Square roots in $mathbb{Z}^{*}_{p^2}$
$begingroup$
For prime $p$, I understand that each $x in mathbb{Z}^{*}_{p}$ has exactly 2 square roots (as there are at least 2, and the quadratic residues are exactly half of the group (the group is cyclic so we take even exponents of a generator g) so there can't be more than 2 by counting (otherwise there is an element that is a root of 2 different elements). I also understand what happens at $x in mathbb{Z}^{*}_{pq}$ where $p$ and $q$ are primes as we can use chinese remainder theorem and compute roots separately and get 4 roots).
I however don't understand what happens at $x in mathbb{Z}^{*}_{p^2}$ as I am unsure if its cyclic so I think my previous argument doesn't work.
thanks.
group-theory number-theory modular-arithmetic
asked Jan 7 at 11:25
ElooEloo
1368
$endgroup$
add a comment |
$begingroup$
For prime $p$, I understand that each $x in mathbb{Z}^{*}_{p}$ has exactly 2 square roots (as there are at least 2, and the quadratic residues are exactly half of the group (the group is cyclic so we take even exponents of a generator g) so there can't be more than 2 by counting (otherwise there is an element that is a root of 2 different elements). I also understand what happens at $x in mathbb{Z}^{*}_{pq}$ where $p$ and $q$ are primes as we can use chinese remainder theorem and compute roots separately and get 4 roots).
I however don't understand what happens at $x in mathbb{Z}^{*}_{p^2}$ as I am unsure if its cyclic so I think my previous argument doesn't work.
thanks.
group-theory number-theory modular-arithmetic
asked Jan 7 at 11:25
ElooEloo
1368
$endgroup$
$begingroup$
Do you really mean $mathbb Z_{p^2}$ (the cyclic group of order $p^2$) or, maybe, $mathbb F_{p^2}$ (the finite field with $p^2$ elements)?
$endgroup$
– W-t-P
Jan 7 at 11:34
2
$begingroup$
Taking your question as it stands, $mathbb Z_{p^2}^ast$ is cyclic, since there are primitive roots modulo $p^2$ (and in fact, modulo $p^n$ with any $nge 1$). If $ginmathbb Z_{p^2}^ast$ is some fixed primitive root, then every element of $mathbb Z_{p^2}^ast$ is uniquely written as $g^k$ with some $0le k<p(p-1)$, and the element is a square if and only if $k$ is even.
$endgroup$
– W-t-P
Jan 7 at 11:39
$begingroup$
(My first intention was to ask what happens for $p=2$ and $p=3$, getting thus also an explicit chance to figure out the notations...)
$endgroup$
– dan_fulea
Jan 7 at 11:41
1
$begingroup$
Your first statement is wrong: every $xinBbb Z_p^*$ has zero or two square roots. Unless $p=2$, in which case every such $x$ (namely, $x=1$) has one square root.
$endgroup$
– TonyK
Jan 7 at 11:54
add a comment |
$begingroup$
For prime $p$, I understand that each $x in mathbb{Z}^{*}_{p}$ has exactly 2 square roots (as there are at least 2, and the quadratic residues are exactly half of the group (the group is cyclic so we take even exponents of a generator g) so there can't be more than 2 by counting (otherwise there is an element that is a root of 2 different elements). I also understand what happens at $x in mathbb{Z}^{*}_{pq}$ where $p$ and $q$ are primes as we can use chinese remainder theorem and compute roots separately and get 4 roots).
I however don't understand what happens at $x in mathbb{Z}^{*}_{p^2}$ as I am unsure if its cyclic so I think my previous argument doesn't work.
thanks.
group-theory number-theory modular-arithmetic
asked Jan 7 at 11:25
ElooEloo
1368
$endgroup$
For prime $p$, I understand that each $x in mathbb{Z}^{*}_{p}$ has exactly 2 square roots (as there are at least 2, and the quadratic residues are exactly half of the group (the group is cyclic so we take even exponents of a generator g) so there can't be more than 2 by counting (otherwise there is an element that is a root of 2 different elements). I also understand what happens at $x in mathbb{Z}^{*}_{pq}$ where $p$ and $q$ are primes as we can use chinese remainder theorem and compute roots separately and get 4 roots).
I however don't understand what happens at $x in mathbb{Z}^{*}_{p^2}$ as I am unsure if its cyclic so I think my previous argument doesn't work.
thanks.
group-theory number-theory modular-arithmetic
group-theory number-theory modular-arithmetic
asked Jan 7 at 11:25
ElooEloo
1368
asked Jan 7 at 11:25
ElooEloo
1368
asked Jan 7 at 11:25
ElooEloo
1368
asked Jan 7 at 11:25
ElooEloo
1368
asked Jan 7 at 11:25
ElooEloo
1368
1368
$begingroup$
Do you really mean $mathbb Z_{p^2}$ (the cyclic group of order $p^2$) or, maybe, $mathbb F_{p^2}$ (the finite field with $p^2$ elements)?
$endgroup$
– W-t-P
Jan 7 at 11:34
2
$begingroup$
Taking your question as it stands, $mathbb Z_{p^2}^ast$ is cyclic, since there are primitive roots modulo $p^2$ (and in fact, modulo $p^n$ with any $nge 1$). If $ginmathbb Z_{p^2}^ast$ is some fixed primitive root, then every element of $mathbb Z_{p^2}^ast$ is uniquely written as $g^k$ with some $0le k<p(p-1)$, and the element is a square if and only if $k$ is even.
