A polynomial that is reducible under every finite field?












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Q: Prove that for any finite field $mathbb{F_q}$, the ring $mathbb{F_q}[x]/(x^9+x^5+x^3+x+1)$ cannot be a field.



Upon first glance I am really not sure where to start. Intuitively, it seems that I should be finding a way to show that $x^9+x^5+x^3+x+1$ is reducible over every finite field.



I have that:
$x^9+x^5+x^3+x+1 = (x^2-x+1)(x^7+x^6-x^4+x^2+2x+1)$ but I don't see where to go from here.



Am I even on the right track? Is there some well-known theorem I've blanked on that would help me out? Thanks in advance!






abstract-algebra ring-theory field-theory finite-fields extension-field







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  • 4




    $begingroup$
    You just factored it over $mathbb{Z}$. You are done. If it's reducible over $mathbb{Z}$ then it's reducible over $mathbb{F}_p$ for every prime $p$.
    $endgroup$
    – Ethan Alwaise
    Dec 9 '16 at 7:50






  • 4




    $begingroup$
    @EthanAlwaise: well, one must be a little careful: $3X^{2}+X$ is reducible as $X(3X+1)$ over $mathbb{Z}$, but is irreducible over $mathbb{F}_{3}$. But since both of the factors in the product above are monic, there are no issues of this kind.
    $endgroup$
    – Alex Wertheim
    Dec 9 '16 at 7:57






  • 1




    $begingroup$
    Ah right, good point.
    $endgroup$
    – Ethan Alwaise
    Dec 9 '16 at 19:23
















1












$begingroup$


Q: Prove that for any finite field $mathbb{F_q}$, the ring $mathbb{F_q}[x]/(x^9+x^5+x^3+x+1)$ cannot be a field.



Upon first glance I am really not sure where to start. Intuitively, it seems that I should be finding a way to show that $x^9+x^5+x^3+x+1$ is reducible over every finite field.



I have that:
$x^9+x^5+x^3+x+1 = (x^2-x+1)(x^7+x^6-x^4+x^2+2x+1)$ but I don't see where to go from here.



Am I even on the right track? Is there some well-known theorem I've blanked on that would help me out? Thanks in advance!






abstract-algebra ring-theory field-theory finite-fields extension-field







share|cite|improve this question









$endgroup$








  • 4




    $begingroup$
    You just factored it over $mathbb{Z}$. You are done. If it's reducible over $mathbb{Z}$ then it's reducible over $mathbb{F}_p$ for every prime $p$.
    $endgroup$
    – Ethan Alwaise
    Dec 9 '16 at 7:50






  • 4




    $begingroup$
    @EthanAlwaise: well, one must be a little careful: $3X^{2}+X$ is reducible as $X(3X+1)$ over $mathbb{Z}$, but is irreducible over $mathbb{F}_{3}$. But since both of the factors in the product above are monic, there are no issues of this kind.
    $endgroup$
    – Alex Wertheim
    Dec 9 '16 at 7:57






  • 1




    $begingroup$
    Ah right, good point.
    $endgroup$
    – Ethan Alwaise
    Dec 9 '16 at 19:23














1












1








1





$begingroup$


Q: Prove that for any finite field $mathbb{F_q}$, the ring $mathbb{F_q}[x]/(x^9+x^5+x^3+x+1)$ cannot be a field.



Upon first glance I am really not sure where to start. Intuitively, it seems that I should be finding a way to show that $x^9+x^5+x^3+x+1$ is reducible over every finite field.



I have that:
$x^9+x^5+x^3+x+1 = (x^2-x+1)(x^7+x^6-x^4+x^2+2x+1)$ but I don't see where to go from here.



Am I even on the right track? Is there some well-known theorem I've blanked on that would help me out? Thanks in advance!






abstract-algebra ring-theory field-theory finite-fields extension-field







share|cite|improve this question









$endgroup$




Q: Prove that for any finite field $mathbb{F_q}$, the ring $mathbb{F_q}[x]/(x^9+x^5+x^3+x+1)$ cannot be a field.



