Exercise 9.37

Proof. (of Lemma.) Let $\pi =a+bi$ a primary prime in $\mathbb{Z}[i]$.

$\bullet$

If $\pi =-q$, where $q\equiv 3(mod4),q>0$ is a rational prime, then $a=-q,b=0$. By definition of the quartic character, $${\chi }_q(i)=i^{}$$

N(q) - 14 = i^ q^2-1 $4$ . $$

Write $-q=a=4k+1,k\in \mathbb{Z}$. Then

$$\begin{array}{llll}\hfill \frac{q^2-1}{4}& =4k^2+2k& \hfill & \\ \hfill & \equiv 2k=\frac{a-1}{2}(mod4).& \hfill & \end{array}$$

Therefore

$${\chi }_{-q}(i)={\chi }_q(i)=i^{\frac{q^2-1}{4}}=i^{}$$

a-12 = ${\left(\frac{1}{i}\right)}^{}$

-a+12= (-i)^ -a+1 $2$ = ( − 1 )

-a+12 i ^ -a+1 $2$ = i

-a+12,

$$since$$

(-1)^ -a+1 $2$ = ( − 1 ) − 2 k = 1 .

Suppose now that $N(\pi )=p$, where $p\equiv 1(mod4)$ is a rational prime. Then

$${\chi }_{\pi }(i)=i^{}$$

N( $\pi )-14=i^{}$

p-14.

$$Since$$

$\pi =a+bi$ is primary, there are two cases.

$\bullet$

If $a\equiv 1(mod4),b\equiv 0(mod4)$, then $a=4A+1,b=4B,A,B\in \mathbb{Z}$. $$\begin{array}{llll}\hfill \frac{p-1}{4}& =\frac{a^2+b^2-1}{4}& \hfill & \\ \hfill & =\frac{16A^2+8A+16B^2}{4}& \hfill & \\ \hfill & =4A^2+2A+4B^2& \hfill & \\ \hfill & \equiv 2A=\frac{a-1}{2}& \hfill & \end{array}$$

Therefore

$${\chi }_{\pi }(i)=i^{\frac{p-1}{4}}=i^{\frac{a-1}{2}}={\left(\frac{1}{i}\right)}^{}$$

-a+12= (-i)^ -a+1 $2$ = ( − 1 )

-a+12 i ^ -a+1 $2$ = i

-a+12,

$$since$$

(-1)^ -a+1 $2$ = ( − 1 ) − 2 A = 1 .

$\bullet$

If $a\equiv 3(mod4),b\equiv 2(mod4)$, then $a=4A-1,B=4B+2,A,B\in \mathbb{Z}$. $$\begin{array}{llll}\hfill \frac{p-1}{4}& =\frac{a^2+b^2-1}{4}& \hfill & \\ \hfill & =\frac{16A^2-8A+16B^2+16B+4}{4}& \hfill & \\ \hfill & =4A^2-2A+4B^{2}4B+1& \hfill & \\ \hfill & \equiv -2A+1=\frac{-a+1}{2}(mod4)& \hfill & \end{array}$$

Therefore ${\chi }_{\pi }(i)={(-1)}^{}$

-a+14 $.$

The equality ${\chi }_{\pi }(i)={(-1)}^{}$

-a+14 $isverifiedforallprimaryprimes$ $\pi$.

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Proof. (of Ex.9.37) Let $\pi =a+ib$ be a primary irreducible in $\mathbb{Z}[i]$.

$\bullet$

If $b=0$, then $\pi =a\in \mathbb{Z}$. As $\pi$ is primary, $\pi =-q,q\equiv 3(mod4)$, where $q$ is a rational prime, so $a=-q,b=0$. By Ex. 9.32 (or its generalization 9.35), $${\chi }_{\pi }(1+i)={\chi }_{-q}(1+i)=i^{(-q-1)∕4}=i^{(a-b-b^2-1)∕4}.$$

$\bullet$

If $b\ne 0$, then $a\wedge b=1$ (see Ex. 9.36), and by Ex. 9.35, $${\chi }_{\pi }(a){\chi }_{\pi }(1+i)=i^{3(a+b-1)∕4}.$$

$\bullet$

If $\pi \equiv 1[4]$, $a\equiv 1[4],b\equiv 0[4]$ : $a=4A+1,b=4B,A,B\in \mathbb{Z}$.

By Ex. 9.34(a),

$${\chi }_{\pi }(a)=i^{(a-1)∕2},{\chi }_{\pi }{(a)}^{-1}={(-i)}^{(a-1)∕2}=i^{(a-1)∕2}.$$

$$\begin{array}{llll}\hfill {\chi }_{\pi }(1+i)& =i^{3\frac{a+b-1}{4}-2\frac{a-1}{4}}& \hfill & \\ \hfill & =i^{\frac{a+3b-1}{4}}& \hfill & \\ \hfill & =i^{\frac{a-b-b^2-1}{4}},& \hfill & \end{array}$$

since $\left(\frac{a+3b-1}{4}\right)-\left(\frac{a-b-b^2-1}{4}\right)=b+\frac{b^2}{4}=4B+4B^2\equiv 0[4]$.

