Is $x^4 + 2x^2 - x + 1$ irreducible in $mathbb Z_7[x]$?












2












$begingroup$


How would one do this?
I know since it doesn't have roots it can only be divisible by irreducible polynomials of degree 2.
How would I prove that it is or it isn't?
There are a lot of polynomials with this conditions.






polynomials finite-fields irreducible-polynomials







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    Just try to solve for the coefficients. if it factored, we'd have $p(x)=(x^2+ax+b)times (x^2+cx+d)$
    $endgroup$
    – lulu
    Jan 2 at 22:12
















2












$begingroup$


How would one do this?
I know since it doesn't have roots it can only be divisible by irreducible polynomials of degree 2.
How would I prove that it is or it isn't?
There are a lot of polynomials with this conditions.






polynomials finite-fields irreducible-polynomials







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    Just try to solve for the coefficients. if it factored, we'd have $p(x)=(x^2+ax+b)times (x^2+cx+d)$
    $endgroup$
    – lulu
    Jan 2 at 22:12














2












2








2





$begingroup$


How would one do this?
I know since it doesn't have roots it can only be divisible by irreducible polynomials of degree 2.
How would I prove that it is or it isn't?
There are a lot of polynomials with this conditions.






polynomials finite-fields irreducible-polynomials







share|cite|improve this question











$endgroup$




How would one do this?
I know since it doesn't have roots it can only be divisible by irreducible polynomials of degree 2.
How would I prove that it is or it isn't?
There are a lot of polynomials with this conditions.






polynomials finite-fields irreducible-polynomials




polynomials finite-fields irreducible-polynomials






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 2 at 22:18









A. Pongrácz

6,0531929




6,0531929










asked Jan 2 at 22:10









J. DionisioJ. Dionisio

11011




11011








  • 3




    $begingroup$
    Just try to solve for the coefficients. if it factored, we'd have $p(x)=(x^2+ax+b)times (x^2+cx+d)$
    $endgroup$
    – lulu
    Jan 2 at 22:12














  • 3




    $begingroup$
    Just try to solve for the coefficients. if it factored, we'd have $p(x)=(x^2+ax+b)times (x^2+cx+d)$
    $endgroup$
    – lulu
    Jan 2 at 22:12








3




3




$begingroup$
Just try to solve for the coefficients. if it factored, we'd have $p(x)=(x^2+ax+b)times (x^2+cx+d)$
$endgroup$
– lulu
Jan 2 at 22:12




$begingroup$
Just try to solve for the coefficients. if it factored, we'd have $p(x)=(x^2+ax+b)times (x^2+cx+d)$
$endgroup$
– lulu
Jan 2 at 22:12










2 Answers
2






active

oldest

votes


















5












$begingroup$

The product of all the monic, irreducible polynomials over $mathbb{F}_7$ wih degree $leq 2$ is given by $x^{49}-x$, so if we prove that $f(x)=(x^2+1)^2-x$ and $x^{48}-1$ are coprime over $mathbb{F}_7[x]$ we are done. Let us compute $x^{48}-1pmod{f(x)}$, for instance through the Brauer chain $x^4to x^8to x^{16}to x^{32}to x^{48}$.
$$ x^4 equiv -2x^2+x-1pmod{f(x)} $$
$$ x^8 equiv 3x^3-3x^2+2x-3pmod{f(x)} $$
$$ x^{16} equiv x^3+3x^2+2x-1pmod{f(x)} $$
$$ x^{32} equiv -x^3+2x^2+x-3pmod{f(x)} $$
$$ x^{48}-1 equiv -x^3-2x^2+x+2 pmod{f(x)}$$
give
$$gcd(x^{48}-1,x^4+2x^2-x+1)=gcd(x^3+2x^2-x-2,x^4+2x^2-x+1) $$
$$ = gcd(x^3+2x^2-x-2,-x-3) $$
but $x=4$ is not a root of $x^3+2x^2-x-2$ since $7nmid 90$ and we have finished.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    See this answer for a generalization - an efficient algorithm often used in practice (a polynomial analog of the Pocklington-Lehmer integer primality test)
    $endgroup$
    – Bill Dubuque
    Jan 2 at 23:13












  • $begingroup$
    That's it. I spent awhile looking for a trick answer, but didn't see any. An alternative way of organizing the calculation would be to first build a list of remainders of $x^0,x^7equiv-1+3x+3x^2+3x^3,x^{14}$ and $x^{21}$ modulo $f(x)$. Raising to seventh power modulo $f(x)$ is linear, so $$x^{49}equiv-1+3x^7+3x^{14}+3x^{21}pmod{f(x)}$$ et cetera. This may help with a higher degree polynomial. I'm afraid I don't have a good feel for the difference in complexity for I usually do this in characteristic two only :-)
    $endgroup$
    – Jyrki Lahtonen
    Jan 2 at 23:19





















