Constructing a field with exactly $81$ elements












4












$begingroup$


I was thinking $frac{mathbb{Z_3}[x]}{(x^3+x+1)} times frac{mathbb{Z_3}[x]}{(x^3+x+1)}$.



$(x^3+x+1)$ is irreducible in $mathbb{Z_3[x]}$ so the quotient will be a field, and a field cross a field looks like it should be a field to me!



How would I do this if working over $mathbb{Z_9}$ and $mathbb{Z_2}$? in $mathbb{Z_9}$ i would have to find an irreducible quadratic, and then quotient out by that, correct? Thanks in advance!!






abstract-algebra number-theory field-theory ideals







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$endgroup$








  • 12




    $begingroup$
    A "field cross a field" is not a field. It can have zero divisors.
    $endgroup$
    – David Peterson
    Sep 30 '18 at 0:11








  • 2




    $begingroup$
    That can’t be a field: a direct product of two rings always has zero divisors: $(a,0)*(0,a) = (0,0)$, even when $aneq 0$. And you can’t do it working over $mathbb{Z}_9$ either, because that has zero divisors.
    $endgroup$
    – Arturo Magidin
    Sep 30 '18 at 0:11








  • 2




    $begingroup$
    Since $81 = 3^4$, try to find an irreducible polynomial of degree 4. Also, notice that $Rtimes S$ has $|R|*|S|$ elements; your cross product has $27times 27 = 729$ elements, not 81.
    $endgroup$
    – Arturo Magidin
    Sep 30 '18 at 0:13












  • $begingroup$
    Is there a process for finding an irreducible polynomial over a given modulus or do you just try things ?? Seems like it would be really tough to find an a degree 4 irreducible polynomial since checking for roots won't be enough
    $endgroup$
    – Math is hard
    Sep 30 '18 at 0:15










  • $begingroup$
    $mathbb Z_9$ is not a field.
    $endgroup$
    – Thomas Andrews
    Sep 30 '18 at 0:22
















4












$begingroup$


I was thinking $frac{mathbb{Z_3}[x]}{(x^3+x+1)} times frac{mathbb{Z_3}[x]}{(x^3+x+1)}$.



$(x^3+x+1)$ is irreducible in $mathbb{Z_3[x]}$ so the quotient will be a field, and a field cross a field looks like it should be a field to me!



How would I do this if working over $mathbb{Z_9}$ and $mathbb{Z_2}$? in $mathbb{Z_9}$ i would have to find an irreducible quadratic, and then quotient out by that, correct? Thanks in advance!!






abstract-algebra number-theory field-theory ideals







share|cite|improve this question











$endgroup$








  • 12




    $begingroup$
    A "field cross a field" is not a field. It can have zero divisors.
    $endgroup$
    – David Peterson
    Sep 30 '18 at 0:11








  • 2




    $begingroup$
    That can’t be a field: a direct product of two rings always has zero divisors: $(a,0)*(0,a) = (0,0)$, even when $aneq 0$. And you can’t do it working over $mathbb{Z}_9$ either, because that has zero divisors.
    $endgroup$
    – Arturo Magidin
    Sep 30 '18 at 0:11








  • 2




    $begingroup$
    Since $81 = 3^4$, try to find an irreducible polynomial of degree 4. Also, notice that $Rtimes S$ has $|R|*|S|$ elements; your cross product has $27times 27 = 729$ elements, not 81.
    $endgroup$
    – Arturo Magidin
    Sep 30 '18 at 0:13












  • $begingroup$
    Is there a process for finding an irreducible polynomial over a given modulus or do you just try things ?? Seems like it would be really tough to find an a degree 4 irreducible polynomial since checking for roots won't be enough
    $endgroup$
    – Math is hard
    Sep 30 '18 at 0:15










  • $begingroup$
    $mathbb Z_9$ is not a field.
    $endgroup$
    – Thomas Andrews
    Sep 30 '18 at 0:22














4












4








4


1



$begingroup$


I was thinking $frac{mathbb{Z_3}[x]}{(x^3+x+1)} times frac{mathbb{Z_3}[x]}{(x^3+x+1)}$.



$(x^3+x+1)$ is irreducible in $mathbb{Z_3[x]}$ so the quotient will be a field, and a field cross a field looks like it should be a field to me!



How would I do this if working over $mathbb{Z_9}$ and $mathbb{Z_2}$? in $mathbb{Z_9}$ i would have to find an irreducible quadratic, and then quotient out by that, correct? Thanks in advance!!






abstract-algebra number-theory field-theory ideals







share|cite|improve this question











$endgroup$




I was thinking $frac{mathbb{Z_3}[x]}{(x^3+x+1)} times frac{mathbb{Z_3}[x]}{(x^3+x+1)}$.



$(x^3+x+1)$ is irreducible in $mathbb{Z_3[x]}$ so the quotient will be a field, and a field cross a field looks like it should be a field to me!