$endgroup$
– W-t-P
Jan 7 at 11:39
$begingroup$
(My first intention was to ask what happens for $p=2$ and $p=3$, getting thus also an explicit chance to figure out the notations...)
$endgroup$
– dan_fulea
Jan 7 at 11:41
1
$begingroup$
Your first statement is wrong: every $xinBbb Z_p^*$ has zero or two square roots. Unless $p=2$, in which case every such $x$ (namely, $x=1$) has one square root.
$endgroup$
– TonyK
Jan 7 at 11:54
add a comment |
$begingroup$
Do you really mean $mathbb Z_{p^2}$ (the cyclic group of order $p^2$) or, maybe, $mathbb F_{p^2}$ (the finite field with $p^2$ elements)?
$endgroup$
– W-t-P
Jan 7 at 11:34
2
$begingroup$
Taking your question as it stands, $mathbb Z_{p^2}^ast$ is cyclic, since there are primitive roots modulo $p^2$ (and in fact, modulo $p^n$ with any $nge 1$). If $ginmathbb Z_{p^2}^ast$ is some fixed primitive root, then every element of $mathbb Z_{p^2}^ast$ is uniquely written as $g^k$ with some $0le k<p(p-1)$, and the element is a square if and only if $k$ is even.
$endgroup$
– W-t-P
Jan 7 at 11:39
$begingroup$
(My first intention was to ask what happens for $p=2$ and $p=3$, getting thus also an explicit chance to figure out the notations...)
$endgroup$
– dan_fulea
Jan 7 at 11:41
1
$begingroup$
Your first statement is wrong: every $xinBbb Z_p^*$ has zero or two square roots. Unless $p=2$, in which case every such $x$ (namely, $x=1$) has one square root.
$endgroup$
– TonyK
Jan 7 at 11:54
$begingroup$
Do you really mean $mathbb Z_{p^2}$ (the cyclic group of order $p^2$) or, maybe, $mathbb F_{p^2}$ (the finite field with $p^2$ elements)?
$endgroup$
– W-t-P
Jan 7 at 11:34
$begingroup$
Do you really mean $mathbb Z_{p^2}$ (the cyclic group of order $p^2$) or, maybe, $mathbb F_{p^2}$ (the finite field with $p^2$ elements)?
$endgroup$
– W-t-P
Jan 7 at 11:34
2
2
$begingroup$
Taking your question as it stands, $mathbb Z_{p^2}^ast$ is cyclic, since there are primitive roots modulo $p^2$ (and in fact, modulo $p^n$ with any $nge 1$). If $ginmathbb Z_{p^2}^ast$ is some fixed primitive root, then every element of $mathbb Z_{p^2}^ast$ is uniquely written as $g^k$ with some $0le k<p(p-1)$, and the element is a square if and only if $k$ is even.
$endgroup$
– W-t-P
Jan 7 at 11:39
$begingroup$
Taking your question as it stands, $mathbb Z_{p^2}^ast$ is cyclic, since there are primitive roots modulo $p^2$ (and in fact, modulo $p^n$ with any $nge 1$). If $ginmathbb Z_{p^2}^ast$ is some fixed primitive root, then every element of $mathbb Z_{p^2}^ast$ is uniquely written as $g^k$ with some $0le k<p(p-1)$, and the element is a square if and only if $k$ is even.
$endgroup$
– W-t-P
Jan 7 at 11:39
$begingroup$
(My first intention was to ask what happens for $p=2$ and $p=3$, getting thus also an explicit chance to figure out the notations...)
$endgroup$
– dan_fulea
Jan 7 at 11:41
$begingroup$
(My first intention was to ask what happens for $p=2$ and $p=3$, getting thus also an explicit chance to figure out the notations...)
$endgroup$
– dan_fulea
Jan 7 at 11:41
1
1
$begingroup$
Your first statement is wrong: every $xinBbb Z_p^*$ has zero or two square roots. Unless $p=2$, in which case every such $x$ (namely, $x=1$) has one square root.
$endgroup$
– TonyK
Jan 7 at 11:54
$begingroup$
Your first statement is wrong: every $xinBbb Z_p^*$ has zero or two square roots. Unless $p=2$, in which case every such $x$ (namely, $x=1$) has one square root.
$endgroup$
– TonyK
Jan 7 at 11:54
add a comment |
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$begingroup$
Do you really mean $mathbb Z_{p^2}$ (the cyclic group of order $p^2$) or, maybe, $mathbb F_{p^2}$ (the finite field with $p^2$ elements)?
$endgroup$
– W-t-P
Jan 7 at 11:34
2
$begingroup$
Taking your question as it stands, $mathbb Z_{p^2}^ast$ is cyclic, since there are primitive roots modulo $p^2$ (and in fact, modulo $p^n$ with any $nge 1$). If $ginmathbb Z_{p^2}^ast$ is some fixed primitive root, then every element of $mathbb Z_{p^2}^ast$ is uniquely written as $g^k$ with some $0le k<p(p-1)$, and the element is a square if and only if $k$ is even.
$endgroup$
– W-t-P
Jan 7 at 11:39
$begingroup$
(My first intention was to ask what happens for $p=2$ and $p=3$, getting thus also an explicit chance to figure out the notations...)
$endgroup$
– dan_fulea
Jan 7 at 11:41
1
$begingroup$
Your first statement is wrong: every $xinBbb Z_p^*$ has zero or two square roots. Unless $p=2$, in which case every such $x$ (namely, $x=1$) has one square root.
$endgroup$
– TonyK
Jan 7 at 11:54