Quotient $mathbf{F}_3[X]/(X^5+1)$












1












$begingroup$



Factor $X^5+1inmathbf{F}_3[X]$ into irreducibles. What does the quotient $mathbf{F}_3[X]/(X^5+1)$ look like?




Since $-1$ is a zero, we divide $X^5+1$ by $X+1$ using long division, to obtain $X^5+1=(X+1)(X^4-X^3+X^2-X+1)$. Now we claim that $X^4-X^3+X^2-X+1$ is irreducible. It has no zeros in $mathbf{F}_3$, and we can check that it can not be written as product of quadratic factors.



I am hesitating about the following part:



$(X^5+1)=(X+1,X^4-X^3+X^2-X+1)$.



By long division, we have $X^4-X^3+X^2-X+1=(X+1)(X^3-2X^2+3X-4)+5$, therefore $(X^5+1)=(X+1,5)$. We have $$frac{mathbf{F}_3[X]}{(X^5+1)}congfrac{mathbf{F}_3[X]}{(X+1,2)}congfrac{mathbf{Z}[X]/(3)}{(X+1,2)}cong frac{mathbf{Z}[X]/(X+1)}{(3,2)}congmathbf{Z}.$$



But $mathbf{Z}$ is not a field, and $frac{mathbf{F}_3[X]}{(X^5+1)}$ is I think. Could someone help me?






abstract-algebra polynomials ring-theory finite-fields irreducible-polynomials







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$endgroup$








  • 2




    $begingroup$
    Note $,5 = -1 in (X^5+1),Rightarrow,X^5+1$ is a unit (invertible) so that can't be true. The problem is the first equality after your "hesitating". Why do you believe that?
    $endgroup$
    – Bill Dubuque
    Dec 14 '18 at 21:26


















1












$begingroup$



Factor $X^5+1inmathbf{F}_3[X]$ into irreducibles. What does the quotient $mathbf{F}_3[X]/(X^5+1)$ look like?




Since $-1$ is a zero, we divide $X^5+1$ by $X+1$ using long division, to obtain $X^5+1=(X+1)(X^4-X^3+X^2-X+1)$. Now we claim that $X^4-X^3+X^2-X+1$ is irreducible. It has no zeros in $mathbf{F}_3$, and we can check that it can not be written as product of quadratic factors.



I am hesitating about the following part:



$(X^5+1)=(X+1,X^4-X^3+X^2-X+1)$.



By long division, we have $X^4-X^3+X^2-X+1=(X+1)(X^3-2X^2+3X-4)+5$, therefore $(X^5+1)=(X+1,5)$. We have $$frac{mathbf{F}_3[X]}{(X^5+1)}congfrac{mathbf{F}_3[X]}{(X+1,2)}congfrac{mathbf{Z}[X]/(3)}{(X+1,2)}cong frac{mathbf{Z}[X]/(X+1)}{(3,2)}congmathbf{Z}.$$



But $mathbf{Z}$ is not a field, and $frac{mathbf{F}_3[X]}{(X^5+1)}$ is I think. Could someone help me?






abstract-algebra polynomials ring-theory finite-fields irreducible-polynomials







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Note $,5 = -1 in (X^5+1),Rightarrow,X^5+1$ is a unit (invertible) so that can't be true. The problem is the first equality after your "hesitating". Why do you believe that?
    $endgroup$
    – Bill Dubuque
    Dec 14 '18 at 21:26
















1












1








1





$begingroup$



Factor $X^5+1inmathbf{F}_3[X]$ into irreducibles. What does the quotient $mathbf{F}_3[X]/(X^5+1)$ look like?




Since $-1$ is a zero, we divide $X^5+1$ by $X+1$ using long division, to obtain $X^5+1=(X+1)(X^4-X^3+X^2-X+1)$. Now we claim that $X^4-X^3+X^2-X+1$ is irreducible. It has no zeros in $mathbf{F}_3$, and we can check that it can not be written as product of quadratic factors.



I am hesitating about the following part:



$(X^5+1)=(X+1,X^4-X^3+X^2-X+1)$.



By long division, we have $X^4-X^3+X^2-X+1=(X+1)(X^3-2X^2+3X-4)+5$, therefore $(X^5+1)=(X+1,5)$. We have $$frac{mathbf{F}_3[X]}{(X^5+1)}congfrac{mathbf{F}_3[X]}{(X+1,2)}congfrac{mathbf{Z}[X]/(3)}{(X+1,2)}cong frac{mathbf{Z}[X]/(X+1)}{(3,2)}congmathbf{Z}.$$



But $mathbf{Z}$ is not a field, and $frac{mathbf{F}_3[X]}{(X^5+1)}$ is I think. Could someone help me?






abstract-algebra polynomials ring-theory finite-fields irreducible-polynomials







share|cite|improve this question











$endgroup$





Factor $X^5+1inmathbf{F}_3[X]$ into irreducibles. What does the quotient $mathbf{F}_3[X]/(X^5+1)$ look like?