How would I do this if working over $mathbb{Z_9}$ and $mathbb{Z_2}$? in $mathbb{Z_9}$ i would have to find an irreducible quadratic, and then quotient out by that, correct? Thanks in advance!!






abstract-algebra number-theory field-theory ideals




abstract-algebra number-theory field-theory ideals






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Sep 30 '18 at 6:15









user21820

40.1k544161




40.1k544161










asked Sep 30 '18 at 0:07









Math is hardMath is hard

822211




822211








  • 12




    $begingroup$
    A "field cross a field" is not a field. It can have zero divisors.
    $endgroup$
    – David Peterson
    Sep 30 '18 at 0:11








  • 2




    $begingroup$
    That can’t be a field: a direct product of two rings always has zero divisors: $(a,0)*(0,a) = (0,0)$, even when $aneq 0$. And you can’t do it working over $mathbb{Z}_9$ either, because that has zero divisors.
    $endgroup$
    – Arturo Magidin
    Sep 30 '18 at 0:11








  • 2




    $begingroup$
    Since $81 = 3^4$, try to find an irreducible polynomial of degree 4. Also, notice that $Rtimes S$ has $|R|*|S|$ elements; your cross product has $27times 27 = 729$ elements, not 81.
    $endgroup$
    – Arturo Magidin
    Sep 30 '18 at 0:13












  • $begingroup$
    Is there a process for finding an irreducible polynomial over a given modulus or do you just try things ?? Seems like it would be really tough to find an a degree 4 irreducible polynomial since checking for roots won't be enough
    $endgroup$
    – Math is hard
    Sep 30 '18 at 0:15










  • $begingroup$
    $mathbb Z_9$ is not a field.
    $endgroup$
    – Thomas Andrews
    Sep 30 '18 at 0:22














  • 12




    $begingroup$
    A "field cross a field" is not a field. It can have zero divisors.
    $endgroup$
    – David Peterson
    Sep 30 '18 at 0:11








  • 2




    $begingroup$
    That can’t be a field: a direct product of two rings always has zero divisors: $(a,0)*(0,a) = (0,0)$, even when $aneq 0$. And you can’t do it working over $mathbb{Z}_9$ either, because that has zero divisors.
    $endgroup$
    – Arturo Magidin
    Sep 30 '18 at 0:11








  • 2




    $begingroup$
    Since $81 = 3^4$, try to find an irreducible polynomial of degree 4. Also, notice that $Rtimes S$ has $|R|*|S|$ elements; your cross product has $27times 27 = 729$ elements, not 81.
    $endgroup$
    – Arturo Magidin
    Sep 30 '18 at 0:13












  • $begingroup$
    Is there a process for finding an irreducible polynomial over a given modulus or do you just try things ?? Seems like it would be really tough to find an a degree 4 irreducible polynomial since checking for roots won't be enough
    $endgroup$
    – Math is hard
    Sep 30 '18 at 0:15










  • $begingroup$
    $mathbb Z_9$ is not a field.
    $endgroup$
    – Thomas Andrews
    Sep 30 '18 at 0:22








12




12




$begingroup$
A "field cross a field" is not a field. It can have zero divisors.
$endgroup$
– David Peterson
Sep 30 '18 at 0:11






$begingroup$
A "field cross a field" is not a field. It can have zero divisors.
$endgroup$
– David Peterson
Sep 30 '18 at 0:11






2




2




$begingroup$
That can’t be a field: a direct product of two rings always has zero divisors: $(a,0)*(0,a) = (0,0)$, even when $aneq 0$. And you can’t do it working over $mathbb{Z}_9$ either, because that has zero divisors.
$endgroup$
– Arturo Magidin
Sep 30 '18 at 0:11






$begingroup$
That can’t be a field: a direct product of two rings always has zero divisors: $(a,0)*(0,a) = (0,0)$, even when $aneq 0$. And you can’t do it working over $mathbb{Z}_9$ either, because that has zero divisors.
$endgroup$
– Arturo Magidin
Sep 30 '18 at 0:11






2




2




$begingroup$
Since $81 = 3^4$, try to find an irreducible polynomial of degree 4. Also, notice that $Rtimes S$ has $|R|*|S|$ elements; your cross product has $27times 27 = 729$ elements, not 81.
$endgroup$
– Arturo Magidin
Sep 30 '18 at 0:13






$begingroup$
Since $81 = 3^4$, try to find an irreducible polynomial of degree 4. Also, notice that $Rtimes S$ has $|R|*|S|$ elements; your cross product has $27times 27 = 729$ elements, not 81.
$endgroup$
– Arturo Magidin
Sep 30 '18 at 0:13














$begingroup$
Is there a process for finding an irreducible polynomial over a given modulus or do you just try things ?? Seems like it would be really tough to find an a degree 4 irreducible polynomial since checking for roots won't be enough
$endgroup$
– Math is hard
Sep 30 '18 at 0:15




$begingroup$
Is there a process for finding an irreducible polynomial over a given modulus or do you just try things ?? Seems like it would be really tough to find an a degree 4 irreducible polynomial since checking for roots won't be enough
$endgroup$
– Math is hard
Sep 30 '18 at 0:15












$begingroup$
$mathbb Z_9$ is not a field.
$endgroup$
– Thomas Andrews
Sep 30 '18 at 0:22




$begingroup$
$mathbb Z_9$ is not a field.
$endgroup$
– Thomas Andrews
Sep 30 '18 at 0:22










4 Answers
4






active

oldest

votes


















4












$begingroup$

A field with $81$ elements must be the splitting field of $x^{81}-x$. It is also an extension of $mathbb F_3$ of degree $4$. Therefore, take any irreducible quartic factor of $x^{81}-x$, for instance, $x^4+x+2$. WA tells you all of them.






share|cite|improve this answer









$endgroup$









  • 3




    $begingroup$
    While this answer is technically correct, it feels incomplete without addressing the confusion about using $mathbb{Z}_{9}$ or $mathbb{Z}_{2}$ to build the field.
    $endgroup$
    – Morgan Rodgers
    Sep 30 '18 at 1:58



















3












$begingroup$

No you can't do $F times F$ because $(a,0) times (0,b) = (0,0)$ would break the field axioms.



