Exercise 3.7.5

Proof. Since $\mathrm{\Delta }p\ge 0$, $b(\mathrm{\Delta },p)$ and $c(\mathrm{\Delta },p)$ are well defined.

Assume first that $p\ne 2$.

$\text{(⇒)}$

Assume that $p$ divides the conductor $f=f(\mathrm{\Delta })$. Then $f=p{f}^{\prime }$ for some integer $f^{\prime }$.

By definition of the conductor, $\mathrm{\Delta }∕f^2$ is a discriminant, thus there are integers $A,B,C$ such that $\mathrm{\Delta }∕f^2=B^2-4\mathit{AC}$. By definition of $C(\mathrm{\Delta },p)$, $\mathrm{\Delta }=b{(\mathrm{\Delta },p)}^2-4\mathit{pc}(\mathrm{\Delta },p)$. Therefore

$$\begin{array}{lll}\hfill \mathrm{\Delta }& =p^2{f{}^{\prime }}^2(B^2-4\mathit{AC})=b{(\mathrm{\Delta },p)}^2-4\mathit{pc}(\mathrm{\Delta },p).& \hfill \text{(1)}\end{array}$$

This equality shows that $p\mid b(\mathrm{\Delta },p)$. Thus $p^2\mid 4\mathit{pc}(\mathrm{\Delta },p)$, so $p\mid c(\mathrm{\Delta },p)$ (here $p$ is odd). This proves that $(p,b(\mathrm{\Delta },p),c(\mathrm{\Delta },p))$ is not primitive.

To conclude, if $(p,b(\mathrm{\Delta },p),c(\mathrm{\Delta },p))$ is primitive, then $p$ does not divide the conductor of $\mathrm{\Delta }$.

$\text{(⇐)}$

Conversely, assume that $(p,b(\mathrm{\Delta },,p),c(\mathrm{\Delta },p))$ is not primitive. Then $$p\mid b(\mathrm{\Delta },p),p\mid c(\mathrm{\Delta },p).$$

Write $b(\mathrm{\Delta },p)=p{b}^{\prime },c(\mathrm{\Delta },p)=p{c}^{\prime }$, where $b^{\prime },c^{\prime }$ are integers.

Since $\mathrm{\Delta }=b{(\mathrm{\Delta },p)}^2-4\mathit{pc}(\mathrm{\Delta },p)$, we obtain $\mathrm{\Delta }=p^2({b{}^{\prime }}^2-4c^{\prime })$ therefore $\mathrm{\Delta }∕p^2={b{}^{\prime }}^2-4c^{\prime }$ is a discriminant, of the form $(1,b^{\prime },c^{\prime })$, and $\mathrm{\Delta }=p^{2}m$, where $m\equiv 0,1(\mathit{mod}4)$.

By definition of the conductor, $\mathrm{\Delta }∕f^2$ is the largest positive integer $f$ such that $\mathrm{\Delta }∕f^2$ is a discriminant, thus $\mathrm{\Delta }=f^2{m}^{\prime }$, where $m^{\prime }\equiv 0,1(\mathit{mod}4)$.

We must show that $p\mid f$. If not, the equality $\mathrm{\Delta }=p^{2}m=f^2{m}^{\prime }$, where $p\nmid f$, shows that $p^2\mid m^{\prime }$, thus $m^{\prime }=p^2{m}^{″}$, where $p^2\equiv 1(\mathit{mod}4)$, and $m^{″}\equiv 0,1(\mathit{mod}4)$. Then $\mathrm{\Delta }∕{(\mathit{fp})}^2=m^{″}$ is a discriminant, in contradiction with the definition of the conductor. This contradiction shows that $p\mid f$.

It remains the case $p=2$.

$\text{(⇒)}$

Assume that $2$ divides $f$. Then $f=2f^{\prime },f^{\prime }\in \mathbb{Z}$. As in the case $p\ne 2$, $\mathrm{\Delta }∕f^2=B^2-4\mathit{AC}$, and the equality (1) becomes $$\begin{array}{lll}\hfill \mathrm{\Delta }=4{f{}^{\prime }}^2(B^2-4\mathit{AC})=b{(\mathrm{\Delta },2)}^2-8c(\mathrm{\Delta },2).& & \hfill \text{(2)}\end{array}$$

Therefore $b(\mathrm{\Delta },2)$ is even, and $\mathrm{\Delta }\equiv 0(\mathit{mod}4)$, and $\mathrm{\Delta }2=0$.

If $\mathrm{\Delta }\equiv 0(\mathit{mod}8)$, then $b(\mathrm{\Delta },p)=0$ (see the definition (3.10)). The equality (2) shows that $f^{\prime }$ or $B$ is even, and $4{f{}^{\prime }}^2{B}^2\equiv -8c(\mathrm{\Delta },2)(\mathit{mod}16)$, thus $0\equiv {f{}^{\prime }}^2{B}^2\equiv -2c(\mathrm{\Delta },2)(\mathit{mod}4)$, so $2\mid c(\mathrm{\Delta },p)$. Therefore $(2,b(\mathrm{\Delta },p),c(\mathrm{\Delta },p))=(2,0,c(\mathrm{\Delta },p))$ is not primitive.

If $\mathrm{\Delta }\equiv 4(\mathit{mod}8)$, then $b(\mathrm{\Delta },p)=2$. The equality (2) shows that $f^{\prime }$ and $B$ are odd, and $4{f{}^{\prime }}^2{B}^2\equiv 4-8c(\mathrm{\Delta },2)(\mathit{mod}16)$, thus $1\equiv {f{}^{\prime }}^2{B}^2\equiv 1-2c(\mathrm{\Delta },2)(\mathit{mod}4)$, so $c(\mathrm{\Delta },p)$ is even. Therefore $(2,b(\mathrm{\Delta },p),c(\mathrm{\Delta },p))=(2,2,c(\mathrm{\Delta },p))$ is not primitive.

$\text{(⇐)}$

Conversely, assume that $(2,b(\mathrm{\Delta },2),c(\mathrm{\Delta },2))$ is not primitive. Then $b(\mathrm{\Delta },p)$ and $c(\mathrm{\Delta },p)$ are even. Write $b(\mathrm{\Delta },p)=2b^{\prime },c(\mathrm{\Delta },p)=2c^{\prime },b^{\prime }{c}^{\prime },\in \mathbb{Z}$.

Then $\mathrm{\Delta }=4({b{}^{\prime }}^2-4c^{\prime })=4m$, where $m\equiv 0,1(\mathit{mod}4)$.

By definition of the conductor, $\mathrm{\Delta }∕f^2$ is the largest positive integer $f$ such that $\mathrm{\Delta }∕f^2$ is a discriminant, thus $\mathrm{\Delta }=f^2{m}^{\prime }$, where $m^{\prime }\equiv 0,1(\mathit{mod}4)$.

We must show that $2\mid f$. If not, the equality $\mathrm{\Delta }=4m=f^2{m}^{\prime }$ shows that $m^{\prime }\equiv 0(\mathit{mod}4)$. Write $m^{\prime }=4m^{″}$. Then $m=f^2{m}^{″}$, where $f$ is odd, thus $m^{″}\equiv m\equiv 0,1(\mathit{mod}4)$. But then $\mathrm{\Delta }∕{(2f)}^2=m^{″}$ is a discriminant, in contradiction with the definition of the conductor $f$. This contradiction shows that $2\mid f$.