Upon first glance I am really not sure where to start. Intuitively, it seems that I should be finding a way to show that $x^9+x^5+x^3+x+1$ is reducible over every finite field.



I have that:
$x^9+x^5+x^3+x+1 = (x^2-x+1)(x^7+x^6-x^4+x^2+2x+1)$ but I don't see where to go from here.



Am I even on the right track? Is there some well-known theorem I've blanked on that would help me out? Thanks in advance!






abstract-algebra ring-theory field-theory finite-fields extension-field




abstract-algebra ring-theory field-theory finite-fields extension-field






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




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asked Dec 9 '16 at 7:45









Jane DoeJane Doe

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  • 4




    $begingroup$
    You just factored it over $mathbb{Z}$. You are done. If it's reducible over $mathbb{Z}$ then it's reducible over $mathbb{F}_p$ for every prime $p$.
    $endgroup$
    – Ethan Alwaise
    Dec 9 '16 at 7:50






  • 4




    $begingroup$
    @EthanAlwaise: well, one must be a little careful: $3X^{2}+X$ is reducible as $X(3X+1)$ over $mathbb{Z}$, but is irreducible over $mathbb{F}_{3}$. But since both of the factors in the product above are monic, there are no issues of this kind.
    $endgroup$
    – Alex Wertheim
    Dec 9 '16 at 7:57






  • 1




    $begingroup$
    Ah right, good point.
    $endgroup$
    – Ethan Alwaise
    Dec 9 '16 at 19:23














  • 4




    $begingroup$
    You just factored it over $mathbb{Z}$. You are done. If it's reducible over $mathbb{Z}$ then it's reducible over $mathbb{F}_p$ for every prime $p$.
    $endgroup$
    – Ethan Alwaise
    Dec 9 '16 at 7:50






  • 4




    $begingroup$
    @EthanAlwaise: well, one must be a little careful: $3X^{2}+X$ is reducible as $X(3X+1)$ over $mathbb{Z}$, but is irreducible over $mathbb{F}_{3}$. But since both of the factors in the product above are monic, there are no issues of this kind.
    $endgroup$
    – Alex Wertheim
    Dec 9 '16 at 7:57






  • 1




    $begingroup$
    Ah right, good point.
    $endgroup$
    – Ethan Alwaise
    Dec 9 '16 at 19:23








4




4




$begingroup$
You just factored it over $mathbb{Z}$. You are done. If it's reducible over $mathbb{Z}$ then it's reducible over $mathbb{F}_p$ for every prime $p$.
$endgroup$
– Ethan Alwaise
Dec 9 '16 at 7:50




$begingroup$
You just factored it over $mathbb{Z}$. You are done. If it's reducible over $mathbb{Z}$ then it's reducible over $mathbb{F}_p$ for every prime $p$.
$endgroup$
– Ethan Alwaise
Dec 9 '16 at 7:50




4




4




$begingroup$
@EthanAlwaise: well, one must be a little careful: $3X^{2}+X$ is reducible as $X(3X+1)$ over $mathbb{Z}$, but is irreducible over $mathbb{F}_{3}$. But since both of the factors in the product above are monic, there are no issues of this kind.
$endgroup$
– Alex Wertheim
Dec 9 '16 at 7:57




$begingroup$
@EthanAlwaise: well, one must be a little careful: $3X^{2}+X$ is reducible as $X(3X+1)$ over $mathbb{Z}$, but is irreducible over $mathbb{F}_{3}$. But since both of the factors in the product above are monic, there are no issues of this kind.
$endgroup$
– Alex Wertheim
Dec 9 '16 at 7:57




1




1




$begingroup$
Ah right, good point.
$endgroup$
– Ethan Alwaise
Dec 9 '16 at 19:23




$begingroup$
Ah right, good point.
$endgroup$
– Ethan Alwaise
Dec 9 '16 at 19:23










1 Answer
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From the comments above.




If $f(x) = x^9 + x^5 + x^3 + x + 1 in mathbb{Z}[x]$, then you have shown that $f = gh$, where $g(x) = x^2 - x + 1 in mathbb{Z}[x]$ and $h(x) = x^7 + x^6 - x^4 + x^2 + 2x + 1 in mathbb{Z}[x]$. Let $p$ be a prime and $bar{f}$, $bar{g}$ and $bar{h}$ denote the images of $f$, $g$ and $h$, respectively, under the canonical map $mathbb{Z} to mathbb{F}_p$. Then, $bar{f} = bar{g} bar{h}$ and $bar{g}$ and $bar{h}$ are nonconstant polynomials in $mathbb{F}_p[x]$. Hence, $bar{f}$ is irreducible over every finite field, as was needed to be shown.






