Number of elements in the ideal of Ring.












1












$begingroup$


We have $$x^9+1 = (x+1)(x^2+x+1)(x^6+x^3+1)$$ is factorization of irreducible polynomials over $GF(2)$ (Galois field).
Then we know that one of its ideal for the ring is $$R = GF(2)[x]/(x^9+1)$$
One of ideals is generated by $$I = (x^2+x+1)(x^6+x^3+1)$$ Then



(i) What is the dimension of this ideal $I$ in $R$?



(ii) Number of elements in this ideal?



(iii) How each element of $I$ looks like?



Any hint or help?






field-theory ideals irreducible-polynomials quotient-spaces







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$endgroup$












  • $begingroup$
    You are doing the same qual as this guy I guess ? math.stackexchange.com/q/3012834/29335
    $endgroup$
    – rschwieb
    Nov 27 '18 at 1:09
















1












$begingroup$


We have $$x^9+1 = (x+1)(x^2+x+1)(x^6+x^3+1)$$ is factorization of irreducible polynomials over $GF(2)$ (Galois field).
Then we know that one of its ideal for the ring is $$R = GF(2)[x]/(x^9+1)$$
One of ideals is generated by $$I = (x^2+x+1)(x^6+x^3+1)$$ Then



(i) What is the dimension of this ideal $I$ in $R$?



(ii) Number of elements in this ideal?



(iii) How each element of $I$ looks like?



Any hint or help?






field-theory ideals irreducible-polynomials quotient-spaces







share|cite|improve this question









$endgroup$












  • $begingroup$
    You are doing the same qual as this guy I guess ? math.stackexchange.com/q/3012834/29335
    $endgroup$
    – rschwieb
    Nov 27 '18 at 1:09














1












1








1





$begingroup$


We have $$x^9+1 = (x+1)(x^2+x+1)(x^6+x^3+1)$$ is factorization of irreducible polynomials over $GF(2)$ (Galois field).
Then we know that one of its ideal for the ring is $$R = GF(2)[x]/(x^9+1)$$
One of ideals is generated by $$I = (x^2+x+1)(x^6+x^3+1)$$ Then



(i) What is the dimension of this ideal $I$ in $R$?



(ii) Number of elements in this ideal?



(iii) How each element of $I$ looks like?



Any hint or help?






field-theory ideals irreducible-polynomials quotient-spaces







share|cite|improve this question









$endgroup$




We have $$x^9+1 = (x+1)(x^2+x+1)(x^6+x^3+1)$$ is factorization of irreducible polynomials over $GF(2)$ (Galois field).
Then we know that one of its ideal for the ring is $$R = GF(2)[x]/(x^9+1)$$
One of ideals is generated by $$I = (x^2+x+1)(x^6+x^3+1)$$ Then



(i) What is the dimension of this ideal $I$ in $R$?



(ii) Number of elements in this ideal?



(iii) How each element of $I$ looks like?



Any hint or help?






field-theory ideals irreducible-polynomials quotient-spaces




field-theory ideals irreducible-polynomials quotient-spaces






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 25 '18 at 13:30









Mittal GMittal G

1,193516




1,193516












  • $begingroup$
    You are doing the same qual as this guy I guess ? math.stackexchange.com/q/3012834/29335
    $endgroup$
    – rschwieb
    Nov 27 '18 at 1:09


















  • $begingroup$
    You are doing the same qual as this guy I guess ? math.stackexchange.com/q/3012834/29335
    $endgroup$
    – rschwieb
    Nov 27 '18 at 1:09
















$begingroup$
You are doing the same qual as this guy I guess ? math.stackexchange.com/q/3012834/29335
$endgroup$
– rschwieb
Nov 27 '18 at 1:09




$begingroup$
You are doing the same qual as this guy I guess ? math.stackexchange.com/q/3012834/29335
$endgroup$
– rschwieb
Nov 27 '18 at 1:09










2 Answers
2






active

oldest

votes


















0












$begingroup$

The ring $R$ has the monomial basis $1, x, x^2, dotsc, x^8$. This is a vector space basis of $R$ as a vector field over $GF(2)$.



