- Thread starter
- #1

- Thread starter matqkks
- Start date

- Thread starter
- #1

- Mar 22, 2013

- 573

(I don't know an answer to the first one, not being a teacher in any respect)

This is an interesting question. I don't recall any direct consequence(s) of it, but thinking of a more general Langland programs, for example, although the consequences are complicated.

The law of quadratic residues is one of the non-trivial results in elementary number theory. Several results of algebraic and analytic number theory lies on this theorem. In fact, there are some open conjectures regarding a smooth generalization of this result, take for example, Hilbert's 9-th problem.

This is an interesting question. I don't recall any direct consequence(s) of it, but thinking of a more general Langland programs, for example, although the consequences are complicated.

The law of quadratic residues is one of the non-trivial results in elementary number theory. Several results of algebraic and analytic number theory lies on this theorem. In fact, there are some open conjectures regarding a smooth generalization of this result, take for example, Hilbert's 9-th problem.

Last edited:

- Feb 15, 2012

- 1,967

Consider the field $\Bbb Z_p$.

Since this *is* a field, its group of units is precisely the set of non-zero elements, $\Bbb Z_p^{\ast}$.

Since this is a finite group, any subset closed under multiplication is a subgroup.

If we denote the set of (non-zero) quadratic resides by $R$, it is clear $R$ is a subgroup of the group of units since if:

$x,y \in R$, we have $x = a^2, y = b^2$ for some $a,b \in \Bbb Z_p^{\ast}$, thus:

$ab \in \Bbb Z_p^{\ast}$ (since any field is, of course, an integral domain), and:

$xy = (a^2)(b^2) = (ab)^2 \implies xy \in R$ (because multiplication in $\Bbb Z_p^{\ast}$ is commutative).

From the above, it should be clear that the map $\phi: x \to x^2$ is a surjective group homomorphism from $\Bbb Z_p^{\ast} \to R$. A natural question to ask, then, is: what is its kernel?

Clearly $1$ is the identity of $R$ ($1 \in R$ since $1 = 1\ast1 = 1^2$). Thus finding the kernel is the same as finding the roots of $x^2 - 1$ in $\Bbb Z_p^{\ast}$. Since $\Bbb Z_p$ is a field, and $x^2 - 1 \in \Bbb Z_p[x]$, we know this equation can have at most 2 roots. In fact, if $p \neq 2$, we have EXACTLY two roots: 1 and -1 (= p-1).

So for an odd prime $p$, group theory tells us that:

$[\Bbb Z_p^{\ast}:R] = |\text{ker}(\phi)| = 2$, that is:

$|R| = \frac{p-1}{2}$.

This, in effect, lets us choose "positive" elements of a finite field: just as in the case of real numbers, the "positive" elements are the ones that be written as the square of another (non-zero) element of the field. The rules for the cosets $R$ and $-R$ are the same as the rules we learned for positive and negative real numbers under "ordinary" multiplication:

$R\ast R = R$

$R \ast -R = -R \ast R = -R$

$-R \ast -R = R$.

This allows us to use "parity" arguments in some cases, saving time and sometimes considerable calculation.

- Mar 10, 2012

- 835

Great expository post!

Consider the field $\Bbb Z_p$.

Since this *is* a field, its group of units is precisely the set of non-zero elements, $\Bbb Z_p^{\ast}$.

Since this is a finite group, any subset closed under multiplication is a subgroup.

If we denote the set of (non-zero) quadratic resides by $R$, it is clear $R$ is a subgroup of the group of units since if:

$x,y \in R$, we have $x = a^2, y = b^2$ for some $a,b \in \Bbb Z_p^{\ast}$, thus:

$ab \in \Bbb Z_p^{\ast}$ (since any field is, of course, an integral domain), and:

$xy = (a^2)(b^2) = (ab)^2 \implies xy \in R$ (because multiplication in $\Bbb Z_p^{\ast}$ is commutative).

From the above, it should be clear that the map $\phi: x \to x^2$ is a surjective group homomorphism from $\Bbb Z_p^{\ast} \to R$. A natural question to ask, then, is: what is its kernel?

Clearly $1$ is the identity of $R$ ($1 \in R$ since $1 = 1\ast1 = 1^2$). Thus finding the kernel is the same as finding the roots of $x^2 - 1$ in $\Bbb Z_p^{\ast}$. Since $\Bbb Z_p$ is a field, and $x^2 - 1 \in \Bbb Z_p[x]$, we know this equation can have at most 2 roots. In fact, if $p \neq 2$, we have EXACTLY two roots: 1 and -1 (= p-1).

So for an odd prime $p$, group theory tells us that:

$[\Bbb Z_p^{\ast}:R] = |\text{ker}(\phi)| = 2$, that is:

$|R| = \frac{p-1}{2}$.

This, in effect, lets us choose "positive" elements of a finite field: just as in the case of real numbers, the "positive" elements are the ones that be written as the square of another (non-zero) element of the field. The rules for the cosets $R$ and $-R$ are the same as the rules we learned for positive and negative real numbers under "ordinary" multiplication:

$R\ast R = R$

$R \ast -R = -R \ast R = -R$

$-R \ast -R = R$.

This allows us to use "parity" arguments in some cases, saving time and sometimes considerable calculation.