Show that for $p$ prime, $mathbb{Z}_p$ is the only finite field with cardinality $p$ (not counting...












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Show that for $p$ prime, $mathbb{Z}_p$ is the only finite field with cardinality $p$ (not counting Isomorphism)




So here my proof. I know that the elements are $p$, in particular $mathbb{Z}_p={ 0,1,2,...,p-1}$. If I suppose there is another field $mathbb{K}_p={ 0,1,2,...,p-1}$ with $p$ elements, I want to show that $mathbb{Z}_p==mathbb{K}_p$.



An element mod $p$ in $mathbb{Z}_p $ exist also in $mathbb{K}_p$ so $mathbb{Z}_p subset mathbb{K}_p$. But it's valid also the vice versa $mathbb{Z}_p supset mathbb{K}_p$. Plus, every operation made using elements $z_i$ can fall into the $mathbb{K}_p$ set and every operation made using elements $k_i$ can fall into the $mathbb{Z}_p$ set.
If those fields are finite, then must be $char(mathbb{Z}_p) = p$ and $char(mathbb{K}_p) = p$, so must be $mathbb{Z}_p==mathbb{K}_p$.



Are those evidence sufficient to prove the assertion?






abstract-algebra elementary-number-theory







share|cite|improve this question









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    0












    $begingroup$



    Show that for $p$ prime, $mathbb{Z}_p$ is the only finite field with cardinality $p$ (not counting Isomorphism)




    So here my proof. I know that the elements are $p$, in particular $mathbb{Z}_p={ 0,1,2,...,p-1}$. If I suppose there is another field $mathbb{K}_p={ 0,1,2,...,p-1}$ with $p$ elements, I want to show that $mathbb{Z}_p==mathbb{K}_p$.



    An element mod $p$ in $mathbb{Z}_p $ exist also in $mathbb{K}_p$ so $mathbb{Z}_p subset mathbb{K}_p$. But it's valid also the vice versa $mathbb{Z}_p supset mathbb{K}_p$. Plus, every operation made using elements $z_i$ can fall into the $mathbb{K}_p$ set and every operation made using elements $k_i$ can fall into the $mathbb{Z}_p$ set.
    If those fields are finite, then must be $char(mathbb{Z}_p) = p$ and $char(mathbb{K}_p) = p$, so must be $mathbb{Z}_p==mathbb{K}_p$.



    Are those evidence sufficient to prove the assertion?






    abstract-algebra elementary-number-theory







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$



      Show that for $p$ prime, $mathbb{Z}_p$ is the only finite field with cardinality $p$ (not counting Isomorphism)




      So here my proof. I know that the elements are $p$, in particular $mathbb{Z}_p={ 0,1,2,...,p-1}$. If I suppose there is another field $mathbb{K}_p={ 0,1,2,...,p-1}$ with $p$ elements, I want to show that $mathbb{Z}_p==mathbb{K}_p$.



      An element mod $p$ in $mathbb{Z}_p $ exist also in $mathbb{K}_p$ so $mathbb{Z}_p subset mathbb{K}_p$. But it's valid also the vice versa $mathbb{Z}_p supset mathbb{K}_p$. Plus, every operation made using elements $z_i$ can fall into the $mathbb{K}_p$ set and every operation made using elements $k_i$ can fall into the $mathbb{Z}_p$ set.
      If those fields are finite, then must be $char(mathbb{Z}_p) = p$ and $char(mathbb{K}_p) = p$, so must be $mathbb{Z}_p==mathbb{K}_p$.



      Are those evidence sufficient to prove the assertion?






      abstract-algebra elementary-number-theory







      share|cite|improve this question









      $endgroup$





      Show that for $p$ prime, $mathbb{Z}_p$ is the only finite field with cardinality $p$ (not counting Isomorphism)




      So here my proof. I know that the elements are $p$, in particular $mathbb{Z}_p={ 0,1,2,...,p-1}$. If I suppose there is another field $mathbb{K}_p={ 0,1,2,...,p-1}$ with $p$ elements, I want to show that $mathbb{Z}_p==mathbb{K}_p$.



      An element mod $p$ in $mathbb{Z}_p $ exist also in $mathbb{K}_p$ so $mathbb{Z}_p subset mathbb{K}_p$. But it's valid also the vice versa $mathbb{Z}_p supset mathbb{K}_p$. Plus, every operation made using elements $z_i$ can fall into the $mathbb{K}_p$ set and every operation made using elements $k_i$ can fall into the $mathbb{Z}_p$ set.
      If those fields are finite, then must be $char(mathbb{Z}_p) = p$ and $char(mathbb{K}_p) = p$, so must be $mathbb{Z}_p==mathbb{K}_p$.



