AKS - proving that $frac{n}{p}$ is introspective
I have a problem with showing that $frac{n}{p}$ is introspective.
Recall that we are in state where a composite integer $n$ fools the AKS test and $pmid n$ a prime number.
First of all, recall the following definitions and facts
$pmid n$ is a prime number and $p>r>1$.- $Rstackrel{text{def}}{=}mathbb{F}_p[X]pmod{X^r-1}$
- $mathcal{P}_nstackrel{text{def}}{=}{fin R | f(X^n)=_Rf(X)^n}$
- $forall 1leq aleq r:X+ain mathcal{P}_n$
- $mathcal{G}stackrel{text{def}}{=}{iinmathbb{N} | (i,r)=1wedge forall finmathcal{P}_n:f(X^i)=_Rf(X)^i}$
- $p,nin mathcal{G}$
We wish to show that $frac{n}{p}in mathcal{G}$. Let $finmathcal{P}_n$ and observe the following
begin{align*}
fbig(X^{frac{n}{p}}big)^p & stackrel{(1)}{equiv} fBig(big(X^{frac{n}{p}}big)^pBig) & pmod{big(X^{frac{n}{p}}big)^r-1,p}\
& equiv f(X^n) & pmod{X^{frac{n}{p}cdot r}-1,p}\
fbig(X^{frac{n}{p}}big)^p & stackrel{(2)}{equiv} f(X^n) & pmod{X^r-1,p}\
& stackrel{(3)}{equiv} f(X)^n & pmod{X^r-1,p}\
& equiv big(f(X)^{frac{n}{p}}big)^p & pmod{X^r-1,p}
end{align*}
Explanations
$pinmathcal{G}$ and using the identity w.r.t. the vairable $Y=X^{frac{n}{p}}$
- $X^r-1mid X^{frac{n}{p}cdot r}-1$
- $ninmathcal{G}$
Concluding we get that
$$fbig(X^{frac{n}{p}}big)^p-big(f(X)^{frac{n}{p}}big)^pequiv 0pmod{X^r-1,p}$$
But, $p>r>1$ all natural numbers, so $p>2$ and in particular it is an odd prime. Therefore
$$Big(fbig(X^{frac{n}{p}}big)-f(X)^{frac{n}{p}}Big)^p=fbig(X^{frac{n}{p}}big)^p+big(-f(X)^{frac{n}{p}}big)^pequiv fbig(X^{frac{n}{p}}big)^p-big(f(X)^{frac{n}{p}}big)^pequiv 0pmod{X^r-1,p}$$
However, $R$ is not an integral domain so we cannot deduce that
$$fbig(X^{frac{n}{p}}big)-f(X)^{frac{n}{p}}equiv 0 pmod{X^r-1,p}$$
Please help, they say in the paper that it follows immediately and I do not understands why.
Edit: I have also noted that
$$f(X^p)^{frac{n}{p}}equiv (f(X)^p)^{frac{n}{p}}equiv f(X^n)equiv f(X)^npmod{X^r-1,p}$$
Note: The version of the paper that I work with
https://www.cse.iitk.ac.in/users/manindra/algebra/primality_v6.pdf
number-theory proof-verification primality-test
add a comment |
I have a problem with showing that $frac{n}{p}$ is introspective.
Recall that we are in state where a composite integer $n$ fools the AKS test and $pmid n$ a prime number.
First of all, recall the following definitions and facts
$pmid n$ is a prime number and $p>r>1$.- $Rstackrel{text{def}}{=}mathbb{F}_p[X]pmod{X^r-1}$
- $mathcal{P}_nstackrel{text{def}}{=}{fin R | f(X^n)=_Rf(X)^n}$
- $forall 1leq aleq r:X+ain mathcal{P}_n$
- $mathcal{G}stackrel{text{def}}{=}{iinmathbb{N} | (i,r)=1wedge forall finmathcal{P}_n:f(X^i)=_Rf(X)^i}$
- $p,nin mathcal{G}$
We wish to show that $frac{n}{p}in mathcal{G}$. Let $finmathcal{P}_n$ and observe the following
begin{align*}
fbig(X^{frac{n}{p}}big)^p & stackrel{(1)}{equiv} fBig(big(X^{frac{n}{p}}big)^pBig) & pmod{big(X^{frac{n}{p}}big)^r-1,p}\
& equiv f(X^n) & pmod{X^{frac{n}{p}cdot r}-1,p}\
fbig(X^{frac{n}{p}}big)^p & stackrel{(2)}{equiv} f(X^n) & pmod{X^r-1,p}\
& stackrel{(3)}{equiv} f(X)^n & pmod{X^r-1,p}\
& equiv big(f(X)^{frac{n}{p}}big)^p & pmod{X^r-1,p}
end{align*}
Explanations
$pinmathcal{G}$ and using the identity w.r.t. the vairable $Y=X^{frac{n}{p}}$
- $X^r-1mid X^{frac{n}{p}cdot r}-1$
- $ninmathcal{G}$
Concluding we get that
$$fbig(X^{frac{n}{p}}big)^p-big(f(X)^{frac{n}{p}}big)^pequiv 0pmod{X^r-1,p}$$
But, $p>r>1$ all natural numbers, so $p>2$ and in particular it is an odd prime. Therefore
$$Big(fbig(X^{frac{n}{p}}big)-f(X)^{frac{n}{p}}Big)^p=fbig(X^{frac{n}{p}}big)^p+big(-f(X)^{frac{n}{p}}big)^pequiv fbig(X^{frac{n}{p}}big)^p-big(f(X)^{frac{n}{p}}big)^pequiv 0pmod{X^r-1,p}$$
However, $R$ is not an integral domain so we cannot deduce that
$$fbig(X^{frac{n}{p}}big)-f(X)^{frac{n}{p}}equiv 0 pmod{X^r-1,p}$$
Please help, they say in the paper that it follows immediately and I do not understands why.
