Suppose that $F$ is a field with $27$ elements. Show that for every element $a in F$, $5a = −a$












0












$begingroup$



Suppose that $F$ is a field with $27$ elements. Show that for every
element $a in F$, $5a = −a$.




I am not able to understand how to approach this problem.






abstract-algebra ring-theory field-theory







share|cite|improve this question











$endgroup$








  • 14




    $begingroup$
    Consider the characteristic.
    $endgroup$
    – Randall
    Jan 2 at 7:16






  • 2




    $begingroup$
    Sambhav Khurana Even $3a=0$
    $endgroup$
    – Michael Rozenberg
    Jan 2 at 7:44










  • $begingroup$
    Can I say that the characteristic must be prime since it's a field and it must divide 27. So 3a=0.?
    $endgroup$
    – Sambhav Khurana
    Jan 2 at 7:56










  • $begingroup$
    Yes, of course! Also, $27$ must be $p^n$, where $p$ is prime.
    $endgroup$
    – Michael Rozenberg
    Jan 2 at 8:00


















0












$begingroup$



Suppose that $F$ is a field with $27$ elements. Show that for every
element $a in F$, $5a = −a$.




I am not able to understand how to approach this problem.






abstract-algebra ring-theory field-theory







share|cite|improve this question











$endgroup$








  • 14




    $begingroup$
    Consider the characteristic.
    $endgroup$
    – Randall
    Jan 2 at 7:16






  • 2




    $begingroup$
    Sambhav Khurana Even $3a=0$
    $endgroup$
    – Michael Rozenberg
    Jan 2 at 7:44










  • $begingroup$
    Can I say that the characteristic must be prime since it's a field and it must divide 27. So 3a=0.?
    $endgroup$
    – Sambhav Khurana
    Jan 2 at 7:56










  • $begingroup$
    Yes, of course! Also, $27$ must be $p^n$, where $p$ is prime.
    $endgroup$
    – Michael Rozenberg
    Jan 2 at 8:00
















0












0








0





$begingroup$



Suppose that $F$ is a field with $27$ elements. Show that for every
element $a in F$, $5a = −a$.




I am not able to understand how to approach this problem.






abstract-algebra ring-theory field-theory







share|cite|improve this question











$endgroup$





Suppose that $F$ is a field with $27$ elements. Show that for every
element $a in F$, $5a = −a$.




I am not able to understand how to approach this problem.






abstract-algebra ring-theory field-theory




abstract-algebra ring-theory field-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 2 at 12:41









egreg

185k1486206




185k1486206










asked Jan 2 at 7:15









Sambhav KhuranaSambhav Khurana

214




214








  • 14




    $begingroup$
    Consider the characteristic.
    $endgroup$
    – Randall
    Jan 2 at 7:16






  • 2




    $begingroup$
    Sambhav Khurana Even $3a=0$
    $endgroup$
    – Michael Rozenberg
    Jan 2 at 7:44










  • $begingroup$
    Can I say that the characteristic must be prime since it's a field and it must divide 27. So 3a=0.?
    $endgroup$
    – Sambhav Khurana
    Jan 2 at 7:56










  • $begingroup$
    Yes, of course! Also, $27$ must be $p^n$, where $p$ is prime.
    $endgroup$
    – Michael Rozenberg
    Jan 2 at 8:00
















  • 14




    $begingroup$
    Consider the characteristic.
    $endgroup$
    – Randall
    Jan 2 at 7:16






  • 2




    $begingroup$
    Sambhav Khurana Even $3a=0$
    $endgroup$
    – Michael Rozenberg
    Jan 2 at 7:44










  • $begingroup$
    Can I say that the characteristic must be prime since it's a field and it must divide 27. So 3a=0.?
    $endgroup$
    – Sambhav Khurana
    Jan 2 at 7:56










  • $begingroup$
    Yes, of course! Also, $27$ must be $p^n$, where $p$ is prime.
    $endgroup$
    – Michael Rozenberg
    Jan 2 at 8:00










14




14




$begingroup$
Consider the characteristic.
$endgroup$
– Randall
Jan 2 at 7:16




$begingroup$
Consider the characteristic.
$endgroup$
– Randall
Jan 2 at 7:16




2




2




$begingroup$
Sambhav Khurana Even $3a=0$
$endgroup$
– Michael Rozenberg
Jan 2 at 7:44




$begingroup$
Sambhav Khurana Even $3a=0$
$endgroup$
– Michael Rozenberg
Jan 2 at 7:44












$begingroup$
Can I say that the characteristic must be prime since it's a field and it must divide 27. So 3a=0.?
$endgroup$
– Sambhav Khurana
Jan 2 at 7:56




$begingroup$
Can I say that the characteristic must be prime since it's a field and it must divide 27. So 3a=0.?
$endgroup$
– Sambhav Khurana
Jan 2 at 7:56












$begingroup$
Yes, of course! Also, $27$ must be $p^n$, where $p$ is prime.
$endgroup$
– Michael Rozenberg
Jan 2 at 8:00






$begingroup$
Yes, of course! Also, $27$ must be $p^n$, where $p$ is prime.
$endgroup$
– Michael Rozenberg
Jan 2 at 8:00












1 Answer
1






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oldest

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1












$begingroup$

Recall that a field $F$ has a characteristic, which is defined as follows:
Define the homomorphism of rings
begin{align}
psi:mathbb{Z}&to F
end{align}
That sends $1$ to $1_F$. Its kernel is principal, say, generated (in particular) by a prime number $p$ or $0$ (if it is injective). That number is called the characteristic of the field $F$.