$\bullet$

If $\pi \equiv 3+2i[4]$, $a\equiv 3[4],b\equiv 2[4]$ : $a=4A-1,b=4B+2,A,B\in \mathbb{Z}$.

By Ex. 9.34(b),

$${\chi }_{\pi }(a)=-i^{(-a-1)∕2},{\chi }_{\pi }{(a)}^{-1}=-i^{(a+1)∕2}=i^{(a-3)∕2},$$

so

$${\chi }_{\pi }(1+i)=i^{(3a+3b-3+2a-6)∕4}=i^{(5a+3b-9)∕4}.$$

Now $\frac{1}{4}[(a-b-b^2-1)-(5a+3b-9)]=\frac{1}{4}(-4a-4b-b^2+8)=-a-b+2-\frac{b^2}{4}=-4A+1-4B-2+2-{(2B+1)}^2\equiv 0[4]$,

thus ${\chi }_{\pi }(1+i)=i^{(a-b-b^2-1)∕4}$.

Conclusion : if $\pi =a+ib$ is primary irreducible, then

$${\chi }_{\pi }(1+i)=i^{(a-b-b^2-1)∕4}$$

Second part : the biquadratic character of 2 (see Ex. 5.25 to 5.28).

Let $p\equiv 1[4]$. Then $p=N(\pi )$, where $\pi =a+bi$ is a primary prime.

We show first that ${\chi }_{\pi }(2)=i^{}$

ab2 $.$

Since $2=i^3{(1+i)}^2$, the first part of the exercise, and the Lemma, give

$$\begin{array}{llll}\hfill {\chi }_{\pi }(2)& ={\chi }_{\pi }{(i)}^3{\chi }_{\pi }{(1+i)}^2& \hfill & \\ \hfill & =i^{}& 3(-a+1)2i^\frac{a-b-b^2-1}{2}\hfill \end{array}$$

= i ^ 1 - a - ( b + 1 ) b 2

Since $\pi$ is primary, $a\equiv b+1\equiv -b+1(mod4)$, therefore

$$\begin{array}{llll}\hfill 1-a-(b+1)\frac{b}{2}& \equiv -b-(b+1)\frac{b}{2}& \hfill & \\ \hfill & \equiv \frac{b}{2}(-b-3)& \hfill & \\ \hfill & \equiv \frac{b}{2}(-b+1)& \hfill & \\ \hfill & \equiv \frac{ab}{2}(mod4),& \hfill & \end{array}$$

so ${\chi }_{\pi }(2)=i^{}$

ab2 $.$

Now we show that $p$ is of the form $p=A^2+64B^2$ if and only if $p\equiv 1(mod4)$ and if $x^4\equiv 2$ has a solution $x\in \mathbb{Z}$.

If $p=A^2+64B^2=A^2+{(8B)}^2$, then the prime number $p$ is a sum of two squares, and $p\ne 2$, therefore $p\equiv 1(mod4)$. Since $p=A^2+64B^2$, $A$ is odd. Put $b=8B$, and $a=A$ if $A\equiv 1(mod4)$, $a=-A$ if $A\equiv -1(mod4)$. Then $\pi =a+bi$ is such that $N(\pi )=a^2+b^2=p$, and $a\equiv 1,b\equiv 0(mod4)$, therefore $\pi$ is a primary prime. Then

$${\chi }_{\pi }(2)=i^{}$$

ab2 = i^4aB = 1.

$$Thereforethereexists$$

$\alpha \in D$ such that $2\equiv {\alpha }^4(mod\pi )$. As $D∕\pi D$ is the set of classes of $0,1,\cdots ,p-1$, there exists $x\in \mathbb{Z}$ such that $x\equiv \alpha (mod\pi )$, so $2\equiv x^4(mod\pi )$.

Then $p=N(\pi )\mid N(x^4-2)={(x^4-2)}^2$, thus $p\mid x^4-2$, in other words $2\equiv x^4(modp)$.

Conversely, suppose that $p\equiv 1(mod4)$ and that $2$ is a biquadratic residue modulo $p$. As $p\equiv 1(mod4)$, $p=\pi \bar{\pi }$, where $\pi =a+bi$ is a primary prime. Since $2\equiv x^4(modp)$ for some $x\in \mathbb{Z}$, then $2\equiv x^4(mod\pi )$, so ${\chi }_{\pi }(2)=1$. Moreover

$$1={\chi }_{\pi }(2)=i^{}$$

ab2.

$$Since$$

a $isodd,$8 ∣b $,therefore$p=A^2 + 64B^2 $,where$A=a, B =b/8 $.$

Conclusion :

$$\exists <!--FUNCTION\; APPLICATION-->(A,B)\in {\mathbb{Z}}^2,p=A^2+64B^2⟺(p\equiv 1[4]and\exists <!--FUNCTION\; APPLICATION-->x\in \mathbb{Z},x^4\equiv 2[p]).$$

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