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I=<(20 + 24$\sqrt{(-7)}$, n> is an ideal. I am trying to find an n$\in$$Z$, so that I$\neq$$Z$$[\sqrt-7$].

So far my idea was to try to see how $\frac{Z}{(20 + 24\sqrt{(-7)})}$ would look like.

With $a \in Z$, $\frac{a}{(20 + 24\sqrt{(-7)})}$=$\frac{20a}{4432}-\frac{24a\sqrt{-7}}{4432}$. (amplified with $20-24\sqrt{-7}$).

I believe this would be above $Z[\sqrt{-7}]$ (generally, this fraction $\in Q[\sqrt{-7}]$). Therefore, when adding the n in the ideal I, I have to make sure the fractions $\frac{20a}{4432} , \frac{24a\sqrt{-7}}{4432}$ shouldn't both be in $Z[\sqrt{-7}]$ $\forall a \in Z$, or, to rephrase, at least one of these is true:

  1. $4432 \nmid 20a$ (mod n) $\forall a\in Z$
  2. $4432 \nmid 24a\sqrt{-7}$ (mod n) $\forall a\in Z$

Is any n , with gcd(4432, n)$\neq$1 good here? (I think, if gcd(4432,n)$\neq$1 then 4432 is not a unit and...the fractions don't become in $Z[\sqrt{-7}]$.

1 Answers1

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I cannot follow your reasoning but the following mine leads to the same conclusion.

$$\begin{align}I=\Bbb Z[\sqrt{-7}]&\iff1\in I \\&\iff\exists a,b,c,d\in\Bbb Z\quad1=\left(20+24\sqrt{-7}\right)\left(a+b\sqrt{-7}\right)+n\left(c+d\sqrt{-7}\right) \\&\iff\exists a,b\in\Bbb Z\quad n\mid 20a-7\cdot24b-1,\quad n\mid20b+24a \\&\iff\exists a,b,e\in\Bbb Z\quad n\mid 20a-168b-1,\quad5b+6a=ne \\&\iff\exists a,b,e,f\in\Bbb Z\quad n\mid 20a-168b-1,\quad b=6f-ne,\quad a=-5f+ne \\&\iff\exists f\in\Bbb Z\quad n\mid 20\left(-5f\right)-168\left(6f\right)-1=-1108f-1 \\&\iff\gcd(1108,n)=1 \\&\iff2\not\mid n\text{ and }227\not\mid n .\end{align}$$

Anne Bauval
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  • I see, thank you. I guess I probably should have checked for $a \in$$Z[\sqrt{-7}]$ instead of just integers, going by that path... Either way, I've never seen a solution like this, so I have at least 1 question. I see you assumed n $\nmid$4, but how did $ne$ disappear in line 6? –  Nov 25 '22 at 10:59
  • I did not "assume $n\nmid4$", I deduced $n$ odd from $n\mid 20a-168b-1.$ And $ne$ disappeared because $n\mid 20\left(-5f\right)-168\left(6f\right)-1+n\cdot{\rm blah}$ is equivalent to $n\mid 20\left(-5f\right)-168\left(6f\right)-1.$ – Anne Bauval Nov 25 '22 at 12:00
  • oh, I see. very interesting solution, and I think I understood how that worked. thank you. Gave your comment the check, I think that closes this thread pretty much? –  Nov 25 '22 at 22:37