Group of units of $mathbb F_p[x]/(x^3-1)$

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Consider the ring $R=mathbb F_p[x]/(x^3-1)$. Find the group of units (as a product of cyclic groups)
for $p=29, 31$.




We have $x^3-1=(x-1)(x^2+x+1)$. If I manage to prove that the ideals $(x-1)$ and $(x^2+x+1)$ are comaximal in $mathbb F_p[x]$, then it will follow by the Chinese remainder theorem that $Rsimeqmathbb F_p[x]/(x-1)times mathbb F_p[x]/(x^2+x+1)$. I don't see how I can obtain $1$ as a linear combination of these generators, so the first question is how to show they are comaximal?



Secondly, $mathbb F_p[x]/(x-1)simeq mathbb F_p$, and the answer depends on the second factor. It is a pain to check whether $x^2+x+1$ has a root for $p=29, 31$, so I was wondering how to find roots of the polynomials quickly or prove that they are irreducible?




abstract-algebra group-theory ring-theory field-theory




share|cite|improve this question















  • 2




    Because $x^2+x+1mid x^3-1$ any root has multiplicative order three (or one but we can ignore that). Because the multiplicative group of $BbbF_p$ is cyclic of order $p-1$ there is such an element in the field iff $3mid p-1$. Meaning that $x^2+x+1$ factors modulo $31$ but is irreducible modulo $29$.
    – Jyrki Lahtonen
    Jul 16 at 4:19






  • 1




    And a hint for comaximality: run the generalized Euclidean algorithm on $x-1$ and $x^2+x+1$ to find polynomials $u(x),v(x)$ such that $u(x)(x+1)+v(x)(x^2+x+1)$ is a constant. If the constant is non-zero you can check that it isn't divisible by either of the choices of $p$. Good luck!
    – Jyrki Lahtonen
    Jul 16 at 4:21






  • 3




    It is, in fact, extremely easy to find a root of $x^2+x+1$ for $p=31$. What squares are closest to $31$?
    – Gerry Myerson
    Jul 16 at 7:30










  • @JyrkiLahtonen Regarding the first part, why can we ignore that a root of $x^2+x+1$ may have order one? For the second part, I got that $(x^2+x+1)-(x+2)(x-1)=3$. Did you mean that from this I can conclude that since $3$ is nonzero in either field and any non-zero element is invertible, hence $(x-1)+(x^2+x+1)=mathbb F_p[x]$?
    – user437309
    Jul 16 at 15:59










  • I guess we can ignore that a root of $x^2+x+1$ may have order one because if it does, then $x=1$, but $1+1+1ne 3$ unless $p=3$.
    – user437309
    Jul 16 at 19:04














up vote
2
down vote

favorite













Consider the ring $R=mathbb F_p[x]/(x^3-1)$. Find the group of units (as a product of cyclic groups)
for $p=29, 31$.




We have $x^3-1=(x-1)(x^2+x+1)$. If I manage to prove that the ideals $(x-1)$ and $(x^2+x+1)$ are comaximal in $mathbb F_p[x]$, then it will follow by the Chinese remainder theorem that $Rsimeqmathbb F_p[x]/(x-1)times mathbb F_p[x]/(x^2+x+1)$. I don't see how I can obtain $1$ as a linear combination of these generators, so the first question is how to show they are comaximal?



Secondly, $mathbb F_p[x]/(x-1)simeq mathbb F_p$, and the answer depends on the second factor. It is a pain to check whether $x^2+x+1$ has a root for $p=29, 31$, so I was wondering how to find roots of the polynomials quickly or prove that they are irreducible?




abstract-algebra group-theory ring-theory field-theory




share|cite|improve this question















  • 2




    Because $x^2+x+1mid x^3-1$ any root has multiplicative order three (or one but we can ignore that). Because the multiplicative group of $BbbF_p$ is cyclic of order $p-1$ there is such an element in the field iff $3mid p-1$. Meaning that $x^2+x+1$ factors modulo $31$ but is irreducible modulo $29$.
    – Jyrki Lahtonen
    Jul 16 at 4:19






  • 1




    And a hint for comaximality: run the generalized Euclidean algorithm on $x-1$ and $x^2+x+1$ to find polynomials $u(x),v(x)$ such that $u(x)(x+1)+v(x)(x^2+x+1)$ is a constant. If the constant is non-zero you can check that it isn't divisible by either of the choices of $p$. Good luck!
    – Jyrki Lahtonen
    Jul 16 at 4:21






  • 3




    It is, in fact, extremely easy to find a root of $x^2+x+1$ for $p=31$. What squares are closest to $31$?
    – Gerry Myerson
    Jul 16 at 7:30