You need to give an irreducible polynomial $P$ of degree $4$ over $mathbb{Z}_3$ so you can give representatives of $mathbb{Z}_3[x]/(P)$ by polynomials of degree $leq 3$. The coefficients of $1$, $x$ $x^2$ and $x^3$ are free to choose so there are a total of $3^4$ elements of the field as desired.



Getting $P$ is more complicated. If you have a guess for an irreducible polynomial check it with Rabin's test






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Seems like it would be tough to find a degree 4 irreducible polynomial over $mathbb{Z_3}$ because checking for roots won't be enough... got any tips?
    $endgroup$
    – Math is hard
    Sep 30 '18 at 0:16






  • 6




    $begingroup$
    @MichaelVaughan: Checking for roots isn’t enough; but then, if it is of degree $4$ and has no roots in $mathbb{Z}_3$, then the only way it can fail to be irreducible is if it is the product of two irreducible quadratics. There aren’t that many irreducible quadratics, so you can check what all the products of two of them are.
    $endgroup$
    – Arturo Magidin
    Sep 30 '18 at 0:21






  • 4




    $begingroup$
    The field extensions of degree 2, 3, 4, .. of $mathbb{Z}_3$ have 8, 26, 80, ... non-zero elements, so the degree 4 extension is the smallest one with elements of order 5 and the polynomial $frac{x^5-1}{x-1}=x^4+x^3+x^2+x+1$ must work.
    $endgroup$
    – random
    Sep 30 '18 at 0:46



















0












$begingroup$

Let's see. What are the irreducible quadratics: $x^2+x+2,,x^2+2x+2,,x^2+1$.



I count $3$ (thanks @Lubin).



So, there are ${3choose 2}+3=6$ combinations to check.



These are the $6$ quartics I get: $x^4+1,,x^4+x^3+x+1,,x^4+2x^3+2x+2 ,,x^4+2x^3+x^2+x+1,,x^4+x^3+x^2+2x+1,,x^4+2x^2+1$.



But there are $2cdot 3^4=162$ quartics to choose from... $81$ of them monic.



So, if I choose a quartic that doesn't have a root and isn't among the $6$ products, it will be irreducible.



So, how about $x^4+2x^3+2$?



Finally, $mathbb F_{3^4}congfrac{mathbb Z_3[X]}{x^4+2x^3+2}$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Man, to me it seems like determining all the irreducible quadratics over $mathbb{Z_3}$ would be tough, but you make it sound like it's so obvious. Did you just go through and check all of them to see which had roots? I guess that's not that bad
    $endgroup$
    – Math is hard
    Sep 30 '18 at 2:34










  • $begingroup$
    Yeah. I was just about to list the $10 $ reducible ones... Then we can choose one of the remaining $152$.
    $endgroup$
    – Chris Custer
    Sep 30 '18 at 2:37










  • $begingroup$
    Wait. I think I miscounted. I included a lot of quartics that have a root.
    $endgroup$
    – Chris Custer
    Sep 30 '18 at 2:48






  • 1




    $begingroup$
    But $2x^2+x+1=2(x^2+2x+2)$. Since there are only three pairs of conjugate points in $Bbb F_9$, there are only three irreducible quadratics over $Bbb F_3$. In fact, there are $81-9=72$ elements of $Bbb F_{81}$ that generate over $Bbb F_3$. These fall into $72/4=18$ quadruples of conjugate points. So there are $18$ irreducible quartics over $Bbb F_3$. It shouldn’t be hard to find one of them.
    $endgroup$
    – Lubin
    Sep 30 '18 at 3:04





















0












$begingroup$

If you don't want to mess around with finding an irreducible quartic, you could construct $mathbb{F}_{81}$ as the quotient of $mathbb{F}_9[x]$ with a quadratic irreducible in $mathbb{F}_9$. $mathbb{F}_9$ can be constructed as the quotient of $mathbb{F}_3[y]$ with a quadratic irreducible in $mathbb{F}_3$, so you just have to find two different irreducible quadratics, which is easy since you just have to verify the quadratic has no roots in the field.