Now, you have to know that every finite field, has positive characteristic $p$ (i.e. the characteristic is not zero) and have $p^n$ elements. This is because as it is finite, via $psi$ we can not have infinitely many sums of $1_F$ that are different pairwise, and if $mathbb{F}_p$ is the field with $p$ elements, $F$ can be seen as a non-trivial $mathbb{F}_p$ vector space, so is isomorphic to $mathbb{F}^n$, for some $n>1$.



Because of the previous results, your field $F$ with $3^3$ elements, has characteristic $3$, so that $3a=0$ for all $ain F$, in particular $6a=0$, so that $5a=-a$.






share|cite|improve this answer











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    $begingroup$

    Recall that a field $F$ has a characteristic, which is defined as follows:
    Define the homomorphism of rings
    begin{align}
    psi:mathbb{Z}&to F
    end{align}
    That sends $1$ to $1_F$. Its kernel is principal, say, generated (in particular) by a prime number $p$ or $0$ (if it is injective). That number is called the characteristic of the field $F$.



    Now, you have to know that every finite field, has positive characteristic $p$ (i.e. the characteristic is not zero) and have $p^n$ elements. This is because as it is finite, via $psi$ we can not have infinitely many sums of $1_F$ that are different pairwise, and if $mathbb{F}_p$ is the field with $p$ elements, $F$ can be seen as a non-trivial $mathbb{F}_p$ vector space, so is isomorphic to $mathbb{F}^n$, for some $n>1$.



    Because of the previous results, your field $F$ with $3^3$ elements, has characteristic $3$, so that $3a=0$ for all $ain F$, in particular $6a=0$, so that $5a=-a$.






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      Recall that a field $F$ has a characteristic, which is defined as follows:
      Define the homomorphism of rings
      begin{align}
      psi:mathbb{Z}&to F
      end{align}
      That sends $1$ to $1_F$. Its kernel is principal, say, generated (in particular) by a prime number $p$ or $0$ (if it is injective). That number is called the characteristic of the field $F$.



      Now, you have to know that every finite field, has positive characteristic $p$ (i.e. the characteristic is not zero) and have $p^n$ elements. This is because as it is finite, via $psi$ we can not have infinitely many sums of $1_F$ that are different pairwise, and if $mathbb{F}_p$ is the field with $p$ elements, $F$ can be seen as a non-trivial $mathbb{F}_p$ vector space, so is isomorphic to $mathbb{F}^n$, for some $n>1$.



      Because of the previous results, your field $F$ with $3^3$ elements, has characteristic $3$, so that $3a=0$ for all $ain F$, in particular $6a=0$, so that $5a=-a$.






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        Recall that a field $F$ has a characteristic, which is defined as follows:
        Define the homomorphism of rings
        begin{align}
        psi:mathbb{Z}&to F
        end{align}
        That sends $1$ to $1_F$. Its kernel is principal, say, generated (in particular) by a prime number $p$ or $0$ (if it is injective). That number is called the characteristic of the field $F$.



        Now, you have to know that every finite field, has positive characteristic $p$ (i.e. the characteristic is not zero) and have $p^n$ elements. This is because as it is finite, via $psi$ we can not have infinitely many sums of $1_F$ that are different pairwise, and if $mathbb{F}_p$ is the field with $p$ elements, $F$ can be seen as a non-trivial $mathbb{F}_p$ vector space, so is isomorphic to $mathbb{F}^n$, for some $n>1$.



        Because of the previous results, your field $F$ with $3^3$ elements, has characteristic $3$, so that $3a=0$ for all $ain F$, in particular $6a=0$, so that $5a=-a$.






        share|cite|improve this answer











        $endgroup$



        Recall that a field $F$ has a characteristic, which is defined as follows:
        Define the homomorphism of rings
        begin{align}
        psi:mathbb{Z}&to F
        end{align}
        That sends $1$ to $1_F$. Its kernel is principal, say, generated (in particular) by a prime number $p$ or $0$ (if it is injective). That number is called the characteristic of the field $F$.



        Now, you have to know that every finite field, has positive characteristic $p$ (i.e. the characteristic is not zero) and have $p^n$ elements. This is because as it is finite, via $psi$ we can not have infinitely many sums of $1_F$ that are different pairwise, and if $mathbb{F}_p$ is the field with $p$ elements, $F$ can be seen as a non-trivial $mathbb{F}_p$ vector space, so is isomorphic to $mathbb{F}^n$, for some $n>1$.



        Because of the previous results, your field $F$ with $3^3$ elements, has characteristic $3$, so that $3a=0$ for all $ain F$, in particular $6a=0$, so that $5a=-a$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 2 at 12:47

























        answered Jan 2 at 12:19









        José Alejandro Aburto AranedaJosé Alejandro Aburto Araneda

        802110




        802110






























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