Edit: I have also noted that
$$f(X^p)^{frac{n}{p}}equiv (f(X)^p)^{frac{n}{p}}equiv f(X^n)equiv f(X)^npmod{X^r-1,p}$$
Note: The version of the paper that I work with
https://www.cse.iitk.ac.in/users/manindra/algebra/primality_v6.pdf
number-theory proof-verification primality-test
add a comment |
I have a problem with showing that $frac{n}{p}$ is introspective.
Recall that we are in state where a composite integer $n$ fools the AKS test and $pmid n$ a prime number.
First of all, recall the following definitions and facts
$pmid n$ is a prime number and $p>r>1$.- $Rstackrel{text{def}}{=}mathbb{F}_p[X]pmod{X^r-1}$
- $mathcal{P}_nstackrel{text{def}}{=}{fin R | f(X^n)=_Rf(X)^n}$
- $forall 1leq aleq r:X+ain mathcal{P}_n$
- $mathcal{G}stackrel{text{def}}{=}{iinmathbb{N} | (i,r)=1wedge forall finmathcal{P}_n:f(X^i)=_Rf(X)^i}$
- $p,nin mathcal{G}$
We wish to show that $frac{n}{p}in mathcal{G}$. Let $finmathcal{P}_n$ and observe the following
begin{align*}
fbig(X^{frac{n}{p}}big)^p & stackrel{(1)}{equiv} fBig(big(X^{frac{n}{p}}big)^pBig) & pmod{big(X^{frac{n}{p}}big)^r-1,p}\
& equiv f(X^n) & pmod{X^{frac{n}{p}cdot r}-1,p}\
fbig(X^{frac{n}{p}}big)^p & stackrel{(2)}{equiv} f(X^n) & pmod{X^r-1,p}\
& stackrel{(3)}{equiv} f(X)^n & pmod{X^r-1,p}\
& equiv big(f(X)^{frac{n}{p}}big)^p & pmod{X^r-1,p}
end{align*}
Explanations
$pinmathcal{G}$ and using the identity w.r.t. the vairable $Y=X^{frac{n}{p}}$
- $X^r-1mid X^{frac{n}{p}cdot r}-1$
- $ninmathcal{G}$
Concluding we get that
$$fbig(X^{frac{n}{p}}big)^p-big(f(X)^{frac{n}{p}}big)^pequiv 0pmod{X^r-1,p}$$
But, $p>r>1$ all natural numbers, so $p>2$ and in particular it is an odd prime. Therefore
$$Big(fbig(X^{frac{n}{p}}big)-f(X)^{frac{n}{p}}Big)^p=fbig(X^{frac{n}{p}}big)^p+big(-f(X)^{frac{n}{p}}big)^pequiv fbig(X^{frac{n}{p}}big)^p-big(f(X)^{frac{n}{p}}big)^pequiv 0pmod{X^r-1,p}$$
However, $R$ is not an integral domain so we cannot deduce that
$$fbig(X^{frac{n}{p}}big)-f(X)^{frac{n}{p}}equiv 0 pmod{X^r-1,p}$$
Please help, they say in the paper that it follows immediately and I do not understands why.
Edit: I have also noted that
$$f(X^p)^{frac{n}{p}}equiv (f(X)^p)^{frac{n}{p}}equiv f(X^n)equiv f(X)^npmod{X^r-1,p}$$
Note: The version of the paper that I work with
https://www.cse.iitk.ac.in/users/manindra/algebra/primality_v6.pdf
number-theory proof-verification primality-test
I have a problem with showing that $frac{n}{p}$ is introspective.