Since $-1$ is a zero, we divide $X^5+1$ by $X+1$ using long division, to obtain $X^5+1=(X+1)(X^4-X^3+X^2-X+1)$. Now we claim that $X^4-X^3+X^2-X+1$ is irreducible. It has no zeros in $mathbf{F}_3$, and we can check that it can not be written as product of quadratic factors.



I am hesitating about the following part:



$(X^5+1)=(X+1,X^4-X^3+X^2-X+1)$.



By long division, we have $X^4-X^3+X^2-X+1=(X+1)(X^3-2X^2+3X-4)+5$, therefore $(X^5+1)=(X+1,5)$. We have $$frac{mathbf{F}_3[X]}{(X^5+1)}congfrac{mathbf{F}_3[X]}{(X+1,2)}congfrac{mathbf{Z}[X]/(3)}{(X+1,2)}cong frac{mathbf{Z}[X]/(X+1)}{(3,2)}congmathbf{Z}.$$



But $mathbf{Z}$ is not a field, and $frac{mathbf{F}_3[X]}{(X^5+1)}$ is I think. Could someone help me?






abstract-algebra polynomials ring-theory finite-fields irreducible-polynomials




abstract-algebra polynomials ring-theory 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 Dec 14 '18 at 21:44









Batominovski

33.1k33293




33.1k33293










asked Dec 14 '18 at 21:12









Heinz DoofenschmirtzHeinz Doofenschmirtz

615620




615620








  • 2




    $begingroup$
    Note $,5 = -1 in (X^5+1),Rightarrow,X^5+1$ is a unit (invertible) so that can't be true. The problem is the first equality after your "hesitating". Why do you believe that?
    $endgroup$
    – Bill Dubuque
    Dec 14 '18 at 21:26
















  • 2




    $begingroup$
    Note $,5 = -1 in (X^5+1),Rightarrow,X^5+1$ is a unit (invertible) so that can't be true. The problem is the first equality after your "hesitating". Why do you believe that?
    $endgroup$
    – Bill Dubuque
    Dec 14 '18 at 21:26










2




2




$begingroup$
Note $,5 = -1 in (X^5+1),Rightarrow,X^5+1$ is a unit (invertible) so that can't be true. The problem is the first equality after your "hesitating". Why do you believe that?
$endgroup$
– Bill Dubuque
Dec 14 '18 at 21:26






$begingroup$
Note $,5 = -1 in (X^5+1),Rightarrow,X^5+1$ is a unit (invertible) so that can't be true. The problem is the first equality after your "hesitating". Why do you believe that?
$endgroup$
– Bill Dubuque
Dec 14 '18 at 21:26












1 Answer
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oldest

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3












$begingroup$

I think the mistake you made is that the ideal $(X^5 + 1) neq (X+1, X^4 - X^3 + X^2 - X + 1)$. Similarly $(X^5 + 1) neq (X + 1,5)$. Long division only gives us "$subset$" in both cases, but not the other direction.



What you want to do is factor $X^5 + 1 = (X + 1) cdot (X^4 - X^3 + X^2 - X + 1)$. Note the ideal generated by $X^5 + 1$ is also the product of the ideal generated by the two factors, but not the sum.



If you want to compute the quotient $mathbb{F}[X] / (X^5 + 1)$, you can now apply the Chinese remainder theorem, that tells us
$$mathbb{F}[X] / (X^5 + 1) cong mathbb{F}[X] / (X+1) times mathbb{F}[X]/(X^4 - X^3+X^2-X+1).$$
Also note that this is not a field, because $X^5+1$ is not irreducible. This is because $X+1$ and $X^4 - X^3 + X^2 - X + 1$ are not in the ideal, i.e. they are not $0$ in the quotient, but if multiplied they give $0$, so they are zero divisors of the quotient.






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    $begingroup$

    I think the mistake you made is that the ideal $(X^5 + 1) neq (X+1, X^4 - X^3 + X^2 - X + 1)$. Similarly $(X^5 + 1) neq (X + 1,5)$. Long division only gives us "$subset$" in both cases, but not the other direction.