Expressing $mathbb{F}_9congmathbb{F}_3[y]/(y^2+1)$, one can check that $x^2+(y+1)$ has no roots in $mathbb{F}_9$, so $mathbb{F}_{81}congmathbb{F}_9[x]/(x^2+y+1)$.






share|cite|improve this answer









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    4 Answers
    4






    active

    oldest

    votes








    4 Answers
    4






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    4












    $begingroup$

    A field with $81$ elements must be the splitting field of $x^{81}-x$. It is also an extension of $mathbb F_3$ of degree $4$. Therefore, take any irreducible quartic factor of $x^{81}-x$, for instance, $x^4+x+2$. WA tells you all of them.






    share|cite|improve this answer









    $endgroup$









    • 3




      $begingroup$
      While this answer is technically correct, it feels incomplete without addressing the confusion about using $mathbb{Z}_{9}$ or $mathbb{Z}_{2}$ to build the field.
      $endgroup$
      – Morgan Rodgers
      Sep 30 '18 at 1:58
















    4












    $begingroup$

    A field with $81$ elements must be the splitting field of $x^{81}-x$. It is also an extension of $mathbb F_3$ of degree $4$. Therefore, take any irreducible quartic factor of $x^{81}-x$, for instance, $x^4+x+2$. WA tells you all of them.






    share|cite|improve this answer









    $endgroup$









    • 3




      $begingroup$
      While this answer is technically correct, it feels incomplete without addressing the confusion about using $mathbb{Z}_{9}$ or $mathbb{Z}_{2}$ to build the field.
      $endgroup$
      – Morgan Rodgers
      Sep 30 '18 at 1:58














    4












    4








    4





    $begingroup$

    A field with $81$ elements must be the splitting field of $x^{81}-x$. It is also an extension of $mathbb F_3$ of degree $4$. Therefore, take any irreducible quartic factor of $x^{81}-x$, for instance, $x^4+x+2$. WA tells you all of them.






    share|cite|improve this answer









    $endgroup$



    A field with $81$ elements must be the splitting field of $x^{81}-x$. It is also an extension of $mathbb F_3$ of degree $4$. Therefore, take any irreducible quartic factor of $x^{81}-x$, for instance, $x^4+x+2$. WA tells you all of them.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Sep 30 '18 at 0:53









    lhflhf

    167k11172404




    167k11172404








    • 3




      $begingroup$
      While this answer is technically correct, it feels incomplete without addressing the confusion about using $mathbb{Z}_{9}$ or $mathbb{Z}_{2}$ to build the field.
      $endgroup$
      – Morgan Rodgers
      Sep 30 '18 at 1:58














    • 3




      $begingroup$
      While this answer is technically correct, it feels incomplete without addressing the confusion about using $mathbb{Z}_{9}$ or $mathbb{Z}_{2}$ to build the field.
      $endgroup$
      – Morgan Rodgers
      Sep 30 '18 at 1:58








    3




    3




    $begingroup$
    While this answer is technically correct, it feels incomplete without addressing the confusion about using $mathbb{Z}_{9}$ or $mathbb{Z}_{2}$ to build the field.
    $endgroup$
    – Morgan Rodgers
    Sep 30 '18 at 1:58




    $begingroup$
    While this answer is technically correct, it feels incomplete without addressing the confusion about using $mathbb{Z}_{9}$ or $mathbb{Z}_{2}$ to build the field.
    $endgroup$
    – Morgan Rodgers
    Sep 30 '18 at 1:58











    3












    $begingroup$

    No you can't do $F times F$ because $(a,0) times (0,b) = (0,0)$ would break the field axioms.



    You need to give an irreducible polynomial $P$ of degree $4$ over $mathbb{Z}_3$ so you can give representatives of $mathbb{Z}_3[x]/(P)$ by polynomials of degree $leq 3$. The coefficients of $1$, $x$ $x^2$ and $x^3$ are free to choose so there are a total of $3^4$ elements of the field as desired.



    Getting $P$ is more complicated. If you have a guess for an irreducible polynomial check it with Rabin's test






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Seems like it would be tough to find a degree 4 irreducible polynomial over $mathbb{Z_3}$ because checking for roots won't be enough... got any tips?
      $endgroup$
      – Math is hard
      Sep 30 '18 at 0:16






    • 6




      $begingroup$
      @MichaelVaughan: Checking for roots isn’t enough; but then, if it is of degree $4$ and has no roots in $mathbb{Z}_3$, then the only way it can fail to be irreducible is if it is the product of two irreducible quadratics. There aren’t that many irreducible quadratics, so you can check what all the products of two of them are.
      $endgroup$
      – Arturo Magidin
      Sep 30 '18 at 0:21






    • 4




      $begingroup$
      The field extensions of degree 2, 3, 4, .. of $mathbb{Z}_3$ have 8, 26, 80, ... non-zero elements, so the degree 4 extension is the smallest one with elements of order 5 and the polynomial $frac{x^5-1}{x-1}=x^4+x^3+x^2+x+1$ must work.
      $endgroup$
      – random
      Sep 30 '18 at 0:46
















    3












    $begingroup$

    No you can't do $F times F$ because $(a,0) times (0,b) = (0,0)$ would break the field axioms.



    You need to give an irreducible polynomial $P$ of degree $4$ over $mathbb{Z}_3$ so you can give representatives of $mathbb{Z}_3[x]/(P)$ by polynomials of degree $leq 3$. The coefficients of $1$, $x$ $x^2$ and $x^3$ are free to choose so there are a total of $3^4$ elements of the field as desired.