      Are those evidence sufficient to prove the assertion?






      abstract-algebra elementary-number-theory




      abstract-algebra elementary-number-theory






      share|cite|improve this question













      share|cite|improve this question











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      share|cite|improve this question










      asked Dec 13 '18 at 14:46









      AlessarAlessar

      313115




      313115






















          1 Answer
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          $begingroup$

          I don't really understand your proof but I don't think that it's correct - you haven't really justified that every element of $mathbb{Z}/(p)$ lies in $mathbb{K}_p$, and then I'm not sure what your next sentence means. I think you have the right idea but haven't quite worded your proof correctly!



          Here are the steps you should follow : suppose $K$ is a field with $p$ elements and multiplicative identity $1_K$. Let $phi,:,mathbb{Z}/(p) rightarrow K$ be the ring homomorphism sending $1 mapsto 1_K$, and show that this must be an isomorpsism






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            $phi : mathbb{Z}_p rightarrow K$ is an homomorphism because $phi(1+z_i) = phi(1)+phi(z_i) = 1_k+k_i$. For the product $phi(1 times z_i) = phi(1) times phi(z_i) = 1_k times k_i = k_i$ so the mapping should be complete, and the proof too. Do I miss something yet? I have great difficulties to use the formalism, sorry
            $endgroup$
            – Alessar
            Dec 13 '18 at 15:18






          • 1




            $begingroup$
            Think about the kernel of $phi$ and what we can say about it, knowing that $mathbb{Z}/(p)$ has prime order.
            $endgroup$
            – ODF
            Dec 13 '18 at 15:23










          • $begingroup$
            We know that if $mathbb{Z_p}$ has order $p$ with $p$ prime, then $Ker phi not = 0$; this reminds me of the algebraic condition or the minimal polynomial. For the algebraic condition, it also mean that is a finite field...?
            $endgroup$
            – Alessar
            Dec 13 '18 at 15:33













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          1 Answer
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          active

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          $begingroup$

          I don't really understand your proof but I don't think that it's correct - you haven't really justified that every element of $mathbb{Z}/(p)$ lies in $mathbb{K}_p$, and then I'm not sure what your next sentence means. I think you have the right idea but haven't quite worded your proof correctly!



          Here are the steps you should follow : suppose $K$ is a field with $p$ elements and multiplicative identity $1_K$. Let $phi,:,mathbb{Z}/(p) rightarrow K$ be the ring homomorphism sending $1 mapsto 1_K$, and show that this must be an isomorpsism






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            $phi : mathbb{Z}_p rightarrow K$ is an homomorphism because $phi(1+z_i) = phi(1)+phi(z_i) = 1_k+k_i$. For the product $phi(1 times z_i) = phi(1) times phi(z_i) = 1_k times k_i = k_i$ so the mapping should be complete, and the proof too. Do I miss something yet? I have great difficulties to use the formalism, sorry
            $endgroup$
            – Alessar
            Dec 13 '18 at 15:18






          • 1




            $begingroup$
            Think about the kernel of $phi$ and what we can say about it, knowing that $mathbb{Z}/(p)$ has prime order.
            $endgroup$
            – ODF
            Dec 13 '18 at 15:23










          • $begingroup$
            We know that if $mathbb{Z_p}$ has order $p$ with $p$ prime, then $Ker phi not = 0$; this reminds me of the algebraic condition or the minimal polynomial. For the algebraic condition, it also mean that is a finite field...?
            $endgroup$
            – Alessar
            Dec 13 '18 at 15:33


















          2












          $begingroup$

          I don't really understand your proof but I don't think that it's correct - you haven't really justified that every element of $mathbb{Z}/(p)$ lies in $mathbb{K}_p$, and then I'm not sure what your next sentence means. I think you have the right idea but haven't quite worded your proof correctly!