  • @JyrkiLahtonen Regarding the first part, why can we ignore that a root of $x^2+x+1$ may have order one? For the second part, I got that $(x^2+x+1)-(x+2)(x-1)=3$. Did you mean that from this I can conclude that since $3$ is nonzero in either field and any non-zero element is invertible, hence $(x-1)+(x^2+x+1)=mathbb F_p[x]$?
    – user437309
    Jul 16 at 15:59










  • I guess we can ignore that a root of $x^2+x+1$ may have order one because if it does, then $x=1$, but $1+1+1ne 3$ unless $p=3$.
    – user437309
    Jul 16 at 19:04












up vote
2
down vote

favorite









up vote
2
down vote

favorite












Consider the ring $R=mathbb F_p[x]/(x^3-1)$. Find the group of units (as a product of cyclic groups)
for $p=29, 31$.




We have $x^3-1=(x-1)(x^2+x+1)$. If I manage to prove that the ideals $(x-1)$ and $(x^2+x+1)$ are comaximal in $mathbb F_p[x]$, then it will follow by the Chinese remainder theorem that $Rsimeqmathbb F_p[x]/(x-1)times mathbb F_p[x]/(x^2+x+1)$. I don't see how I can obtain $1$ as a linear combination of these generators, so the first question is how to show they are comaximal?



Secondly, $mathbb F_p[x]/(x-1)simeq mathbb F_p$, and the answer depends on the second factor. It is a pain to check whether $x^2+x+1$ has a root for $p=29, 31$, so I was wondering how to find roots of the polynomials quickly or prove that they are irreducible?




abstract-algebra group-theory ring-theory field-theory




share|cite|improve this question












Consider the ring $R=mathbb F_p[x]/(x^3-1)$. Find the group of units (as a product of cyclic groups)
for $p=29, 31$.




We have $x^3-1=(x-1)(x^2+x+1)$. If I manage to prove that the ideals $(x-1)$ and $(x^2+x+1)$ are comaximal in $mathbb F_p[x]$, then it will follow by the Chinese remainder theorem that $Rsimeqmathbb F_p[x]/(x-1)times mathbb F_p[x]/(x^2+x+1)$. I don't see how I can obtain $1$ as a linear combination of these generators, so the first question is how to show they are comaximal?



Secondly, $mathbb F_p[x]/(x-1)simeq mathbb F_p$, and the answer depends on the second factor. It is a pain to check whether $x^2+x+1$ has a root for $p=29, 31$, so I was wondering how to find roots of the polynomials quickly or prove that they are irreducible?





abstract-algebra group-theory ring-theory field-theory





share|cite|improve this question










share|cite|improve this question




share|cite|improve this question









asked Jul 16 at 4:10









user437309

556212




556212







  • 2




    Because $x^2+x+1mid x^3-1$ any root has multiplicative order three (or one but we can ignore that). Because the multiplicative group of $BbbF_p$ is cyclic of order $p-1$ there is such an element in the field iff $3mid p-1$. Meaning that $x^2+x+1$ factors modulo $31$ but is irreducible modulo $29$.
    – Jyrki Lahtonen
    Jul 16 at 4:19






  • 1




    And a hint for comaximality: run the generalized Euclidean algorithm on $x-1$ and $x^2+x+1$ to find polynomials $u(x),v(x)$ such that $u(x)(x+1)+v(x)(x^2+x+1)$ is a constant. If the constant is non-zero you can check that it isn't divisible by either of the choices of $p$. Good luck!
    – Jyrki Lahtonen
    Jul 16 at 4:21






  • 3




    It is, in fact, extremely easy to find a root of $x^2+x+1$ for $p=31$. What squares are closest to $31$?
    – Gerry Myerson
    Jul 16 at 7:30










  • @JyrkiLahtonen Regarding the first part, why can we ignore that a root of $x^2+x+1$ may have order one? For the second part, I got that $(x^2+x+1)-(x+2)(x-1)=3$. Did you mean that from this I can conclude that since $3$ is nonzero in either field and any non-zero element is invertible, hence $(x-1)+(x^2+x+1)=mathbb F_p[x]$?
    – user437309
    Jul 16 at 15:59










  • I guess we can ignore that a root of $x^2+x+1$ may have order one because if it does, then $x=1$, but $1+1+1ne 3$ unless $p=3$.
    – user437309
    Jul 16 at 19:04












