Exercise 3.7.3

Question

Show that all integral forms $(a,b,c)$ whose discriminant is a fundamental discriminant are primitive.

Richard Ganaye · Accepted Answer

ewcommand{\Q}{\mathbb{Q}}

ewcommand{\Z}{\mathbb{Z}}

ewcommand{\N}{\mathbb{N}}

ewcommand{\R}{\mathbb{R}}

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

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

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

{\bf Lemma.}  Let $\Delta 
e 0$ be an integer. Then $\Delta$ is a discriminant if and only if there is some integer quadratic form $f= (a,b,c)$ such that $\Delta = b^2 - 4ac = \mathrm{discr}(f)$.

\medskip
{\it Proof of lemma.}
\begin{enumerate}
\item[$(\Leftarrow)$] If $\Delta = b^2 -4ac$, then $\Delta \equiv b^2 \equiv 0,1 \pmod 4$.
\item[$(\Rightarrow)$] If $\Delta \equiv 0 \pmod 4$, then $\Delta$ is the discriminant of the integer form $(1,0, \Delta/4)$, and if $\Delta \equiv 1 \pmod 4$, then $\Delta$ is the discriminant of $(1,1, (1-\Delta)/4)$.
\end{enumerate}
\qed
\begin{proof}(Ex.3.7.3.)
 Let $f =(a,b,c)$ be a form with discriminant $\Delta$, where $\Delta$ is a fundamental discriminant. Reasoning by contradiction,  assume that $f = (a,b,c)$ is not primitive. Then there exists $d>1$ such that $f = (dA, d B, dC),\ A,B,C \in \Z$. Thus 
$\Delta= d^2(B^2-4AC)$, where $d>1$. Therefore $\Delta/d^2 = B^2 - 4AC$ is the discriminant of $(A,B,C)$. Using the Lemma, this proves that the conductor satisfies
$$f(\Delta) \geq d>1,$$
and so $\Delta$ is not fundamental.

All integral forms $(a,b,c)$ whose discriminant is a fundamental discriminant are primitive.
\end{proof}

\bigskip
Note: The converse is true.

\medskip
{\bf Proposition.} \ Let $\Delta$ be an integer. Then $\Delta$ is fundamental if and only if every form $f$ of discriminant $\Delta$ is primitive.
\begin{proof}
The direct part is proved in Exercise 3.7.3.

Assume now that every form of discriminant $\Delta$ is primitive.

Reasoning by contradiction, suppose that there is some integer $f>1$ such that $f ^2 \mid \Delta$ and $\Delta/f^2$ is a discriminant. Using the Lemma, whe know that there is some form $g = (a,b,c)$ such that $\Delta/f^2 = \mathrm{discr}(g) = b^2 - 4ac$. Then $ \Delta = f^2(b^2 - 4ac)$ is the discriminant of the form $(fa,fb,fc)$, which is not primitive. This contradiction proves that there is no $f >1$ such that $\Delta/f^2$ is a discriminant. Therefore $\Delta$ is a fundamental discriminant.
\end{proof}

Exercise 3.7.3

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Exercise 3.7.3

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