    What you want to do is factor $X^5 + 1 = (X + 1) cdot (X^4 - X^3 + X^2 - X + 1)$. Note the ideal generated by $X^5 + 1$ is also the product of the ideal generated by the two factors, but not the sum.



    If you want to compute the quotient $mathbb{F}[X] / (X^5 + 1)$, you can now apply the Chinese remainder theorem, that tells us
    $$mathbb{F}[X] / (X^5 + 1) cong mathbb{F}[X] / (X+1) times mathbb{F}[X]/(X^4 - X^3+X^2-X+1).$$
    Also note that this is not a field, because $X^5+1$ is not irreducible. This is because $X+1$ and $X^4 - X^3 + X^2 - X + 1$ are not in the ideal, i.e. they are not $0$ in the quotient, but if multiplied they give $0$, so they are zero divisors of the quotient.






    share|cite|improve this answer











    $endgroup$


















      3












      $begingroup$

      I think the mistake you made is that the ideal $(X^5 + 1) neq (X+1, X^4 - X^3 + X^2 - X + 1)$. Similarly $(X^5 + 1) neq (X + 1,5)$. Long division only gives us "$subset$" in both cases, but not the other direction.



      What you want to do is factor $X^5 + 1 = (X + 1) cdot (X^4 - X^3 + X^2 - X + 1)$. Note the ideal generated by $X^5 + 1$ is also the product of the ideal generated by the two factors, but not the sum.



      If you want to compute the quotient $mathbb{F}[X] / (X^5 + 1)$, you can now apply the Chinese remainder theorem, that tells us
      $$mathbb{F}[X] / (X^5 + 1) cong mathbb{F}[X] / (X+1) times mathbb{F}[X]/(X^4 - X^3+X^2-X+1).$$
      Also note that this is not a field, because $X^5+1$ is not irreducible. This is because $X+1$ and $X^4 - X^3 + X^2 - X + 1$ are not in the ideal, i.e. they are not $0$ in the quotient, but if multiplied they give $0$, so they are zero divisors of the quotient.






      share|cite|improve this answer











      $endgroup$
















        3












        3








        3





        $begingroup$

        I think the mistake you made is that the ideal $(X^5 + 1) neq (X+1, X^4 - X^3 + X^2 - X + 1)$. Similarly $(X^5 + 1) neq (X + 1,5)$. Long division only gives us "$subset$" in both cases, but not the other direction.



        What you want to do is factor $X^5 + 1 = (X + 1) cdot (X^4 - X^3 + X^2 - X + 1)$. Note the ideal generated by $X^5 + 1$ is also the product of the ideal generated by the two factors, but not the sum.



        If you want to compute the quotient $mathbb{F}[X] / (X^5 + 1)$, you can now apply the Chinese remainder theorem, that tells us
        $$mathbb{F}[X] / (X^5 + 1) cong mathbb{F}[X] / (X+1) times mathbb{F}[X]/(X^4 - X^3+X^2-X+1).$$
        Also note that this is not a field, because $X^5+1$ is not irreducible. This is because $X+1$ and $X^4 - X^3 + X^2 - X + 1$ are not in the ideal, i.e. they are not $0$ in the quotient, but if multiplied they give $0$, so they are zero divisors of the quotient.






        share|cite|improve this answer











        $endgroup$



        I think the mistake you made is that the ideal $(X^5 + 1) neq (X+1, X^4 - X^3 + X^2 - X + 1)$. Similarly $(X^5 + 1) neq (X + 1,5)$. Long division only gives us "$subset$" in both cases, but not the other direction.



        What you want to do is factor $X^5 + 1 = (X + 1) cdot (X^4 - X^3 + X^2 - X + 1)$. Note the ideal generated by $X^5 + 1$ is also the product of the ideal generated by the two factors, but not the sum.



        If you want to compute the quotient $mathbb{F}[X] / (X^5 + 1)$, you can now apply the Chinese remainder theorem, that tells us
        $$mathbb{F}[X] / (X^5 + 1) cong mathbb{F}[X] / (X+1) times mathbb{F}[X]/(X^4 - X^3+X^2-X+1).$$
        Also note that this is not a field, because $X^5+1$ is not irreducible. This is because $X+1$ and $X^4 - X^3 + X^2 - X + 1$ are not in the ideal, i.e. they are not $0$ in the quotient, but if multiplied they give $0$, so they are zero divisors of the quotient.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 14 '18 at 22:17

























        answered Dec 14 '18 at 22:09









        red_trumpetred_trumpet

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        991319






























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