    Getting $P$ is more complicated. If you have a guess for an irreducible polynomial check it with Rabin's test






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Seems like it would be tough to find a degree 4 irreducible polynomial over $mathbb{Z_3}$ because checking for roots won't be enough... got any tips?
      $endgroup$
      – Math is hard
      Sep 30 '18 at 0:16






    • 6




      $begingroup$
      @MichaelVaughan: Checking for roots isn’t enough; but then, if it is of degree $4$ and has no roots in $mathbb{Z}_3$, then the only way it can fail to be irreducible is if it is the product of two irreducible quadratics. There aren’t that many irreducible quadratics, so you can check what all the products of two of them are.
      $endgroup$
      – Arturo Magidin
      Sep 30 '18 at 0:21






    • 4




      $begingroup$
      The field extensions of degree 2, 3, 4, .. of $mathbb{Z}_3$ have 8, 26, 80, ... non-zero elements, so the degree 4 extension is the smallest one with elements of order 5 and the polynomial $frac{x^5-1}{x-1}=x^4+x^3+x^2+x+1$ must work.
      $endgroup$
      – random
      Sep 30 '18 at 0:46














    3












    3








    3





    $begingroup$

    No you can't do $F times F$ because $(a,0) times (0,b) = (0,0)$ would break the field axioms.



    You need to give an irreducible polynomial $P$ of degree $4$ over $mathbb{Z}_3$ so you can give representatives of $mathbb{Z}_3[x]/(P)$ by polynomials of degree $leq 3$. The coefficients of $1$, $x$ $x^2$ and $x^3$ are free to choose so there are a total of $3^4$ elements of the field as desired.



    Getting $P$ is more complicated. If you have a guess for an irreducible polynomial check it with Rabin's test






    share|cite|improve this answer











    $endgroup$



    No you can't do $F times F$ because $(a,0) times (0,b) = (0,0)$ would break the field axioms.



    You need to give an irreducible polynomial $P$ of degree $4$ over $mathbb{Z}_3$ so you can give representatives of $mathbb{Z}_3[x]/(P)$ by polynomials of degree $leq 3$. The coefficients of $1$, $x$ $x^2$ and $x^3$ are free to choose so there are a total of $3^4$ elements of the field as desired.



    Getting $P$ is more complicated. If you have a guess for an irreducible polynomial check it with Rabin's test







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Sep 30 '18 at 0:21

























    answered Sep 30 '18 at 0:14









    AHusainAHusain

    2,8532916




    2,8532916












    • $begingroup$
      Seems like it would be tough to find a degree 4 irreducible polynomial over $mathbb{Z_3}$ because checking for roots won't be enough... got any tips?
      $endgroup$
      – Math is hard
      Sep 30 '18 at 0:16






    • 6




      $begingroup$
      @MichaelVaughan: Checking for roots isn’t enough; but then, if it is of degree $4$ and has no roots in $mathbb{Z}_3$, then the only way it can fail to be irreducible is if it is the product of two irreducible quadratics. There aren’t that many irreducible quadratics, so you can check what all the products of two of them are.
      $endgroup$
      – Arturo Magidin
      Sep 30 '18 at 0:21






    • 4




      $begingroup$
      The field extensions of degree 2, 3, 4, .. of $mathbb{Z}_3$ have 8, 26, 80, ... non-zero elements, so the degree 4 extension is the smallest one with elements of order 5 and the polynomial $frac{x^5-1}{x-1}=x^4+x^3+x^2+x+1$ must work.
      $endgroup$
      – random
      Sep 30 '18 at 0:46


















    • $begingroup$
      Seems like it would be tough to find a degree 4 irreducible polynomial over $mathbb{Z_3}$ because checking for roots won't be enough... got any tips?
      $endgroup$
      – Math is hard
      Sep 30 '18 at 0:16






    • 6




      $begingroup$
      @MichaelVaughan: Checking for roots isn’t enough; but then, if it is of degree $4$ and has no roots in $mathbb{Z}_3$, then the only way it can fail to be irreducible is if it is the product of two irreducible quadratics. There aren’t that many irreducible quadratics, so you can check what all the products of two of them are.
      $endgroup$
      – Arturo Magidin
      Sep 30 '18 at 0:21






    • 4




      $begingroup$
      The field extensions of degree 2, 3, 4, .. of $mathbb{Z}_3$ have 8, 26, 80, ... non-zero elements, so the degree 4 extension is the smallest one with elements of order 5 and the polynomial $frac{x^5-1}{x-1}=x^4+x^3+x^2+x+1$ must work.
      $endgroup$
      – random
      Sep 30 '18 at 0:46
















    $begingroup$
    Seems like it would be tough to find a degree 4 irreducible polynomial over $mathbb{Z_3}$ because checking for roots won't be enough... got any tips?
    $endgroup$
    – Math is hard
    Sep 30 '18 at 0:16




    $begingroup$
    Seems like it would be tough to find a degree 4 irreducible polynomial over $mathbb{Z_3}$ because checking for roots won't be enough... got any tips?
    $endgroup$
    – Math is hard
    Sep 30 '18 at 0:16




    6




    6




    $begingroup$
    @MichaelVaughan: Checking for roots isn’t enough; but then, if it is of degree $4$ and has no roots in $mathbb{Z}_3$, then the only way it can fail to be irreducible is if it is the product of two irreducible quadratics. There aren’t that many irreducible quadratics, so you can check what all the products of two of them are.
    $endgroup$
    – Arturo Magidin
    Sep 30 '18 at 0:21




