AKS - proving that $frac{n}{p}$ is introspective












2














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





  1. $pinmathcal{G}$ and using the identity w.r.t. the vairable $Y=X^{frac{n}{p}}$

  2. $X^r-1mid X^{frac{n}{p}cdot r}-1$

  3. $ninmathcal{G}$


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







share|cite|improve this question





























    2














    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





    1. $pinmathcal{G}$ and using the identity w.r.t. the vairable $Y=X^{frac{n}{p}}$

    2. $X^r-1mid X^{frac{n}{p}cdot r}-1$

    3. $ninmathcal{G}$


    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







    share|cite|improve this question



























      2












      2








      2







      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





      1. $pinmathcal{G}$ and using the identity w.r.t. the vairable $Y=X^{frac{n}{p}}$

      2. $X^r-1mid X^{frac{n}{p}cdot r}-1$

      3. $ninmathcal{G}$


      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







      share|cite|improve this question















      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





      1. $pinmathcal{G}$ and using the identity w.r.t. the vairable $Y=X^{frac{n}{p}}$

      2. $X^r-1mid X^{frac{n}{p}cdot r}-1$

      3. $ninmathcal{G}$


      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






      share|cite|improve this question















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      edited Dec 1 at 16:30

























      asked Nov 29 at 21:20









      Don Fanucci

      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 $R$.






          share|cite|improve this answer





















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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 $R$.






            share|cite|improve this answer


























              0














              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 $R$.






              share|cite|improve this answer
























                0












                0








                0






                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 $R$.






                share|cite|improve this answer












                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 $R$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 4 at 18:41









                Don Fanucci

                1,093419




                1,093419






























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