Recall that we are in state where a composite integer $n$ fools the AKS test and $pmid n$ a prime number.
First of all, recall the following definitions and facts
$pmid n$ is a prime number and $p>r>1$.- $Rstackrel{text{def}}{=}mathbb{F}_p[X]pmod{X^r-1}$
- $mathcal{P}_nstackrel{text{def}}{=}{fin R | f(X^n)=_Rf(X)^n}$
- $forall 1leq aleq r:X+ain mathcal{P}_n$
- $mathcal{G}stackrel{text{def}}{=}{iinmathbb{N} | (i,r)=1wedge forall finmathcal{P}_n:f(X^i)=_Rf(X)^i}$
- $p,nin mathcal{G}$
We wish to show that $frac{n}{p}in mathcal{G}$. Let $finmathcal{P}_n$ and observe the following
begin{align*}
fbig(X^{frac{n}{p}}big)^p & stackrel{(1)}{equiv} fBig(big(X^{frac{n}{p}}big)^pBig) & pmod{big(X^{frac{n}{p}}big)^r-1,p}\
& equiv f(X^n) & pmod{X^{frac{n}{p}cdot r}-1,p}\
fbig(X^{frac{n}{p}}big)^p & stackrel{(2)}{equiv} f(X^n) & pmod{X^r-1,p}\
& stackrel{(3)}{equiv} f(X)^n & pmod{X^r-1,p}\
& equiv big(f(X)^{frac{n}{p}}big)^p & pmod{X^r-1,p}
end{align*}
Explanations
$pinmathcal{G}$ and using the identity w.r.t. the vairable $Y=X^{frac{n}{p}}$
- $X^r-1mid X^{frac{n}{p}cdot r}-1$
- $ninmathcal{G}$
Concluding we get that
$$fbig(X^{frac{n}{p}}big)^p-big(f(X)^{frac{n}{p}}big)^pequiv 0pmod{X^r-1,p}$$
But, $p>r>1$ all natural numbers, so $p>2$ and in particular it is an odd prime. Therefore
$$Big(fbig(X^{frac{n}{p}}big)-f(X)^{frac{n}{p}}Big)^p=fbig(X^{frac{n}{p}}big)^p+big(-f(X)^{frac{n}{p}}big)^pequiv fbig(X^{frac{n}{p}}big)^p-big(f(X)^{frac{n}{p}}big)^pequiv 0pmod{X^r-1,p}$$
However, $R$ is not an integral domain so we cannot deduce that
$$fbig(X^{frac{n}{p}}big)-f(X)^{frac{n}{p}}equiv 0 pmod{X^r-1,p}$$
Please help, they say in the paper that it follows immediately and I do not understands why.
Edit: I have also noted that
$$f(X^p)^{frac{n}{p}}equiv (f(X)^p)^{frac{n}{p}}equiv f(X^n)equiv f(X)^npmod{X^r-1,p}$$
Note: The version of the paper that I work with
https://www.cse.iitk.ac.in/users/manindra/algebra/primality_v6.pdf
number-theory proof-verification primality-test
number-theory proof-verification primality-test
edited Dec 1 '18 at 16:30
Don Fanucci
asked Nov 29 '18 at 21:20
Don FanucciDon Fanucci
1,093419
1,093419
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Finally, the solution is simple, I think.
We factor $X^r-1$ into irreducible elements (using UFD) and take out the $p$ exponent, trivially (the resulting quotient rings are ID).
Finally, we use the CRT for rings in order to lose the exponent in our original ring $R$.
add a comment |
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Finally, the solution is simple, I think.
We factor $X^r-1$ into irreducible elements (using UFD) and take out the $p$ exponent, trivially (the resulting quotient rings are ID).
Finally, we use the CRT for rings in order to lose the exponent in our original ring $R$.
add a comment |
Finally, the solution is simple, I think.
We factor $X^r-1$ into irreducible elements (using UFD) and take out the $p$ exponent, trivially (the resulting quotient rings are ID).
Finally, we use the CRT for rings in order to lose the exponent in our original ring $R$.
add a comment |
Finally, the solution is simple, I think.
We factor $X^r-1$ into irreducible elements (using UFD) and take out the $p$ exponent, trivially (the resulting quotient rings are ID).
Finally, we use the CRT for rings in order to lose the exponent in our original ring $R$.
Finally, the solution is simple, I think.
We factor $X^r-1$ into irreducible elements (using UFD) and take out the $p$ exponent, trivially (the resulting quotient rings are ID).
Finally, we use the CRT for rings in order to lose the exponent in our original ring $R$.
answered Dec 4 '18 at 18:41
Don FanucciDon Fanucci
1,093419
1,093419
add a comment |
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