    $begingroup$
    @MichaelVaughan: Checking for roots isn’t enough; but then, if it is of degree $4$ and has no roots in $mathbb{Z}_3$, then the only way it can fail to be irreducible is if it is the product of two irreducible quadratics. There aren’t that many irreducible quadratics, so you can check what all the products of two of them are.
    $endgroup$
    – Arturo Magidin
    Sep 30 '18 at 0:21




    4




    4




    $begingroup$
    The field extensions of degree 2, 3, 4, .. of $mathbb{Z}_3$ have 8, 26, 80, ... non-zero elements, so the degree 4 extension is the smallest one with elements of order 5 and the polynomial $frac{x^5-1}{x-1}=x^4+x^3+x^2+x+1$ must work.
    $endgroup$
    – random
    Sep 30 '18 at 0:46




    $begingroup$
    The field extensions of degree 2, 3, 4, .. of $mathbb{Z}_3$ have 8, 26, 80, ... non-zero elements, so the degree 4 extension is the smallest one with elements of order 5 and the polynomial $frac{x^5-1}{x-1}=x^4+x^3+x^2+x+1$ must work.
    $endgroup$
    – random
    Sep 30 '18 at 0:46











    0












    $begingroup$

    Let's see. What are the irreducible quadratics: $x^2+x+2,,x^2+2x+2,,x^2+1$.



    I count $3$ (thanks @Lubin).



    So, there are ${3choose 2}+3=6$ combinations to check.



    These are the $6$ quartics I get: $x^4+1,,x^4+x^3+x+1,,x^4+2x^3+2x+2 ,,x^4+2x^3+x^2+x+1,,x^4+x^3+x^2+2x+1,,x^4+2x^2+1$.



    But there are $2cdot 3^4=162$ quartics to choose from... $81$ of them monic.



    So, if I choose a quartic that doesn't have a root and isn't among the $6$ products, it will be irreducible.



    So, how about $x^4+2x^3+2$?



    Finally, $mathbb F_{3^4}congfrac{mathbb Z_3[X]}{x^4+2x^3+2}$.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Man, to me it seems like determining all the irreducible quadratics over $mathbb{Z_3}$ would be tough, but you make it sound like it's so obvious. Did you just go through and check all of them to see which had roots? I guess that's not that bad
      $endgroup$
      – Math is hard
      Sep 30 '18 at 2:34










    • $begingroup$
      Yeah. I was just about to list the $10 $ reducible ones... Then we can choose one of the remaining $152$.
      $endgroup$
      – Chris Custer
      Sep 30 '18 at 2:37










    • $begingroup$
      Wait. I think I miscounted. I included a lot of quartics that have a root.
      $endgroup$
      – Chris Custer
      Sep 30 '18 at 2:48






    • 1




      $begingroup$
      But $2x^2+x+1=2(x^2+2x+2)$. Since there are only three pairs of conjugate points in $Bbb F_9$, there are only three irreducible quadratics over $Bbb F_3$. In fact, there are $81-9=72$ elements of $Bbb F_{81}$ that generate over $Bbb F_3$. These fall into $72/4=18$ quadruples of conjugate points. So there are $18$ irreducible quartics over $Bbb F_3$. It shouldn’t be hard to find one of them.
      $endgroup$
      – Lubin
      Sep 30 '18 at 3:04


















    0












    $begingroup$

    Let's see. What are the irreducible quadratics: $x^2+x+2,,x^2+2x+2,,x^2+1$.



    I count $3$ (thanks @Lubin).



    So, there are ${3choose 2}+3=6$ combinations to check.



    These are the $6$ quartics I get: $x^4+1,,x^4+x^3+x+1,,x^4+2x^3+2x+2 ,,x^4+2x^3+x^2+x+1,,x^4+x^3+x^2+2x+1,,x^4+2x^2+1$.



    But there are $2cdot 3^4=162$ quartics to choose from... $81$ of them monic.



    So, if I choose a quartic that doesn't have a root and isn't among the $6$ products, it will be irreducible.



    So, how about $x^4+2x^3+2$?



    Finally, $mathbb F_{3^4}congfrac{mathbb Z_3[X]}{x^4+2x^3+2}$.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Man, to me it seems like determining all the irreducible quadratics over $mathbb{Z_3}$ would be tough, but you make it sound like it's so obvious. Did you just go through and check all of them to see which had roots? I guess that's not that bad
      $endgroup$
      – Math is hard
      Sep 30 '18 at 2:34










    • $begingroup$
      Yeah. I was just about to list the $10 $ reducible ones... Then we can choose one of the remaining $152$.
      $endgroup$
      – Chris Custer
      Sep 30 '18 at 2:37










    • $begingroup$
      Wait. I think I miscounted. I included a lot of quartics that have a root.
      $endgroup$
      – Chris Custer
      Sep 30 '18 at 2:48






    • 1




      $begingroup$
      But $2x^2+x+1=2(x^2+2x+2)$. Since there are only three pairs of conjugate points in $Bbb F_9$, there are only three irreducible quadratics over $Bbb F_3$. In fact, there are $81-9=72$ elements of $Bbb F_{81}$ that generate over $Bbb F_3$. These fall into $72/4=18$ quadruples of conjugate points. So there are $18$ irreducible quartics over $Bbb F_3$. It shouldn’t be hard to find one of them.
      $endgroup$
      – Lubin
      Sep 30 '18 at 3:04
















    0












    0








    0





    $begingroup$

    Let's see. What are the irreducible quadratics: $x^2+x+2,,x^2+2x+2,,x^2+1$.



