0
$\begingroup$

7 is a Heegner number. Therefore the integer ring $O_K$ corresponding to $K=\mathbb{Q}[\sqrt{-7}]$ is a unique factorization domain. Now, it is easy to show that $\mathbb{Z}[\sqrt{-7}]\subset O_K$, i.e. $O_K \neq \mathbb{Z}$, because $x=\sqrt{-7}$ is a solution to $x^2+7=0$. On the other hand, $8=2^3=(1+\sqrt{-7})(1-\sqrt{-7})$. Why is this not a non-unique pair of factorizations? If $O_K=\mathbb{Z}[\sqrt{-7}]$, its pretty easy to show that $2$ and $1\pm\sqrt{-7}$ are not further factorable (resorting to the fact that the square of the modulus must be integral).

But even if we look at all $\mathbb{Q}[\sqrt{-7}]$, we can show that 2 cannot be factored except trivially.

Where am I going wrong????

Edit: "GH from MO"'s answer is really all you need. But, my second last line above, asserting "2 cannot be factored except trivially" - was wrong. All else above is correct. Last line was not wrong-headed though, the key was to look outside of $\mathbb{Z}[\sqrt{-7}]$, because $O_K$ is indeed larger.

$\endgroup$
  • $\begingroup$ Given this "MathOverflow is for mathematicians to ask each other questions about their research. See Math.StackExchange to ask general questions in mathematics.", this is indeed off-topic. Can mods move it to Math.StackExchange possibly? $\endgroup$ – David I. McIntosh Mar 20 '18 at 12:43
11
$\begingroup$

The ring of integers of $\mathbb{Q}(\sqrt{-7})$ is not $\mathbb{Z}[\sqrt{-7}]$ but $\mathbb{Z}\left[\frac{1+\sqrt{-7}}{2}\right]$. In this ring, the ideal $(2)$ factors into prime ideals as follows: $$ (2)=\left(\frac{1+\sqrt{-7}}{2}\right)\left(\frac{1-\sqrt{-7}}{2}\right). $$ For further information, see here.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Arrrrggg. I thought it might be that the integral ring was larger, and I looked for a factorization of 2 - hence my second last line. Should have been easy (and now that I think about it, I only needed to divide my factorization equation by 4 to get it!!!). Now I will have to go back and see what stupid mistake I made. Thanks. $\endgroup$ – David I. McIntosh Mar 18 '18 at 23:08

Not the answer you're looking for? Browse other questions tagged or ask your own question.