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I am doing Problem 5.10 of Marcus where it is given that $m$ is a square-free negative integer and that $\mathcal{O}_K$ is a PID where $K = \Bbb{Q}(\sqrt{m})$. Now in part (b) of this problem he asks:

5.10 (b): Suppose $p$ is an odd prime such that $m < 4p$. Show that $m$ is a non-square mod $p$.

Is there a typo in this question? Firstly I am confused because $m$ in question 5.10 is supposed to be a negative integer. Then sure any negative integer will be less than $4p$ for any given odd prime $p$. Further if we take $p = 3$ and the negative integer $m = -2$ clearly $-2 = 1\mod{3}$ that is a square. It is also clear that $\mathcal{O}_K$ where $K= \Bbb{Q}(\sqrt{-2})$ is a PID.

Are there supposed to be some further hypothesis on $m$ and $p$ or is the problem just plain wrong? Or is there a typo in the inequality $m < 4p$?


Edit: Brett Frankel has requested that I add part (a) of the problem (which I have done) :

5.10(a): Show that $m \equiv 5\mod{8}$ except when $m \equiv -1,-2,$ or $-7$. (Consider a prime lying over $2$).

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What is part (a)? With some more context one may be able to solve the problem and figure out what's supposed to be going on. –  Brett Frankel Jan 26 '13 at 3:20
@BrettFrankel I have added (a) of the problem. Please see my question for the edit. –  fpqc Jan 26 '13 at 3:28
@BrettFrankel Hmm perhaps I am beginning to think Marcus meant in (b): Given $m \equiv 5 \mod{8}$ and $m \neq -1,-2,-7$ show that...( ) –  fpqc Jan 26 '13 at 3:29
I think you want to use the facts that either $m\equiv5\pmod 8$ or $m\in\{-1,-2,-7\}$ and then either $p<4|m|$ or $|m|<4|p|$. The inequality may serve to limit the number of times you have to apply quadratic reciprocity to compute the answer. –  Brett Frankel Jan 26 '13 at 3:36
@BrettFrankel The second inequality $m < 4p$ to me is redundant because $m$ is already negative. The typos in this text sometimes drive me crazy. –  fpqc Jan 26 '13 at 3:39
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1 Answer

up vote 2 down vote accepted

I think what Marcus meant was: Suppose $\mathcal{O}_K$ is a PID. Then prove under the assumption that $p < \operatorname{disc}\mathcal{O}_K/4$, we have $m$ a non-square mod $p$.

This is proved easily as follows. If $p < \operatorname{disc}\mathcal{O}_K/4$ then there is no element of norm $p$ in $\mathcal{O}_K$, thus no non-trivial factor of $p$ is principal. This forces $m$ to be a quadratic non-residue mod $p$. Otherwise if $m \equiv n^2 \mod{p}$ then

$$p\mathcal{O}_K = \begin{cases} (p,\sqrt{m}), &\text{if $p |m$} \\ (p,n +\sqrt{m})(p,n-\sqrt{m}),& \text{if $p\nmid m$} \end{cases}$$

which contradicts $\mathcal{O}_K$ being a PID.

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