    I count $3$ (thanks @Lubin).



    So, there are ${3choose 2}+3=6$ combinations to check.



    These are the $6$ quartics I get: $x^4+1,,x^4+x^3+x+1,,x^4+2x^3+2x+2 ,,x^4+2x^3+x^2+x+1,,x^4+x^3+x^2+2x+1,,x^4+2x^2+1$.



    But there are $2cdot 3^4=162$ quartics to choose from... $81$ of them monic.



    So, if I choose a quartic that doesn't have a root and isn't among the $6$ products, it will be irreducible.



    So, how about $x^4+2x^3+2$?



    Finally, $mathbb F_{3^4}congfrac{mathbb Z_3[X]}{x^4+2x^3+2}$.






    share|cite|improve this answer











    $endgroup$



    Let's see. What are the irreducible quadratics: $x^2+x+2,,x^2+2x+2,,x^2+1$.



    I count $3$ (thanks @Lubin).



    So, there are ${3choose 2}+3=6$ combinations to check.



    These are the $6$ quartics I get: $x^4+1,,x^4+x^3+x+1,,x^4+2x^3+2x+2 ,,x^4+2x^3+x^2+x+1,,x^4+x^3+x^2+2x+1,,x^4+2x^2+1$.



    But there are $2cdot 3^4=162$ quartics to choose from... $81$ of them monic.



    So, if I choose a quartic that doesn't have a root and isn't among the $6$ products, it will be irreducible.



    So, how about $x^4+2x^3+2$?



    Finally, $mathbb F_{3^4}congfrac{mathbb Z_3[X]}{x^4+2x^3+2}$.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Sep 30 '18 at 4:13

























    answered Sep 30 '18 at 1:05









    Chris CusterChris Custer

    14.3k3827




    14.3k3827












    • $begingroup$
      Man, to me it seems like determining all the irreducible quadratics over $mathbb{Z_3}$ would be tough, but you make it sound like it's so obvious. Did you just go through and check all of them to see which had roots? I guess that's not that bad
      $endgroup$
      – Math is hard
      Sep 30 '18 at 2:34










    • $begingroup$
      Yeah. I was just about to list the $10 $ reducible ones... Then we can choose one of the remaining $152$.
      $endgroup$
      – Chris Custer
      Sep 30 '18 at 2:37










    • $begingroup$
      Wait. I think I miscounted. I included a lot of quartics that have a root.
      $endgroup$
      – Chris Custer
      Sep 30 '18 at 2:48






    • 1




      $begingroup$
      But $2x^2+x+1=2(x^2+2x+2)$. Since there are only three pairs of conjugate points in $Bbb F_9$, there are only three irreducible quadratics over $Bbb F_3$. In fact, there are $81-9=72$ elements of $Bbb F_{81}$ that generate over $Bbb F_3$. These fall into $72/4=18$ quadruples of conjugate points. So there are $18$ irreducible quartics over $Bbb F_3$. It shouldn’t be hard to find one of them.
      $endgroup$
      – Lubin
      Sep 30 '18 at 3:04




















    • $begingroup$
      Man, to me it seems like determining all the irreducible quadratics over $mathbb{Z_3}$ would be tough, but you make it sound like it's so obvious. Did you just go through and check all of them to see which had roots? I guess that's not that bad
      $endgroup$
      – Math is hard
      Sep 30 '18 at 2:34










    • $begingroup$
      Yeah. I was just about to list the $10 $ reducible ones... Then we can choose one of the remaining $152$.
      $endgroup$
      – Chris Custer
      Sep 30 '18 at 2:37










    • $begingroup$
      Wait. I think I miscounted. I included a lot of quartics that have a root.
      $endgroup$
      – Chris Custer
      Sep 30 '18 at 2:48






    • 1




      $begingroup$
      But $2x^2+x+1=2(x^2+2x+2)$. Since there are only three pairs of conjugate points in $Bbb F_9$, there are only three irreducible quadratics over $Bbb F_3$. In fact, there are $81-9=72$ elements of $Bbb F_{81}$ that generate over $Bbb F_3$. These fall into $72/4=18$ quadruples of conjugate points. So there are $18$ irreducible quartics over $Bbb F_3$. It shouldn’t be hard to find one of them.
      $endgroup$
      – Lubin
      Sep 30 '18 at 3:04


















    $begingroup$
    Man, to me it seems like determining all the irreducible quadratics over $mathbb{Z_3}$ would be tough, but you make it sound like it's so obvious. Did you just go through and check all of them to see which had roots? I guess that's not that bad
    $endgroup$
    – Math is hard
    Sep 30 '18 at 2:34