3












$begingroup$

You can also try to factor it as a product of two polynomials of degree $2$. Such a factorization must have the form



begin{align}x^4 + 2x^2 - x + 1 & = (x^2+ax+b)(x^2-ax+b^{-1})\
& =x^4+(b+b^{-1}-a^2)x^2+(ab^{-1}-ab)x+1end{align}
Hence the relations
$$b+b^{-1}-a^2=2, ab^{-1}-ab=-1,$$
which can be easily seen to have no solution in $mathbb{Z}_7$.






share|cite|improve this answer









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    2 Answers
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    2 Answers
    2






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    active

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    5












    $begingroup$

    The product of all the monic, irreducible polynomials over $mathbb{F}_7$ wih degree $leq 2$ is given by $x^{49}-x$, so if we prove that $f(x)=(x^2+1)^2-x$ and $x^{48}-1$ are coprime over $mathbb{F}_7[x]$ we are done. Let us compute $x^{48}-1pmod{f(x)}$, for instance through the Brauer chain $x^4to x^8to x^{16}to x^{32}to x^{48}$.
    $$ x^4 equiv -2x^2+x-1pmod{f(x)} $$
    $$ x^8 equiv 3x^3-3x^2+2x-3pmod{f(x)} $$
    $$ x^{16} equiv x^3+3x^2+2x-1pmod{f(x)} $$
    $$ x^{32} equiv -x^3+2x^2+x-3pmod{f(x)} $$
    $$ x^{48}-1 equiv -x^3-2x^2+x+2 pmod{f(x)}$$
    give
    $$gcd(x^{48}-1,x^4+2x^2-x+1)=gcd(x^3+2x^2-x-2,x^4+2x^2-x+1) $$
    $$ = gcd(x^3+2x^2-x-2,-x-3) $$
    but $x=4$ is not a root of $x^3+2x^2-x-2$ since $7nmid 90$ and we have finished.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      See this answer for a generalization - an efficient algorithm often used in practice (a polynomial analog of the Pocklington-Lehmer integer primality test)
      $endgroup$
      – Bill Dubuque
      Jan 2 at 23:13












    • $begingroup$
      That's it. I spent awhile looking for a trick answer, but didn't see any. An alternative way of organizing the calculation would be to first build a list of remainders of $x^0,x^7equiv-1+3x+3x^2+3x^3,x^{14}$ and $x^{21}$ modulo $f(x)$. Raising to seventh power modulo $f(x)$ is linear, so $$x^{49}equiv-1+3x^7+3x^{14}+3x^{21}pmod{f(x)}$$ et cetera. This may help with a higher degree polynomial. I'm afraid I don't have a good feel for the difference in complexity for I usually do this in characteristic two only :-)
      $endgroup$
      – Jyrki Lahtonen
      Jan 2 at 23:19


















    5












    $begingroup$

    The product of all the monic, irreducible polynomials over $mathbb{F}_7$ wih degree $leq 2$ is given by $x^{49}-x$, so if we prove that $f(x)=(x^2+1)^2-x$ and $x^{48}-1$ are coprime over $mathbb{F}_7[x]$ we are done. Let us compute $x^{48}-1pmod{f(x)}$, for instance through the Brauer chain $x^4to x^8to x^{16}to x^{32}to x^{48}$.
    $$ x^4 equiv -2x^2+x-1pmod{f(x)} $$
    $$ x^8 equiv 3x^3-3x^2+2x-3pmod{f(x)} $$
    $$ x^{16} equiv x^3+3x^2+2x-1pmod{f(x)} $$
    $$ x^{32} equiv -x^3+2x^2+x-3pmod{f(x)} $$
    $$ x^{48}-1 equiv -x^3-2x^2+x+2 pmod{f(x)}$$
    give
    $$gcd(x^{48}-1,x^4+2x^2-x+1)=gcd(x^3+2x^2-x-2,x^4+2x^2-x+1) $$
    $$ = gcd(x^3+2x^2-x-2,-x-3) $$
    but $x=4$ is not a root of $x^3+2x^2-x-2$ since $7nmid 90$ and we have finished.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      See this answer for a generalization - an efficient algorithm often used in practice (a polynomial analog of the Pocklington-Lehmer integer primality test)
      $endgroup$
      – Bill Dubuque
      Jan 2 at 23:13