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    $begingroup$

    From the comments above.




    If $f(x) = x^9 + x^5 + x^3 + x + 1 in mathbb{Z}[x]$, then you have shown that $f = gh$, where $g(x) = x^2 - x + 1 in mathbb{Z}[x]$ and $h(x) = x^7 + x^6 - x^4 + x^2 + 2x + 1 in mathbb{Z}[x]$. Let $p$ be a prime and $bar{f}$, $bar{g}$ and $bar{h}$ denote the images of $f$, $g$ and $h$, respectively, under the canonical map $mathbb{Z} to mathbb{F}_p$. Then, $bar{f} = bar{g} bar{h}$ and $bar{g}$ and $bar{h}$ are nonconstant polynomials in $mathbb{F}_p[x]$. Hence, $bar{f}$ is irreducible over every finite field, as was needed to be shown.






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      2












      $begingroup$

      From the comments above.




      If $f(x) = x^9 + x^5 + x^3 + x + 1 in mathbb{Z}[x]$, then you have shown that $f = gh$, where $g(x) = x^2 - x + 1 in mathbb{Z}[x]$ and $h(x) = x^7 + x^6 - x^4 + x^2 + 2x + 1 in mathbb{Z}[x]$. Let $p$ be a prime and $bar{f}$, $bar{g}$ and $bar{h}$ denote the images of $f$, $g$ and $h$, respectively, under the canonical map $mathbb{Z} to mathbb{F}_p$. Then, $bar{f} = bar{g} bar{h}$ and $bar{g}$ and $bar{h}$ are nonconstant polynomials in $mathbb{F}_p[x]$. Hence, $bar{f}$ is irreducible over every finite field, as was needed to be shown.






      share|cite|improve this answer











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        $begingroup$

        From the comments above.




        If $f(x) = x^9 + x^5 + x^3 + x + 1 in mathbb{Z}[x]$, then you have shown that $f = gh$, where $g(x) = x^2 - x + 1 in mathbb{Z}[x]$ and $h(x) = x^7 + x^6 - x^4 + x^2 + 2x + 1 in mathbb{Z}[x]$. Let $p$ be a prime and $bar{f}$, $bar{g}$ and $bar{h}$ denote the images of $f$, $g$ and $h$, respectively, under the canonical map $mathbb{Z} to mathbb{F}_p$. Then, $bar{f} = bar{g} bar{h}$ and $bar{g}$ and $bar{h}$ are nonconstant polynomials in $mathbb{F}_p[x]$. Hence, $bar{f}$ is irreducible over every finite field, as was needed to be shown.






        share|cite|improve this answer











        $endgroup$



        From the comments above.




        If $f(x) = x^9 + x^5 + x^3 + x + 1 in mathbb{Z}[x]$, then you have shown that $f = gh$, where $g(x) = x^2 - x + 1 in mathbb{Z}[x]$ and $h(x) = x^7 + x^6 - x^4 + x^2 + 2x + 1 in mathbb{Z}[x]$. Let $p$ be a prime and $bar{f}$, $bar{g}$ and $bar{h}$ denote the images of $f$, $g$ and $h$, respectively, under the canonical map $mathbb{Z} to mathbb{F}_p$. Then, $bar{f} = bar{g} bar{h}$ and $bar{g}$ and $bar{h}$ are nonconstant polynomials in $mathbb{F}_p[x]$. Hence, $bar{f}$ is irreducible over every finite field, as was needed to be shown.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        answered Nov 30 '18 at 10:44


























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