Timeline for What's the "best" proof of quadratic reciprocity?
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| Jul 6, 2022 at 16:37 | history | edited | darij grinberg | CC BY-SA 4.0 |
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| Apr 27, 2020 at 10:40 | comment | added | Franz Lemmermeyer | Try $a = b = c = 1$ and $d = 2$. The proof can be found in Caldero & Germoni, Histoire h\'edonistes de groupes et g\'eom\'etries,, p. 185 ff. They use that over finite fields, symmetric matrices with the same rank and the same determinant are congruent. | |
| Apr 22, 2020 at 22:44 | comment | added | caffeinemachine | @FranzLemmermeyer I am unable to follow the step where it says that 'since $\det(A)=1$, $A$ is congruent to the identity matrix.' I considered the following example. Let $F=\mathbb F_3$ and let $A=\begin{bmatrix}2& 0\\ 0& 2\end{bmatrix}$. Then $A$ has determinant $1$ in $F$. But if $A=P^tP$ for some $P=\begin{bmatrix}a& b\\ c& d\end{bmatrix}$, then we would have that there are $a^2+b^2=c^2+d^2=2, ab+cd=0$. But the first equation forces $a=b=c=d=1$, and thus $ab+cd=2\neq 0$. So $A$ is not congruent to $I$. | |
| Jul 25, 2011 at 17:11 | comment | added | Qiaochu Yuan | For odd primes! I guess there is probably a straightforward variant for $p = 2$. | |
| Jul 25, 2011 at 16:00 | history | answered | Franz Lemmermeyer | CC BY-SA 3.0 |