Legendre symbol: An exact sequence point of view.

The aim of this blog is deduce the Legendre symbol by the exact sequence.

$$\begin{array}{}\text{(1)}& \mathbb{Z}/(p-1)\mathbb{Z}\stackrel{2}{\to }\mathbb{Z}/(p-1)\mathbb{Z}\stackrel{\pi }{\to }\mathbb{Z}/2\mathbb{Z}\stackrel{2}{\to }0\end{array}$$

Indeed, the Legendre symbol is given by $\pi$$\pi$.

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Let $U_p$$U_p$ be the multiplicative group of ${\mathbb{F}}_p$$\mathbb F_p$, here $p\ge 3.$$p\ge 3.$

We know that the primitive root of $U_p$$U_p$ exists, hence $U_p\cong C_{p-1}\cong \mathbb{Z}/(p-1)\mathbb{Z}$$U_p\cong C_{p-1}\cong\mathbb Z/(p-1)\mathbb Z$.

Notice that $p-1$$p-1$ is an even number, hence, if ${\mathrm{ord}}_2(p-1)=n$$\mathrm{ord}_2(p-1)=n$

Now let us deal with the quadratic residues.

Definition.

An integer $a\in U_p$$a\in U_p$ is a quadratic residue if $x^2\equiv a\mathrm{mod}p$$x^2\equiv a\mod p$ has a solution; otherwise $a$$a$ is a quadratic non-residue.

Let $g$$g$ be a primitive root of $U_P$$U_P$.

Using the isomorphism

$$\begin{array}{}\text{(2)}& {\mathrm{ind}}_g:U_p\to \mathbb{Z}/(p-1)\mathbb{Z}\end{array}$$

We get

$$\begin{array}{}\text{(3)}& x^2\equiv a\mathrm{mod}p⟺2{\mathrm{ind}}_g(x)\equiv {\mathrm{ind}}_g(a)\mathrm{mod}p-1\end{array}$$

Let $y={\mathrm{ind}}_g(x)$$y=\mathrm{ind}_g(x)$, we could rewrite $(3)$$(3)$ as:

$$\begin{array}{}\text{(4)}& 2y\equiv {\mathrm{ind}}_g(a)\mathrm{mod}p-1\end{array}$$

Notice that $\mathbb{Z}/(p-1)\mathbb{Z}$$\mathbb{Z}/(p-1)\mathbb{Z}$ is a $\mathbb{Z}-$$\mathbb{Z}-$module, and $2(x):=2x$$2(x):=2x$ is an endomorphism of $\mathbb{Z}/(p-1)\mathbb{Z}$$\mathbb Z/(p-1)\mathbb Z$

Remark. View $\mathbb{Z}/m\mathbb{Z}$$\mathbb Z/ m\mathbb Z$ as a ring.

$$\begin{array}{}\text{(5)}& \mathbb{Z}/m\mathbb{Z}\cong {\mathrm{End}}_{\mathrm{Ab}}(\mathbb{Z}/m\mathbb{Z})\end{array}$$

The bijiection is given by

$$\begin{array}{}\text{(6)}& f\mapsto f(1)◻\end{array}$$

Hence $a\in U_P$$a\in U_P$ is a quadratic residue if and only if ${\mathrm{ind}}_g(a)$$\mathrm{ind}_g(a)$ in the image of $2$$2$. That is, $0,2,4,...,p-1$$0,2,4,...,p-1$.

We have the following exact sequence:

$$\begin{array}{}\text{(7)}& \mathbb{Z}/(p-1)\mathbb{Z}\stackrel{2}{\to }\mathbb{Z}/(p-1)\mathbb{Z}\stackrel{\pi }{\to }\mathbb{Z}/2\mathbb{Z}\stackrel{2}{\to }0\end{array}$$

That is,

$$\begin{array}{}\text{(8)}& U_p\stackrel{(-{)}^2}{\to }{U}_p\stackrel{{\pi }^{\prime }}{\to }\{1,-1\}\stackrel{(-{)}^2}{\to }0\end{array}$$

Hence $a$$a$ is a quadratic residue iff ${\pi }^{\prime }(a)=1$$\pi'(a)=1$.

How could we construct ${\pi }^{\prime }$$\pi'$ here?

Let ${\pi }^{\prime }=(-{)}^{\frac{p-1}{2}}$$\pi'=(-)^\frac{p-1}{2}$ sounds a good choice.

Since

$$\begin{array}{}\text{(9)}& 2\circ \frac{p-1}{2}(x)=0=2\circ \pi (x)\end{array}$$

In other word,

$$\begin{array}{}\text{(10)}& ((a{)}^{\frac{p-1}{2}}{)}^2=1={\pi }^{\prime 2}(a)\end{array}$$

and

$$\begin{array}{}\text{(11)}& u^2=1⟺u=\pm 1\end{array}$$

Since $\begin{array}{c}(-{)}^{\frac{p-1}{2}}\ne 1\end{array}$$(-)^{\frac{p-1}{2}}\ne 1\\$, here $1$$1$ is the identity map, $\begin{array}{c}(-{)}^{\frac{p-1}{2}}\end{array}$$(-)^{\frac{p-1}{2}}\\$ is a surjective group homomorphism onto $\{1,-1\}.$$\{1,-1\}.$

By the first isomorphism theorem

$$\begin{array}{}\text{(12)}& \begin{array}{c}{a}^{\frac{p-1}{2}}={\pi }^{\prime }(a)\end{array}\end{array}$$

Hence

$$\begin{array}{}\text{(13)}& a\text{ is a quadratic residue}⟺a^{\frac{p-1}{2}}\equiv 1\mathrm{mod}p\end{array}$$

Definition.

The Legendre symbol is given by the group homomorphism:

$$\begin{array}{}\text{(14)}& \left(\frac{a}{p}\right)\equiv a^{\frac{p-1}{2}}\mathrm{mod}p\end{array}$$

Proposition.

$$\begin{array}{}\text{(15)}& \sum _{a\in U_p}\left(\frac{a}{p}\right)=0\end{array}$$

Proof.

Obviously $|{\pi }^{\prime -1}(1)|=|{\pi }^{\prime -1}(-1)|.◻$$|\pi'^{-1}(1)|=|\pi'^{-1}(-1)|.\quad\square$