You want to find a $GF(2)$ basis for $I$. This can be done by taking your generator $f = (x^2 + x + 1)(x^6 + x^3 + 1)$ and computing the elements
$$f, xf, x^2f, x^3f,dotsc$$
Those clearly generate $I$ over $GF(2)$, so you want to choose a basis from this generating system. To do this, write the elements in the monomial basis $1, x, x^2, dotsc, x^8$ of $R$, i.e. $x^kf = sum_i a_{i,k} x^i$ and look at the vectors
$$v_k = left[begin{matrix}a_{0,k}\vdots\a_{8,k}end{matrix}right].$$
For which $l$ does the system ${v_0,v_1,dots,v_l}$ become linearly dependent? Then ${v_0,dots,v_{l-1}}$ is your basis.



This directly gives you the dimension of $I$, so it has $2^l$ elements.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Can we say that $l=8$?
    $endgroup$
    – Mittal G
    Nov 25 '18 at 15:23



















0












$begingroup$

You’re asking for the ideal generated by the product $h(x)=(x^2+x+1)(x^6+x^3+1)$, in the ring $R=Bbb F_2[x]/(x^9+1)=Bbb F_2/bigl((x+1)h(x)bigr)$. Now by direct inspection, $h(x)=sum_{j=0}^8x^j$, and you easily check that $xhequiv hpmod{(x^9+1)}$. It follows from this that $h^2equiv hpmod{(x^9+1)}$, from which you conclude that your ideal is one-dimensional over $Bbb F_2$: its only elements are $0$ and $h$.






share|cite|improve this answer









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    2 Answers
    2






    active

    oldest

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    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

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    active

    oldest

    votes









    0












    $begingroup$

    The ring $R$ has the monomial basis $1, x, x^2, dotsc, x^8$. This is a vector space basis of $R$ as a vector field over $GF(2)$.



    You want to find a $GF(2)$ basis for $I$. This can be done by taking your generator $f = (x^2 + x + 1)(x^6 + x^3 + 1)$ and computing the elements
    $$f, xf, x^2f, x^3f,dotsc$$
    Those clearly generate $I$ over $GF(2)$, so you want to choose a basis from this generating system. To do this, write the elements in the monomial basis $1, x, x^2, dotsc, x^8$ of $R$, i.e. $x^kf = sum_i a_{i,k} x^i$ and look at the vectors
    $$v_k = left[begin{matrix}a_{0,k}\vdots\a_{8,k}end{matrix}right].$$
    For which $l$ does the system ${v_0,v_1,dots,v_l}$ become linearly dependent? Then ${v_0,dots,v_{l-1}}$ is your basis.



    This directly gives you the dimension of $I$, so it has $2^l$ elements.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Can we say that $l=8$?
      $endgroup$
      – Mittal G
      Nov 25 '18 at 15:23
















    0












    $begingroup$

    The ring $R$ has the monomial basis $1, x, x^2, dotsc, x^8$. This is a vector space basis of $R$ as a vector field over $GF(2)$.



    You want to find a $GF(2)$ basis for $I$. This can be done by taking your generator $f = (x^2 + x + 1)(x^6 + x^3 + 1)$ and computing the elements
    $$f, xf, x^2f, x^3f,dotsc$$
    Those clearly generate $I$ over $GF(2)$, so you want to choose a basis from this generating system. To do this, write the elements in the monomial basis $1, x, x^2, dotsc, x^8$ of $R$, i.e. $x^kf = sum_i a_{i,k} x^i$ and look at the vectors
    $$v_k = left[begin{matrix}a_{0,k}\vdots\a_{8,k}end{matrix}right].$$
    For which $l$ does the system ${v_0,v_1,dots,v_l}$ become linearly dependent? Then ${v_0,dots,v_{l-1}}$ is your basis.



    This directly gives you the dimension of $I$, so it has $2^l$ elements.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Can we say that $l=8$?
      $endgroup$
      – Mittal G
      Nov 25 '18 at 15:23














    0












    0








    0





    $begingroup$

    The ring $R$ has the monomial basis $1, x, x^2, dotsc, x^8$. This is a vector space basis of $R$ as a vector field over $GF(2)$.



    You want to find a $GF(2)$ basis for $I$. This can be done by taking your generator $f = (x^2 + x + 1)(x^6 + x^3 + 1)$ and computing the elements
    $$f, xf, x^2f, x^3f,dotsc$$
    Those clearly generate $I$ over $GF(2)$, so you want to choose a basis from this generating system. To do this, write the elements in the monomial basis $1, x, x^2, dotsc, x^8$ of $R$, i.e. $x^kf = sum_i a_{i,k} x^i$ and look at the vectors
    $$v_k = left[begin{matrix}a_{0,k}\vdots\a_{8,k}end{matrix}right].$$
    For which $l$ does the system ${v_0,v_1,dots,v_l}$ become linearly dependent? Then ${v_0,dots,v_{l-1}}$ is your basis.



