Let $K$ be a quadratic number field, $D$ its discriminant. Let $ax^2 + bxy + cy^2$ be an integral binary quadratic form such that $D = b^2 - 4ac$. It seems that gcd$(a, b, c) = 1$(see this question). How do you prove this if it is true?
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asked Nov 29, 2013 at 18:50
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2 Answers
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By writing down the ring of integers of $\mathbb{Q}(\sqrt{d})$ explicitly ($d$ square-free), you see that the discriminant of $\mathbb{Q}(\sqrt{d})$ is $d$ if $d\equiv 1\pmod 4$, and $4d$ otherwise. Since $d$ is square-free, you must have $gcd(a,b,c)=1$ or 2 (since the square of this gcd divides the discriminant). But if it's 2, then on the one hand, the discriminant is divisible by 4, so $d$ is not $1\pmod4$. On the other hand $b^2-4ac\equiv 4\pmod{16}$, so $\frac{b^2-4ac}{4}\equiv 1\pmod{4}$. This is a contradiction.
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$\begingroup$ Could you explain why $b^2-4ac\equiv 4\pmod{16}$? Regards, $\endgroup$ Nov 30, 2013 at 15:42
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1$\begingroup$ If $2|a,c$, then $16|4ac$. Also, we are assuming that $2|b$, i.e. $4|b^2$, and we cannot have $4|b$, since discriminants are not divisible by 16. Finally, the odd part $b'$ of $b$ satisfies $b'^2\equiv 1\pmod4$, so $b^2$ cannot be 12 mod 16. If you put that together, you get the claim. $\endgroup$– Alex B.Nov 30, 2013 at 18:10
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Let $k =$ gcd$(a,b,c)$. Then $ax^2 + bxy + cy^2 = k(a'x^2 + b'xy + c'y^2)$, where $a = ka', b = kb', c = kc'$. Then $D = k^2D'$, where $D' = b'^2 - 4a'c'$. Since $D' \equiv b'^2$ (mod $4$), $D' \equiv 0$ (mod $4$) or $D' \equiv 1$ (mod $4$). By this question, $D'$ can be written uniquely as $D' = f^2 d$, where $f$ is a positive integer and $d$ is the discriminant of a unique quadratic number field. Hence $D = k^2f^2d$. Since $D$ is a discriminant of a quadratic number field, by using the question again, $D = d$ and $k = 1$.
answered Dec 2, 2013 at 2:54