Exercise 2.9.11

Question

Determine all proper representations of $3$ by $f= (1,1,1)$.

Richard Ganaye · Accepted Answer

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ewcommand{\C}{\mathbb{C}}

ewcommand{\F}{\mathbb{F}}

ewcommand{\U}{\mathbb{U}}

\begin{proof}
The discriminant of $f$ is $-3$. All $\Gamma$-orbits of a form $(3, \cdot, \cdot)$ have a representative $(3, B, C)$ with $-2 \leq B \leq 3$. We must find all pairs $(B,C)$ such that $B^2 - 12 C = -3$ and $-2 \leq B \leq 3$. This implies $B \equiv 0 \pmod 3$, thus $B = 0$ or $B = 3$. $B = 0$ gives no solution, and $B = 3$ gives the solution $g = (3, 3, 1)$. Thus $(3,3,1) \cdot \Gamma$ is the only $\Gamma$-orbit of forms $(3,\cdot, \cdot)$ properly equivalent to $(1,1,1)$.

By exercise 2.9.10, 
$$g = (3,3,1) = f \cdot \begin{pmatrix} 1 & 0 \ 1 & 1 \end{pmatrix},$$
so the first column $(1,1)$ is a proper representation of $3 = f(1,1)$.

By exercise 2.9.9, the group $\mathrm{Aut}^+(f)$ is cyclic of order 6, generated by 
$$A =  \begin{pmatrix} 0 & -1 \ 1 & 1 \end{pmatrix}.$$
Therefore, all proper representations of $3$ by $f= (1,1,1)$ are given by $A^k\cdot (1,1),\ 0 \leq k <6$, that is
$$(1,1), (-1,2), (-2,1), (-1,-1), (1,-2),(2,-1).$$
\end{proof}

Exercise 2.9.11

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Exercise 2.9.11

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