Invertible elements of $mathbb{Z}_3[x] / (x^4+x^3-1)^3$












4












$begingroup$


Let $F=mathbb{Z}/3mathbb{Z}$, $h(x)=x^4+x^3-1$, $R = F[x]/(h(x)^3)$.



I know $R$ has $4$ ideals and $1$ maximal ideal. Let $M$ be the maximal ideal $(h(x))/(h(x)^3)$



I need to find the number of invertible elements of $R$ and in order to do so I need the number of elements in $M$. I think this number is $3^8$ since we must have (I'm not sure about that):



$|R| = |R/M||M|$ ($R/M$ is the quotient additive group)



$|R|$ = $mathrm{3}^{12}$



$|R/M| = |F[x]/(h(x))| = 3^4$ (is that so?)



Then:



$mathrm{3}^{12} = 3^4 cdot |M|$ so $|M| = 3^8$



But according to the solution of the exercice, $|M| = 3^9$. Where did I go wrong?










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  • $begingroup$
    Another argument would be that $M simeq F[x] / langle h^2 rangle$ (the right-to-left map would be induced by multiplication by $h$) and the right hand side has size $3^8$ agreeing with your calculation.
    $endgroup$
    – Daniel Schepler
    Jan 17 at 19:17










  • $begingroup$
    I suspect the solution on the paper is wrong but, most importantly, is my procedure valid?
    $endgroup$
    – MattP9
    Jan 17 at 19:20
















4












$begingroup$


Let $F=mathbb{Z}/3mathbb{Z}$, $h(x)=x^4+x^3-1$, $R = F[x]/(h(x)^3)$.



I know $R$ has $4$ ideals and $1$ maximal ideal. Let $M$ be the maximal ideal $(h(x))/(h(x)^3)$



I need to find the number of invertible elements of $R$ and in order to do so I need the number of elements in $M$. I think this number is $3^8$ since we must have (I'm not sure about that):



$|R| = |R/M||M|$ ($R/M$ is the quotient additive group)



$|R|$ = $mathrm{3}^{12}$



$|R/M| = |F[x]/(h(x))| = 3^4$ (is that so?)



Then:



$mathrm{3}^{12} = 3^4 cdot |M|$ so $|M| = 3^8$



But according to the solution of the exercice, $|M| = 3^9$. Where did I go wrong?










share|cite|improve this question











$endgroup$












  • $begingroup$
    Another argument would be that $M simeq F[x] / langle h^2 rangle$ (the right-to-left map would be induced by multiplication by $h$) and the right hand side has size $3^8$ agreeing with your calculation.
    $endgroup$
    – Daniel Schepler
    Jan 17 at 19:17










  • $begingroup$
    I suspect the solution on the paper is wrong but, most importantly, is my procedure valid?
    $endgroup$
    – MattP9
    Jan 17 at 19:20














4












4








4





$begingroup$


Let $F=mathbb{Z}/3mathbb{Z}$, $h(x)=x^4+x^3-1$, $R = F[x]/(h(x)^3)$.



I know $R$ has $4$ ideals and $1$ maximal ideal. Let $M$ be the maximal ideal $(h(x))/(h(x)^3)$



I need to find the number of invertible elements of $R$ and in order to do so I need the number of elements in $M$. I think this number is $3^8$ since we must have (I'm not sure about that):



$|R| = |R/M||M|$ ($R/M$ is the quotient additive group)



$|R|$ = $mathrm{3}^{12}$



$|R/M| = |F[x]/(h(x))| = 3^4$ (is that so?)



Then:



$mathrm{3}^{12} = 3^4 cdot |M|$ so $|M| = 3^8$



But according to the solution of the exercice, $|M| = 3^9$. Where did I go wrong?










share|cite|improve this question











$endgroup$




Let $F=mathbb{Z}/3mathbb{Z}$, $h(x)=x^4+x^3-1$, $R = F[x]/(h(x)^3)$.



I know $R$ has $4$ ideals and $1$ maximal ideal. Let $M$ be the maximal ideal $(h(x))/(h(x)^3)$



I need to find the number of invertible elements of $R$ and in order to do so I need the number of elements in $M$. I think this number is $3^8$ since we must have (I'm not sure about that):



$|R| = |R/M||M|$ ($R/M$ is the quotient additive group)



$|R|$ = $mathrm{3}^{12}$



$|R/M| = |F[x]/(h(x))| = 3^4$ (is that so?)



