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Say L = $\mathbb{Q}$($\sqrt{m}$) be a quadratic number field generated by square-free number 'm'. Say 'p' is an odd prime ideal in ring of integers of L, which means that 'p' does not occur in factorization of $\langle 2 \rangle$ in $\mathcal{O}_{L}$.

Now say a,b are two elements in ring of integers such that ideal generated by 'a' and 'b' are coprime to 'p'. Like we did in integers, can I surmise that, one of 'a', b or ab must be a square modulo 'p'. This would be the case if equivalently defined Legendre Symbol in $\mathcal{O}_{L}$ would follow following rule for mentioned 'a', 'b' and odd 'p'.

($\frac{ab}{p}$) = ($\frac{a}{p}$).($\frac{b}{p}$)

My question is , is above thing true?

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    $\begingroup$ You might be looking for Hilbert's reciprocity law (for the quadratic Hilbert symbol). $\endgroup$ – Eric Towers Nov 2 '19 at 20:02
  • $\begingroup$ This is an elementary question on (characters of) finite cyclic groups; it has nothing to do with reciprocity. $\endgroup$ – franz lemmermeyer Nov 3 '19 at 11:22
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Actually, I was just reading Franz Lemmermeyer's chapter $12$ on Quadratic Reciprocity in Number Fields, which answers your question! It works indeed similarly.

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