Exercise 5.36

Question

\def\imp{\Rightarrow}
\def\gcro{\mbox{[\hspace{-.15em}[}}
\def\dcro{\mbox{]\hspace{-.15em}]}}

ewcommand{\D}{\mathrm{d}}

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{e}{\,\mathrm{Re}\,}

ewcommand{\ord}{\mathrm{ord}}

ewcommand{
}{\mathrm{N}}

ewcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}}

Show that part (c) of Proposition 5.2.2 is true if $a$ is negative and $b$ is positive (both still odd).

Richard Ganaye · Accepted Answer

\def\imp{\Rightarrow} \def\gcro{\mbox{[\hspace{-.15em}[}} \def\dcro{\mbox{]\hspace{-.15em}]}} ewcommand{\D}{\mathrm{d}} 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{ e}{\,\mathrm{Re}\,} ewcommand{\ord}{\mathrm{ord}} ewcommand{ }{\mathrm{N}} ewcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}} \bigskip As said by Adam Michalik, the Jacobi symbol $\legendre{a}{b}$ only defined for positive $b$, so the question, which concerns $\legendre{a}{b} \legendre{b}{a}, a<0$ makes no sense. To give sense to this question, we must substitute the Kronecker symbol to the Jacobi symbol. The Kronecker symbol (not defined in Ireland-Rosen) is the usual extension of Jacobi symbol (see for instance [Henri Cohen] A course in computational algebraic number theory, [Henri Cohen] Number theory (vol. 1), or [Harvey Cohn] Advanced number theory). We define Kronecker (or Kronecker-Jacobi) symbol $\legendre{a}{b}$ for any $a$ and $b$ in $\Z$ in the following way. \begin{enumerate} \item[(1)] If $b=0$, then $\legendre{a}{0} = 1$ if $a = \pm 1$, and$\legendre{a}{0} = 0$ otherwise. \item[(2)] For $b e 0$, write $b = \prod p$, where the $p$ are not necessarily distinct primes (including $2$), or $p = -1$ to take care of the sign. Then we set $$\legendre{a}{b} = \prod \legendre{a}{p},$$ where $\legendre{a}{p}$ is the Legendre symbol defined above for $p>2$, and where we define $$\legendre{a}{2} = \begin{cases} 0 & ext{if $a$ is even}\ (-1)^{(a^2-1)/8} & ext{if $a$ is odd,} \end{cases} $$ and also $$\legendre{a}{-1} = \begin{cases} 1 & ext{if $a\geq 0$}\ -1 & ext{if $a<0$} \end{cases} $$ \end{enumerate} \begin{proof} Suppose that $a <0, b>0$, both odd. Let $a = -A,A>0, A = p_1p_2\cdots p_k$, where the $p_i$ are not necessarily distinct primes. Then \begin{align*} \legendre{a}{b}\legendre{b}{a} &= \legendre{-A}{b}\legendre{b}{-A} \ \legendre{-A}{b} &= \legendre{-1}{b} \legendre{A}{b} =(-1)^{(b-1)/2} \legendre{A}{b}\ \legendre{b}{-A} &= \legendre{b}{-1} \legendre{b}{p_1}\cdots \legendre{b}{p_k} = \legendre{b}{A}. \end{align*} Therefore, by Prop. 5.2.2, as $A,b$ are odd and positive, \begin{align*} \legendre{a}{b}\legendre{b}{a} &= (-1)^{ \frac{b-1}{2} } \legendre{A}{b} \legendre{b}{A}\ &=(-1)^{ \frac{b-1}{2} } (-1)^{\frac{A-1}{2}\frac{b-1}{2}}\ &=(-1) ^{\frac{b-1}{2} [1 + \frac{-a-1}{2}]}\ &=(-1)^{\frac{b-1}{2}\frac{1-a}{2} }\ &=(-1)^{\frac{b-1}{2}\frac{a-1}{2} }\ \end{align*} So the law of quadratic reciprocity remains valid for the Kronecker symbol when $a$ is negative ($b>0$, $a,b$ both odd). \end{proof}

Exercise 5.36

Answers

Comments

Exercise 5.36

Answers

Comments

Add answer