    This directly gives you the dimension of $I$, so it has $2^l$ elements.






    share|cite|improve this answer









    $endgroup$



    The ring $R$ has the monomial basis $1, x, x^2, dotsc, x^8$. This is a vector space basis of $R$ as a vector field over $GF(2)$.



    You want to find a $GF(2)$ basis for $I$. This can be done by taking your generator $f = (x^2 + x + 1)(x^6 + x^3 + 1)$ and computing the elements
    $$f, xf, x^2f, x^3f,dotsc$$
    Those clearly generate $I$ over $GF(2)$, so you want to choose a basis from this generating system. To do this, write the elements in the monomial basis $1, x, x^2, dotsc, x^8$ of $R$, i.e. $x^kf = sum_i a_{i,k} x^i$ and look at the vectors
    $$v_k = left[begin{matrix}a_{0,k}\vdots\a_{8,k}end{matrix}right].$$
    For which $l$ does the system ${v_0,v_1,dots,v_l}$ become linearly dependent? Then ${v_0,dots,v_{l-1}}$ is your basis.



    This directly gives you the dimension of $I$, so it has $2^l$ elements.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Nov 25 '18 at 13:43









    red_trumpetred_trumpet

    841219




    841219












    • $begingroup$
      Can we say that $l=8$?
      $endgroup$
      – Mittal G
      Nov 25 '18 at 15:23


















    • $begingroup$
      Can we say that $l=8$?
      $endgroup$
      – Mittal G
      Nov 25 '18 at 15:23
















    $begingroup$
    Can we say that $l=8$?
    $endgroup$
    – Mittal G
    Nov 25 '18 at 15:23




    $begingroup$
    Can we say that $l=8$?
    $endgroup$
    – Mittal G
    Nov 25 '18 at 15:23











    0












    $begingroup$

    You’re asking for the ideal generated by the product $h(x)=(x^2+x+1)(x^6+x^3+1)$, in the ring $R=Bbb F_2[x]/(x^9+1)=Bbb F_2/bigl((x+1)h(x)bigr)$. Now by direct inspection, $h(x)=sum_{j=0}^8x^j$, and you easily check that $xhequiv hpmod{(x^9+1)}$. It follows from this that $h^2equiv hpmod{(x^9+1)}$, from which you conclude that your ideal is one-dimensional over $Bbb F_2$: its only elements are $0$ and $h$.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      You’re asking for the ideal generated by the product $h(x)=(x^2+x+1)(x^6+x^3+1)$, in the ring $R=Bbb F_2[x]/(x^9+1)=Bbb F_2/bigl((x+1)h(x)bigr)$. Now by direct inspection, $h(x)=sum_{j=0}^8x^j$, and you easily check that $xhequiv hpmod{(x^9+1)}$. It follows from this that $h^2equiv hpmod{(x^9+1)}$, from which you conclude that your ideal is one-dimensional over $Bbb F_2$: its only elements are $0$ and $h$.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        You’re asking for the ideal generated by the product $h(x)=(x^2+x+1)(x^6+x^3+1)$, in the ring $R=Bbb F_2[x]/(x^9+1)=Bbb F_2/bigl((x+1)h(x)bigr)$. Now by direct inspection, $h(x)=sum_{j=0}^8x^j$, and you easily check that $xhequiv hpmod{(x^9+1)}$. It follows from this that $h^2equiv hpmod{(x^9+1)}$, from which you conclude that your ideal is one-dimensional over $Bbb F_2$: its only elements are $0$ and $h$.






        share|cite|improve this answer









        $endgroup$



        You’re asking for the ideal generated by the product $h(x)=(x^2+x+1)(x^6+x^3+1)$, in the ring $R=Bbb F_2[x]/(x^9+1)=Bbb F_2/bigl((x+1)h(x)bigr)$. Now by direct inspection, $h(x)=sum_{j=0}^8x^j$, and you easily check that $xhequiv hpmod{(x^9+1)}$. It follows from this that $h^2equiv hpmod{(x^9+1)}$, from which you conclude that your ideal is one-dimensional over $Bbb F_2$: its only elements are $0$ and $h$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 25 '18 at 16:34









        LubinLubin

        44k44585




        44k44585






























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