to prove this without using the $gcd$ argument? – John Oct 29 '20 at 00:19

3 Answers3

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Ideals are closed under norm: $\,w\in I\Rightarrow N(w) = w\bar w \in I.\,$ Here $\,w = 1+\sqrt{11}\,$ has norm ${-}10$ so $\,I\supset (10,n) = (10,\,n\bmod 7) = (10,7) = 1,\,$ so $\,I = (1).\,$ Similarly every ideal contains the gcd of the norms of all generators (a special case of the method of simpler multiples), e.g. here.

Bill Dubuque
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Hint: Note that the ideal also contains the element $$(1+\sqrt{11})(1-\sqrt{11})=1-11^2=-10.$$

Servaes
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Assume $n$ is positive. Then we have that $n=10k+7$ for some $k \in \mathbb{Z}$ and $(1-\sqrt{11})(1+\sqrt{11})=-10$. Thus it follows that $\gcd(n, 10)=1$, thus $1 \in I$.

user26857
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ben huni
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  • $n=10k+7$ thus $3n=3 \times 10k +2 \times 10 +1$, so $(3k+2)10+1=3n$. Finally the result is that $3n+(3k+2)(-10)=1$. Sorry I was not explicit. :) – ben huni Nov 05 '20 at 11:47