    $begingroup$
    Man, to me it seems like determining all the irreducible quadratics over $mathbb{Z_3}$ would be tough, but you make it sound like it's so obvious. Did you just go through and check all of them to see which had roots? I guess that's not that bad
    $endgroup$
    – Math is hard
    Sep 30 '18 at 2:34












    $begingroup$
    Yeah. I was just about to list the $10 $ reducible ones... Then we can choose one of the remaining $152$.
    $endgroup$
    – Chris Custer
    Sep 30 '18 at 2:37




    $begingroup$
    Yeah. I was just about to list the $10 $ reducible ones... Then we can choose one of the remaining $152$.
    $endgroup$
    – Chris Custer
    Sep 30 '18 at 2:37












    $begingroup$
    Wait. I think I miscounted. I included a lot of quartics that have a root.
    $endgroup$
    – Chris Custer
    Sep 30 '18 at 2:48




    $begingroup$
    Wait. I think I miscounted. I included a lot of quartics that have a root.
    $endgroup$
    – Chris Custer
    Sep 30 '18 at 2:48




    1




    1




    $begingroup$
    But $2x^2+x+1=2(x^2+2x+2)$. Since there are only three pairs of conjugate points in $Bbb F_9$, there are only three irreducible quadratics over $Bbb F_3$. In fact, there are $81-9=72$ elements of $Bbb F_{81}$ that generate over $Bbb F_3$. These fall into $72/4=18$ quadruples of conjugate points. So there are $18$ irreducible quartics over $Bbb F_3$. It shouldn’t be hard to find one of them.
    $endgroup$
    – Lubin
    Sep 30 '18 at 3:04






    $begingroup$
    But $2x^2+x+1=2(x^2+2x+2)$. Since there are only three pairs of conjugate points in $Bbb F_9$, there are only three irreducible quadratics over $Bbb F_3$. In fact, there are $81-9=72$ elements of $Bbb F_{81}$ that generate over $Bbb F_3$. These fall into $72/4=18$ quadruples of conjugate points. So there are $18$ irreducible quartics over $Bbb F_3$. It shouldn’t be hard to find one of them.
    $endgroup$
    – Lubin
    Sep 30 '18 at 3:04













    0












    $begingroup$

    If you don't want to mess around with finding an irreducible quartic, you could construct $mathbb{F}_{81}$ as the quotient of $mathbb{F}_9[x]$ with a quadratic irreducible in $mathbb{F}_9$. $mathbb{F}_9$ can be constructed as the quotient of $mathbb{F}_3[y]$ with a quadratic irreducible in $mathbb{F}_3$, so you just have to find two different irreducible quadratics, which is easy since you just have to verify the quadratic has no roots in the field.



    Expressing $mathbb{F}_9congmathbb{F}_3[y]/(y^2+1)$, one can check that $x^2+(y+1)$ has no roots in $mathbb{F}_9$, so $mathbb{F}_{81}congmathbb{F}_9[x]/(x^2+y+1)$.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      If you don't want to mess around with finding an irreducible quartic, you could construct $mathbb{F}_{81}$ as the quotient of $mathbb{F}_9[x]$ with a quadratic irreducible in $mathbb{F}_9$. $mathbb{F}_9$ can be constructed as the quotient of $mathbb{F}_3[y]$ with a quadratic irreducible in $mathbb{F}_3$, so you just have to find two different irreducible quadratics, which is easy since you just have to verify the quadratic has no roots in the field.



      Expressing $mathbb{F}_9congmathbb{F}_3[y]/(y^2+1)$, one can check that $x^2+(y+1)$ has no roots in $mathbb{F}_9$, so $mathbb{F}_{81}congmathbb{F}_9[x]/(x^2+y+1)$.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        If you don't want to mess around with finding an irreducible quartic, you could construct $mathbb{F}_{81}$ as the quotient of $mathbb{F}_9[x]$ with a quadratic irreducible in $mathbb{F}_9$. $mathbb{F}_9$ can be constructed as the quotient of $mathbb{F}_3[y]$ with a quadratic irreducible in $mathbb{F}_3$, so you just have to find two different irreducible quadratics, which is easy since you just have to verify the quadratic has no roots in the field.



        Expressing $mathbb{F}_9congmathbb{F}_3[y]/(y^2+1)$, one can check that $x^2+(y+1)$ has no roots in $mathbb{F}_9$, so $mathbb{F}_{81}congmathbb{F}_9[x]/(x^2+y+1)$.






        share|cite|improve this answer









        $endgroup$



        If you don't want to mess around with finding an irreducible quartic, you could construct $mathbb{F}_{81}$ as the quotient of $mathbb{F}_9[x]$ with a quadratic irreducible in $mathbb{F}_9$. $mathbb{F}_9$ can be constructed as the quotient of $mathbb{F}_3[y]$ with a quadratic irreducible in $mathbb{F}_3$, so you just have to find two different irreducible quadratics, which is easy since you just have to verify the quadratic has no roots in the field.



        Expressing $mathbb{F}_9congmathbb{F}_3[y]/(y^2+1)$, one can check that $x^2+(y+1)$ has no roots in $mathbb{F}_9$, so $mathbb{F}_{81}congmathbb{F}_9[x]/(x^2+y+1)$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 5 at 18:38









        azaghalazaghal

        736




        736






























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