Element of given order in a finite field












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$begingroup$


Find an element of multiplicative order 4 and an element of order 5 in $F_{121}$ defined by $x^{2} +x +7$ ∈ $Z_{11}$.



The most obvious way to go about this seems to find a generator and raise it to a quarter the order of the field, thus producing an element that is equal to 1 when raised to the power of 4, according to an analog of Fermat's Little theorem. But since the polynomial's coefficients are over $Z_{11}$, I can't seem to find an obvious generator, and the reduction mod the quadratic seems cumbersome. Is there a more efficient elegant way to gleam elements of a desired order from this finite field?






abstract-algebra field-theory finite-fields







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    2












    $begingroup$


    Find an element of multiplicative order 4 and an element of order 5 in $F_{121}$ defined by $x^{2} +x +7$ ∈ $Z_{11}$.



    The most obvious way to go about this seems to find a generator and raise it to a quarter the order of the field, thus producing an element that is equal to 1 when raised to the power of 4, according to an analog of Fermat's Little theorem. But since the polynomial's coefficients are over $Z_{11}$, I can't seem to find an obvious generator, and the reduction mod the quadratic seems cumbersome. Is there a more efficient elegant way to gleam elements of a desired order from this finite field?






    abstract-algebra field-theory finite-fields







    share|cite|improve this question









    $endgroup$















      2












      2








      2





      $begingroup$


      Find an element of multiplicative order 4 and an element of order 5 in $F_{121}$ defined by $x^{2} +x +7$ ∈ $Z_{11}$.



      The most obvious way to go about this seems to find a generator and raise it to a quarter the order of the field, thus producing an element that is equal to 1 when raised to the power of 4, according to an analog of Fermat's Little theorem. But since the polynomial's coefficients are over $Z_{11}$, I can't seem to find an obvious generator, and the reduction mod the quadratic seems cumbersome. Is there a more efficient elegant way to gleam elements of a desired order from this finite field?






      abstract-algebra field-theory finite-fields







      share|cite|improve this question









      $endgroup$




      Find an element of multiplicative order 4 and an element of order 5 in $F_{121}$ defined by $x^{2} +x +7$ ∈ $Z_{11}$.



      The most obvious way to go about this seems to find a generator and raise it to a quarter the order of the field, thus producing an element that is equal to 1 when raised to the power of 4, according to an analog of Fermat's Little theorem. But since the polynomial's coefficients are over $Z_{11}$, I can't seem to find an obvious generator, and the reduction mod the quadratic seems cumbersome. Is there a more efficient elegant way to gleam elements of a desired order from this finite field?






      abstract-algebra field-theory finite-fields




      abstract-algebra field-theory finite-fields






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      user613048user613048

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          I don't think there's a magic shortcut to this kind of question in general, but the specific cases we have here are simple enough that there are some tricks. For an element of order $5$, we can note that $5$ divides $11-1$, so there is actually an element of order $5$ in $mathbb{F}_{11}$, which is easy to find by trial and error. A bit more systematically, we can see that the subgroup of squares in $mathbb{F}_{11}^*$ is cyclic of order $5$ so any square besides $0$ and $1$ must have order $5$.



          For an element of order $4$, we can save some work by noticing that an element of order $4$ is just a square root of $-1$. So, we just need to solve for $a,binmathbb{F}_{11}$ such that $(a+bx)^2=-1$. Expanding out $(a+bx)^2$ using $x^2+x+7=0$ we get $$(a^2-7b^2)+(2ab-b^2)x=-1$$ and so we need $$a^2-7b^2=-1$$ and $$2ab-b^2=0.$$ The second equation gives $b=0$ or $2a=b$. The first case does not work (since there is no square root of $-1$ in $mathbb{F}_7$); in the second case the first equation simplifies to $$5a^2=1$$ which we can easily solve in $mathbb{F}_{11}$ to get $a=pm 3$. So, the elements of order $4$ are $pm(3+6x)$.






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            $begingroup$

            I don't think there's a magic shortcut to this kind of question in general, but the specific cases we have here are simple enough that there are some tricks. For an element of order $5$, we can note that $5$ divides $11-1$, so there is actually an element of order $5$ in $mathbb{F}_{11}$, which is easy to find by trial and error. A bit more systematically, we can see that the subgroup of squares in $mathbb{F}_{11}^*$ is cyclic of order $5$ so any square besides $0$ and $1$ must have order $5$.



