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?

• You might be looking for Hilbert's reciprocity law (for the quadratic Hilbert symbol). – Eric Towers Nov 2 '19 at 20:02
• This is an elementary question on (characters of) finite cyclic groups; it has nothing to do with reciprocity. – franz lemmermeyer Nov 3 '19 at 11:22

Actually, I was just reading Franz Lemmermeyer's chapter $$12$$ on Quadratic Reciprocity in Number Fields, which answers your question! It works indeed similarly.