Exercise 3.5.11 ($\mathrm{gcd}(a,b,c)$ is an invariant.)

Question

Suppose that $ax^2 + bxy + cy^2 \sim Ax^2 + Bxy + Cy^2$. Show that $g.c.d.(a,b,c) = g.c.d.(A, B ,C)$.

Richard Ganaye · Accepted Answer

\begin{proof} Let $f,g$ be the quadratic forms $f(x,y) = ax^2 + bxy + cy^2,\ g(x,y) = Ax^2 + Bxy + Cy^2$. Write $ \delta = \mathrm{gcd}(a,b,c)$, and $\Delta = \mathrm{gcd}(A,B,C)$.
Then $g = f\cdot A$, where
\begin{align*}
A &= \begin{pmatrix} m_{11} & m_{12} \ m_{21} & m_{22} \end{pmatrix} \in \Gamma,
\end{align*}
so that 
\begin{align*}
g(x,y)  &= f(m_{11} x + m_{12} y, m_{21} x + m_{22y}) \
&= a(m_{11} x + m_{12} y)^2 + b (m_{11} x + m_{12} y) (m_{21} x + m_{22y}) + (m_{21} x + m_{22y})^2.
\end{align*}

By equations (3.7),
\begin{align*}
A &= a m_{11}^2 + b m_{11}m_{21} +cm_{21}^2, & (3.7a)\
B &= 2am_{11} m_{12} + b(m_{11}m_{22} + m_{12}m_{21}) + 2c m_{21} m_{22},& (3.7b)\
C &= am_{12}^2 + b m_{12}m_{22} + c m_{22}^2. &(3.7c)
\end{align*}
Since $\delta \mid a, \delta \mid b, \delta \mid c$, then $ \delta \mid A$ by (3.7a), $ \delta \mid B$ by (3.7b), $ \delta \mid C$ by (3.7c). Therefore $\delta \mid \Delta$.

Consider now
\begin{align*}
A ^{-1} &= \begin{pmatrix} m'_{11} & m'_{12} \ {m'}_{21} & {m'}_{22} \end{pmatrix}.
\end{align*}
Since $\Gamma$ is a group, $A^{-1} \in \Gamma$, so $m'_{11}, m'_{12}, m'_{21}, m'_{22}$ are integers, and $\det(A^{-1}) = 1$.

Moreover, for all $A,B \in \Gamma$,
\begin{align*}
(f \cdot A) \cdot B = f \cdot (AB),\qquad f \cdot I = f.
\end{align*}
(This says that $\Gamma$ acts on the right on the set of binary quadratic forms with integers coefficients.)

Therefore $g \cdot A^{-1} = f$, so that the equations (3.7) give
\begin{align*}
a &= A {m'_{11}}^2 + B {m'}_{11}m'_{21} +C{m'_{21}}^2, & (3.7a')\
b &= 2Am'_{11} m'_{12} + B(m'_{11}m'_{22} + m'_{12}m'_{21}) + C m'_{21} m'_{22},& (3.7b')\
c &= A{m'_{12}}^2 + B {m'}_{12}m'_{22} + C {m'_{22}}^2. &(3.7c')
\end{align*}
The same reasoning shows that $\Delta \mid \delta$.

Since $\delta \mid \Delta$ and $\Delta \mid \delta$, where $\delta \geq 0, \Delta \geq 0$, we obtain $\delta = \Delta$, that is
$$\mathrm{gcd}(a,b,c) = \mathrm{gcd}(A,B,C).$$
\end{proof}

Exercise 3.5.11 ($\mathrm{gcd}(a,b,c)$ is an invariant.)

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Exercise 3.5.11 ($\mathrm{gcd}(a,b,c)$ is an invariant.)

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