            For an element of order $4$, we can save some work by noticing that an element of order $4$ is just a square root of $-1$. So, we just need to solve for $a,binmathbb{F}_{11}$ such that $(a+bx)^2=-1$. Expanding out $(a+bx)^2$ using $x^2+x+7=0$ we get $$(a^2-7b^2)+(2ab-b^2)x=-1$$ and so we need $$a^2-7b^2=-1$$ and $$2ab-b^2=0.$$ The second equation gives $b=0$ or $2a=b$. The first case does not work (since there is no square root of $-1$ in $mathbb{F}_7$); in the second case the first equation simplifies to $$5a^2=1$$ which we can easily solve in $mathbb{F}_{11}$ to get $a=pm 3$. So, the elements of order $4$ are $pm(3+6x)$.






            share|cite|improve this answer









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              4












              $begingroup$

              I don't think there's a magic shortcut to this kind of question in general, but the specific cases we have here are simple enough that there are some tricks. For an element of order $5$, we can note that $5$ divides $11-1$, so there is actually an element of order $5$ in $mathbb{F}_{11}$, which is easy to find by trial and error. A bit more systematically, we can see that the subgroup of squares in $mathbb{F}_{11}^*$ is cyclic of order $5$ so any square besides $0$ and $1$ must have order $5$.



              For an element of order $4$, we can save some work by noticing that an element of order $4$ is just a square root of $-1$. So, we just need to solve for $a,binmathbb{F}_{11}$ such that $(a+bx)^2=-1$. Expanding out $(a+bx)^2$ using $x^2+x+7=0$ we get $$(a^2-7b^2)+(2ab-b^2)x=-1$$ and so we need $$a^2-7b^2=-1$$ and $$2ab-b^2=0.$$ The second equation gives $b=0$ or $2a=b$. The first case does not work (since there is no square root of $-1$ in $mathbb{F}_7$); in the second case the first equation simplifies to $$5a^2=1$$ which we can easily solve in $mathbb{F}_{11}$ to get $a=pm 3$. So, the elements of order $4$ are $pm(3+6x)$.






              share|cite|improve this answer









              $endgroup$
















                4












                4








                4





                $begingroup$

                I don't think there's a magic shortcut to this kind of question in general, but the specific cases we have here are simple enough that there are some tricks. For an element of order $5$, we can note that $5$ divides $11-1$, so there is actually an element of order $5$ in $mathbb{F}_{11}$, which is easy to find by trial and error. A bit more systematically, we can see that the subgroup of squares in $mathbb{F}_{11}^*$ is cyclic of order $5$ so any square besides $0$ and $1$ must have order $5$.



                For an element of order $4$, we can save some work by noticing that an element of order $4$ is just a square root of $-1$. So, we just need to solve for $a,binmathbb{F}_{11}$ such that $(a+bx)^2=-1$. Expanding out $(a+bx)^2$ using $x^2+x+7=0$ we get $$(a^2-7b^2)+(2ab-b^2)x=-1$$ and so we need $$a^2-7b^2=-1$$ and $$2ab-b^2=0.$$ The second equation gives $b=0$ or $2a=b$. The first case does not work (since there is no square root of $-1$ in $mathbb{F}_7$); in the second case the first equation simplifies to $$5a^2=1$$ which we can easily solve in $mathbb{F}_{11}$ to get $a=pm 3$. So, the elements of order $4$ are $pm(3+6x)$.






                share|cite|improve this answer









                $endgroup$



                I don't think there's a magic shortcut to this kind of question in general, but the specific cases we have here are simple enough that there are some tricks. For an element of order $5$, we can note that $5$ divides $11-1$, so there is actually an element of order $5$ in $mathbb{F}_{11}$, which is easy to find by trial and error. A bit more systematically, we can see that the subgroup of squares in $mathbb{F}_{11}^*$ is cyclic of order $5$ so any square besides $0$ and $1$ must have order $5$.



                For an element of order $4$, we can save some work by noticing that an element of order $4$ is just a square root of $-1$. So, we just need to solve for $a,binmathbb{F}_{11}$ such that $(a+bx)^2=-1$. Expanding out $(a+bx)^2$ using $x^2+x+7=0$ we get $$(a^2-7b^2)+(2ab-b^2)x=-1$$ and so we need $$a^2-7b^2=-1$$ and $$2ab-b^2=0.$$ The second equation gives $b=0$ or $2a=b$. The first case does not work (since there is no square root of $-1$ in $mathbb{F}_7$); in the second case the first equation simplifies to $$5a^2=1$$ which we can easily solve in $mathbb{F}_{11}$ to get $a=pm 3$. So, the elements of order $4$ are $pm(3+6x)$.







                share|cite|improve this answer












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                answered 4 hours ago









                Eric WofseyEric Wofsey

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