    • $begingroup$
      That's it. I spent awhile looking for a trick answer, but didn't see any. An alternative way of organizing the calculation would be to first build a list of remainders of $x^0,x^7equiv-1+3x+3x^2+3x^3,x^{14}$ and $x^{21}$ modulo $f(x)$. Raising to seventh power modulo $f(x)$ is linear, so $$x^{49}equiv-1+3x^7+3x^{14}+3x^{21}pmod{f(x)}$$ et cetera. This may help with a higher degree polynomial. I'm afraid I don't have a good feel for the difference in complexity for I usually do this in characteristic two only :-)
      $endgroup$
      – Jyrki Lahtonen
      Jan 2 at 23:19
















    5












    5








    5





    $begingroup$

    The product of all the monic, irreducible polynomials over $mathbb{F}_7$ wih degree $leq 2$ is given by $x^{49}-x$, so if we prove that $f(x)=(x^2+1)^2-x$ and $x^{48}-1$ are coprime over $mathbb{F}_7[x]$ we are done. Let us compute $x^{48}-1pmod{f(x)}$, for instance through the Brauer chain $x^4to x^8to x^{16}to x^{32}to x^{48}$.
    $$ x^4 equiv -2x^2+x-1pmod{f(x)} $$
    $$ x^8 equiv 3x^3-3x^2+2x-3pmod{f(x)} $$
    $$ x^{16} equiv x^3+3x^2+2x-1pmod{f(x)} $$
    $$ x^{32} equiv -x^3+2x^2+x-3pmod{f(x)} $$
    $$ x^{48}-1 equiv -x^3-2x^2+x+2 pmod{f(x)}$$
    give
    $$gcd(x^{48}-1,x^4+2x^2-x+1)=gcd(x^3+2x^2-x-2,x^4+2x^2-x+1) $$
    $$ = gcd(x^3+2x^2-x-2,-x-3) $$
    but $x=4$ is not a root of $x^3+2x^2-x-2$ since $7nmid 90$ and we have finished.






    share|cite|improve this answer









    $endgroup$



    The product of all the monic, irreducible polynomials over $mathbb{F}_7$ wih degree $leq 2$ is given by $x^{49}-x$, so if we prove that $f(x)=(x^2+1)^2-x$ and $x^{48}-1$ are coprime over $mathbb{F}_7[x]$ we are done. Let us compute $x^{48}-1pmod{f(x)}$, for instance through the Brauer chain $x^4to x^8to x^{16}to x^{32}to x^{48}$.
    $$ x^4 equiv -2x^2+x-1pmod{f(x)} $$
    $$ x^8 equiv 3x^3-3x^2+2x-3pmod{f(x)} $$
    $$ x^{16} equiv x^3+3x^2+2x-1pmod{f(x)} $$
    $$ x^{32} equiv -x^3+2x^2+x-3pmod{f(x)} $$
    $$ x^{48}-1 equiv -x^3-2x^2+x+2 pmod{f(x)}$$
    give
    $$gcd(x^{48}-1,x^4+2x^2-x+1)=gcd(x^3+2x^2-x-2,x^4+2x^2-x+1) $$
    $$ = gcd(x^3+2x^2-x-2,-x-3) $$
    but $x=4$ is not a root of $x^3+2x^2-x-2$ since $7nmid 90$ and we have finished.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jan 2 at 22:39









    Jack D'AurizioJack D'Aurizio

    292k33284672




    292k33284672












    • $begingroup$
      See this answer for a generalization - an efficient algorithm often used in practice (a polynomial analog of the Pocklington-Lehmer integer primality test)
      $endgroup$
      – Bill Dubuque
      Jan 2 at 23:13












    • $begingroup$
      That's it. I spent awhile looking for a trick answer, but didn't see any. An alternative way of organizing the calculation would be to first build a list of remainders of $x^0,x^7equiv-1+3x+3x^2+3x^3,x^{14}$ and $x^{21}$ modulo $f(x)$. Raising to seventh power modulo $f(x)$ is linear, so $$x^{49}equiv-1+3x^7+3x^{14}+3x^{21}pmod{f(x)}$$ et cetera. This may help with a higher degree polynomial. I'm afraid I don't have a good feel for the difference in complexity for I usually do this in characteristic two only :-)
      $endgroup$
      – Jyrki Lahtonen
      Jan 2 at 23:19




















    • $begingroup$
      See this answer for a generalization - an efficient algorithm often used in practice (a polynomial analog of the Pocklington-Lehmer integer primality test)
      $endgroup$
      – Bill Dubuque
      Jan 2 at 23:13