Then:



$mathrm{3}^{12} = 3^4 cdot |M|$ so $|M| = 3^8$



But according to the solution of the exercice, $|M| = 3^9$. Where did I go wrong?







ring-theory galois-theory ideals finite-fields irreducible-polynomials






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edited Jan 17 at 20:15









user289143

1,069313




1,069313










asked Jan 17 at 19:00









MattP9MattP9

304




304












  • $begingroup$
    Another argument would be that $M simeq F[x] / langle h^2 rangle$ (the right-to-left map would be induced by multiplication by $h$) and the right hand side has size $3^8$ agreeing with your calculation.
    $endgroup$
    – Daniel Schepler
    Jan 17 at 19:17










  • $begingroup$
    I suspect the solution on the paper is wrong but, most importantly, is my procedure valid?
    $endgroup$
    – MattP9
    Jan 17 at 19:20


















  • $begingroup$
    Another argument would be that $M simeq F[x] / langle h^2 rangle$ (the right-to-left map would be induced by multiplication by $h$) and the right hand side has size $3^8$ agreeing with your calculation.
    $endgroup$
    – Daniel Schepler
    Jan 17 at 19:17










  • $begingroup$
    I suspect the solution on the paper is wrong but, most importantly, is my procedure valid?
    $endgroup$
    – MattP9
    Jan 17 at 19:20
















$begingroup$
Another argument would be that $M simeq F[x] / langle h^2 rangle$ (the right-to-left map would be induced by multiplication by $h$) and the right hand side has size $3^8$ agreeing with your calculation.
$endgroup$
– Daniel Schepler
Jan 17 at 19:17




$begingroup$
Another argument would be that $M simeq F[x] / langle h^2 rangle$ (the right-to-left map would be induced by multiplication by $h$) and the right hand side has size $3^8$ agreeing with your calculation.
$endgroup$
– Daniel Schepler
Jan 17 at 19:17












$begingroup$
I suspect the solution on the paper is wrong but, most importantly, is my procedure valid?
$endgroup$
– MattP9
Jan 17 at 19:20




$begingroup$
I suspect the solution on the paper is wrong but, most importantly, is my procedure valid?
$endgroup$
– MattP9
Jan 17 at 19:20










1 Answer
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$begingroup$

Your procedure is entirely valid; you can verify that $|R/M|=3^4$ by simply noting that $R/M$ is a vector space over $F$ with basis ${1,x,x^2,x^3}$. In fact, whatever argument you use to prove that $|R|=3^{12}$ undoubtedly also proves that $|R/M|=3^4$. Indeed it follows that $|M|=3^8$, and the solution given to you is wrong.






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    1 Answer
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    $begingroup$

    Your procedure is entirely valid; you can verify that $|R/M|=3^4$ by simply noting that $R/M$ is a vector space over $F$ with basis ${1,x,x^2,x^3}$. In fact, whatever argument you use to prove that $|R|=3^{12}$ undoubtedly also proves that $|R/M|=3^4$. Indeed it follows that $|M|=3^8$, and the solution given to you is wrong.






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      Your procedure is entirely valid; you can verify that $|R/M|=3^4$ by simply noting that $R/M$ is a vector space over $F$ with basis ${1,x,x^2,x^3}$. In fact, whatever argument you use to prove that $|R|=3^{12}$ undoubtedly also proves that $|R/M|=3^4$. Indeed it follows that $|M|=3^8$, and the solution given to you is wrong.






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        Your procedure is entirely valid; you can verify that $|R/M|=3^4$ by simply noting that $R/M$ is a vector space over $F$ with basis ${1,x,x^2,x^3}$. In fact, whatever argument you use to prove that $|R|=3^{12}$ undoubtedly also proves that $|R/M|=3^4$. Indeed it follows that $|M|=3^8$, and the solution given to you is wrong.






        share|cite|improve this answer









        $endgroup$



        Your procedure is entirely valid; you can verify that $|R/M|=3^4$ by simply noting that $R/M$ is a vector space over $F$ with basis ${1,x,x^2,x^3}$. In fact, whatever argument you use to prove that $|R|=3^{12}$ undoubtedly also proves that $|R/M|=3^4$. Indeed it follows that $|M|=3^8$, and the solution given to you is wrong.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 17 at 19:30









        ServaesServaes

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        30.6k342101






























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