Does $a\not\in( \Bbb{F}_q ^{\times})^2$ and $b\not\in( \Bbb{F}_q^{\times})^2$ imply $ab\in (\Bbb{F}_q^{\times})^2$?

Question

Let $a \in \Bbb{F}_p$. $a\not\in (\Bbb{F}_p^{\times})^2$ and $b\not\in (\Bbb{F}_p^{\times})^2$ implies $ab\in (\Bbb{F}_p^{\times})^2$ because $\left (\frac{a}{p}\right)\left (\frac{b}{p}\right)=\left (\frac{ab}{p}\right)$ holds.

Let $a \in \Bbb{F}_q$. But does $a\not\in (\Bbb{F}_q^{\times})^2$ and $b\not\in (\Bbb{F}_q^{\times})^2$ imply $ab\in (\Bbb{F}_q ^{\times})^2$, where $q=p^r$($r \ge 2$ is positive integer) ?

I have never heard generalization of Legendre symbol into general finite field, so this may have some counterexamples.

Answer 1

Yes. As the homomorphism map $x\mapsto x^2$ from $\mathbb F_q$ to itself has kernel $\{\pm 1\}$, we know the image has precisely $|\mathbb F_q^{\times}|/2$ elements (assuming $2\nmid q$, otherwise the problem is trivial). In other words, $\mathbb F_q^{\times}/(\mathbb F_q^{\times})^2\simeq \mathbb Z/2\mathbb Z$ as abelian groups. Hence if $a,b\not\in(\mathbb F_q^{\times})^2$, in other words, their images in the quotient group both correspond to $1$ in $\mathbb Z/2\mathbb Z$, then $ab$ must correspond to $0$, that is, $ab$ is in the kernel $(\mathbb F_q^{\times})^2$.

Put in another way, $\mathbb F_q\rightarrow\mathbb F_q^{\times}/(\mathbb F_q^{\times})^2\simeq (\{\pm 1\}, \text{multiplication})$ is precisely the generalization of the Legendre symbol.

Answer 2

The term quadratic residue is sometimes used for elements in a finite field $\mathbb F_q$ that have square roots in the field. Now, all the nonzero elements in a finite field can be expressed as powers of a primitive element $\alpha$.

• If the characteristic $p$ of the field is greater than $2$, then all the quadratic residues in $\mathbb F_q$ are of the form $\alpha^{2i}$, while all the quadratic nonresidues are of the form $\alpha^{2i+1}$. The product $ab$ of two quadratic nonresidues $a = \alpha^{2i+1}$ and $b = \alpha^{2k+1}$ is $\alpha^{2(i+k+1)}$ which is a quadratic residue.
• If the characteristic $p$ of the field is $2$, then all the nonzero elements of the field are quadratic residues because $q-1$ is an odd number, and thus $\alpha^{2i+1}$ has (unique) square root $\alpha^{i+\frac{q+1}{2}}$ in the field.

Answer 3

This holds for all odd prime powers $q$ because $\mathbf F_q^\times$ is cyclic of even order $q-1$.

In a cyclic group $G$ of order $n$, there is a unique subgroup $H_d$ of order $d$ for each $d$ dividing $n$, and $H_d = \ker(g \mapsto g^d)$, so the set $P_d = \{g^d : g \in G\}$ of $d$th powers in $G$ is a subgroup of order $n/d$ and index $n/(n/d) = d$.

In particular, if $G$ is cyclic of even order then the set $S = P_2$ of squares in $G$ is a subgroup of index $2$, so $G/S$ has order $2$. In a group of order $2$, the nontrivial element has order $2$, so if $a$ and $b$ are nonsquares in $G$ then $a$ and $b$ are nontrivial in $G/S$, so $a = b$ in $G/S$. Thus in $G/S$, we have $ab = a^2 = 1$, so in $G$ the product $ab$ is in $S$: the product of two nonsquares in $G$ is a square in $G$.

If $q$ is a power of $2$ then all elements of $\mathbf F_q$ are squares, so there are no nonsquares.

For odd prime powers $q$ that are not prime, you have not heard of a generalization of the Legendre symbol for $\mathbf F_q$ because that generalization just has no tidy name or notation. It is called the quadratic character on $\mathbf F_q^\times$: it is the unique character on the group $\mathbf F_q^\times$ with order $2$.