    • $begingroup$
      That's it. I spent awhile looking for a trick answer, but didn't see any. An alternative way of organizing the calculation would be to first build a list of remainders of $x^0,x^7equiv-1+3x+3x^2+3x^3,x^{14}$ and $x^{21}$ modulo $f(x)$. Raising to seventh power modulo $f(x)$ is linear, so $$x^{49}equiv-1+3x^7+3x^{14}+3x^{21}pmod{f(x)}$$ et cetera. This may help with a higher degree polynomial. I'm afraid I don't have a good feel for the difference in complexity for I usually do this in characteristic two only :-)
      $endgroup$
      – Jyrki Lahtonen
      Jan 2 at 23:19


















    $begingroup$
    See this answer for a generalization - an efficient algorithm often used in practice (a polynomial analog of the Pocklington-Lehmer integer primality test)
    $endgroup$
    – Bill Dubuque
    Jan 2 at 23:13






    $begingroup$
    See this answer for a generalization - an efficient algorithm often used in practice (a polynomial analog of the Pocklington-Lehmer integer primality test)
    $endgroup$
    – Bill Dubuque
    Jan 2 at 23:13














    $begingroup$
    That's it. I spent awhile looking for a trick answer, but didn't see any. An alternative way of organizing the calculation would be to first build a list of remainders of $x^0,x^7equiv-1+3x+3x^2+3x^3,x^{14}$ and $x^{21}$ modulo $f(x)$. Raising to seventh power modulo $f(x)$ is linear, so $$x^{49}equiv-1+3x^7+3x^{14}+3x^{21}pmod{f(x)}$$ et cetera. This may help with a higher degree polynomial. I'm afraid I don't have a good feel for the difference in complexity for I usually do this in characteristic two only :-)
    $endgroup$
    – Jyrki Lahtonen
    Jan 2 at 23:19






    $begingroup$
    That's it. I spent awhile looking for a trick answer, but didn't see any. An alternative way of organizing the calculation would be to first build a list of remainders of $x^0,x^7equiv-1+3x+3x^2+3x^3,x^{14}$ and $x^{21}$ modulo $f(x)$. Raising to seventh power modulo $f(x)$ is linear, so $$x^{49}equiv-1+3x^7+3x^{14}+3x^{21}pmod{f(x)}$$ et cetera. This may help with a higher degree polynomial. I'm afraid I don't have a good feel for the difference in complexity for I usually do this in characteristic two only :-)
    $endgroup$
    – Jyrki Lahtonen
    Jan 2 at 23:19













    3












    $begingroup$

    You can also try to factor it as a product of two polynomials of degree $2$. Such a factorization must have the form



    begin{align}x^4 + 2x^2 - x + 1 & = (x^2+ax+b)(x^2-ax+b^{-1})\
    & =x^4+(b+b^{-1}-a^2)x^2+(ab^{-1}-ab)x+1end{align}
    Hence the relations
    $$b+b^{-1}-a^2=2, ab^{-1}-ab=-1,$$
    which can be easily seen to have no solution in $mathbb{Z}_7$.






    share|cite|improve this answer









    $endgroup$


















      3












      $begingroup$

      You can also try to factor it as a product of two polynomials of degree $2$. Such a factorization must have the form



      begin{align}x^4 + 2x^2 - x + 1 & = (x^2+ax+b)(x^2-ax+b^{-1})\
      & =x^4+(b+b^{-1}-a^2)x^2+(ab^{-1}-ab)x+1end{align}
      Hence the relations
      $$b+b^{-1}-a^2=2, ab^{-1}-ab=-1,$$
      which can be easily seen to have no solution in $mathbb{Z}_7$.






      share|cite|improve this answer









      $endgroup$
















        3












        3








        3





        $begingroup$

        You can also try to factor it as a product of two polynomials of degree $2$. Such a factorization must have the form



        begin{align}x^4 + 2x^2 - x + 1 & = (x^2+ax+b)(x^2-ax+b^{-1})\
        & =x^4+(b+b^{-1}-a^2)x^2+(ab^{-1}-ab)x+1end{align}
        Hence the relations
        $$b+b^{-1}-a^2=2, ab^{-1}-ab=-1,$$
        which can be easily seen to have no solution in $mathbb{Z}_7$.






        share|cite|improve this answer









        $endgroup$



        You can also try to factor it as a product of two polynomials of degree $2$. Such a factorization must have the form



        begin{align}x^4 + 2x^2 - x + 1 & = (x^2+ax+b)(x^2-ax+b^{-1})\
        & =x^4+(b+b^{-1}-a^2)x^2+(ab^{-1}-ab)x+1end{align}
        Hence the relations
        $$b+b^{-1}-a^2=2, ab^{-1}-ab=-1,$$
        which can be easily seen to have no solution in $mathbb{Z}_7$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 2 at 22:57









        moutheticsmouthetics

        52137




        52137






























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