  • 2




    Because $x^2+x+1mid x^3-1$ any root has multiplicative order three (or one but we can ignore that). Because the multiplicative group of $BbbF_p$ is cyclic of order $p-1$ there is such an element in the field iff $3mid p-1$. Meaning that $x^2+x+1$ factors modulo $31$ but is irreducible modulo $29$.
    – Jyrki Lahtonen
    Jul 16 at 4:19






  • 1




    And a hint for comaximality: run the generalized Euclidean algorithm on $x-1$ and $x^2+x+1$ to find polynomials $u(x),v(x)$ such that $u(x)(x+1)+v(x)(x^2+x+1)$ is a constant. If the constant is non-zero you can check that it isn't divisible by either of the choices of $p$. Good luck!
    – Jyrki Lahtonen
    Jul 16 at 4:21






  • 3




    It is, in fact, extremely easy to find a root of $x^2+x+1$ for $p=31$. What squares are closest to $31$?
    – Gerry Myerson
    Jul 16 at 7:30










  • @JyrkiLahtonen Regarding the first part, why can we ignore that a root of $x^2+x+1$ may have order one? For the second part, I got that $(x^2+x+1)-(x+2)(x-1)=3$. Did you mean that from this I can conclude that since $3$ is nonzero in either field and any non-zero element is invertible, hence $(x-1)+(x^2+x+1)=mathbb F_p[x]$?
    – user437309
    Jul 16 at 15:59










  • I guess we can ignore that a root of $x^2+x+1$ may have order one because if it does, then $x=1$, but $1+1+1ne 3$ unless $p=3$.
    – user437309
    Jul 16 at 19:04







2




2




Because $x^2+x+1mid x^3-1$ any root has multiplicative order three (or one but we can ignore that). Because the multiplicative group of $BbbF_p$ is cyclic of order $p-1$ there is such an element in the field iff $3mid p-1$. Meaning that $x^2+x+1$ factors modulo $31$ but is irreducible modulo $29$.
– Jyrki Lahtonen
Jul 16 at 4:19




Because $x^2+x+1mid x^3-1$ any root has multiplicative order three (or one but we can ignore that). Because the multiplicative group of $BbbF_p$ is cyclic of order $p-1$ there is such an element in the field iff $3mid p-1$. Meaning that $x^2+x+1$ factors modulo $31$ but is irreducible modulo $29$.
– Jyrki Lahtonen
Jul 16 at 4:19




1




1




And a hint for comaximality: run the generalized Euclidean algorithm on $x-1$ and $x^2+x+1$ to find polynomials $u(x),v(x)$ such that $u(x)(x+1)+v(x)(x^2+x+1)$ is a constant. If the constant is non-zero you can check that it isn't divisible by either of the choices of $p$. Good luck!
– Jyrki Lahtonen
Jul 16 at 4:21




And a hint for comaximality: run the generalized Euclidean algorithm on $x-1$ and $x^2+x+1$ to find polynomials $u(x),v(x)$ such that $u(x)(x+1)+v(x)(x^2+x+1)$ is a constant. If the constant is non-zero you can check that it isn't divisible by either of the choices of $p$. Good luck!
– Jyrki Lahtonen
Jul 16 at 4:21




3




3




It is, in fact, extremely easy to find a root of $x^2+x+1$ for $p=31$. What squares are closest to $31$?
– Gerry Myerson
Jul 16 at 7:30




It is, in fact, extremely easy to find a root of $x^2+x+1$ for $p=31$. What squares are closest to $31$?
– Gerry Myerson
Jul 16 at 7:30












@JyrkiLahtonen Regarding the first part, why can we ignore that a root of $x^2+x+1$ may have order one? For the second part, I got that $(x^2+x+1)-(x+2)(x-1)=3$. Did you mean that from this I can conclude that since $3$ is nonzero in either field and any non-zero element is invertible, hence $(x-1)+(x^2+x+1)=mathbb F_p[x]$?
– user437309
Jul 16 at 15:59




@JyrkiLahtonen Regarding the first part, why can we ignore that a root of $x^2+x+1$ may have order one? For the second part, I got that $(x^2+x+1)-(x+2)(x-1)=3$. Did you mean that from this I can conclude that since $3$ is nonzero in either field and any non-zero element is invertible, hence $(x-1)+(x^2+x+1)=mathbb F_p[x]$?
– user437309
Jul 16 at 15:59












I guess we can ignore that a root of $x^2+x+1$ may have order one because if it does, then $x=1$, but $1+1+1ne 3$ unless $p=3$.
– user437309
Jul 16 at 19:04




I guess we can ignore that a root of $x^2+x+1$ may have order one because if it does, then $x=1$, but $1+1+1ne 3$ unless $p=3$.
– user437309
Jul 16 at 19:04















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