          Here are the steps you should follow : suppose $K$ is a field with $p$ elements and multiplicative identity $1_K$. Let $phi,:,mathbb{Z}/(p) rightarrow K$ be the ring homomorphism sending $1 mapsto 1_K$, and show that this must be an isomorpsism






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            $phi : mathbb{Z}_p rightarrow K$ is an homomorphism because $phi(1+z_i) = phi(1)+phi(z_i) = 1_k+k_i$. For the product $phi(1 times z_i) = phi(1) times phi(z_i) = 1_k times k_i = k_i$ so the mapping should be complete, and the proof too. Do I miss something yet? I have great difficulties to use the formalism, sorry
            $endgroup$
            – Alessar
            Dec 13 '18 at 15:18






          • 1




            $begingroup$
            Think about the kernel of $phi$ and what we can say about it, knowing that $mathbb{Z}/(p)$ has prime order.
            $endgroup$
            – ODF
            Dec 13 '18 at 15:23










          • $begingroup$
            We know that if $mathbb{Z_p}$ has order $p$ with $p$ prime, then $Ker phi not = 0$; this reminds me of the algebraic condition or the minimal polynomial. For the algebraic condition, it also mean that is a finite field...?
            $endgroup$
            – Alessar
            Dec 13 '18 at 15:33
















          2












          2








          2





          $begingroup$

          I don't really understand your proof but I don't think that it's correct - you haven't really justified that every element of $mathbb{Z}/(p)$ lies in $mathbb{K}_p$, and then I'm not sure what your next sentence means. I think you have the right idea but haven't quite worded your proof correctly!



          Here are the steps you should follow : suppose $K$ is a field with $p$ elements and multiplicative identity $1_K$. Let $phi,:,mathbb{Z}/(p) rightarrow K$ be the ring homomorphism sending $1 mapsto 1_K$, and show that this must be an isomorpsism






          share|cite|improve this answer









          $endgroup$



          I don't really understand your proof but I don't think that it's correct - you haven't really justified that every element of $mathbb{Z}/(p)$ lies in $mathbb{K}_p$, and then I'm not sure what your next sentence means. I think you have the right idea but haven't quite worded your proof correctly!



          Here are the steps you should follow : suppose $K$ is a field with $p$ elements and multiplicative identity $1_K$. Let $phi,:,mathbb{Z}/(p) rightarrow K$ be the ring homomorphism sending $1 mapsto 1_K$, and show that this must be an isomorpsism







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 13 '18 at 14:53









          ODFODF

          1,486510




          1,486510












          • $begingroup$
            $phi : mathbb{Z}_p rightarrow K$ is an homomorphism because $phi(1+z_i) = phi(1)+phi(z_i) = 1_k+k_i$. For the product $phi(1 times z_i) = phi(1) times phi(z_i) = 1_k times k_i = k_i$ so the mapping should be complete, and the proof too. Do I miss something yet? I have great difficulties to use the formalism, sorry
            $endgroup$
            – Alessar
            Dec 13 '18 at 15:18






          • 1




            $begingroup$
            Think about the kernel of $phi$ and what we can say about it, knowing that $mathbb{Z}/(p)$ has prime order.
            $endgroup$
            – ODF
            Dec 13 '18 at 15:23










          • $begingroup$
            We know that if $mathbb{Z_p}$ has order $p$ with $p$ prime, then $Ker phi not = 0$; this reminds me of the algebraic condition or the minimal polynomial. For the algebraic condition, it also mean that is a finite field...?
            $endgroup$
            – Alessar
            Dec 13 '18 at 15:33




















          • $begingroup$
            $phi : mathbb{Z}_p rightarrow K$ is an homomorphism because $phi(1+z_i) = phi(1)+phi(z_i) = 1_k+k_i$. For the product $phi(1 times z_i) = phi(1) times phi(z_i) = 1_k times k_i = k_i$ so the mapping should be complete, and the proof too. Do I miss something yet? I have great difficulties to use the formalism, sorry
            $endgroup$
            – Alessar
            Dec 13 '18 at 15:18






          • 1




            $begingroup$
            Think about the kernel of $phi$ and what we can say about it, knowing that $mathbb{Z}/(p)$ has prime order.
            $endgroup$
            – ODF
            Dec 13 '18 at 15:23










          • $begingroup$
            We know that if $mathbb{Z_p}$ has order $p$ with $p$ prime, then $Ker phi not = 0$; this reminds me of the algebraic condition or the minimal polynomial. For the algebraic condition, it also mean that is a finite field...?
            $endgroup$
            – Alessar
            Dec 13 '18 at 15:33


















          $begingroup$
          $phi : mathbb{Z}_p rightarrow K$ is an homomorphism because $phi(1+z_i) = phi(1)+phi(z_i) = 1_k+k_i$. For the product $phi(1 times z_i) = phi(1) times phi(z_i) = 1_k times k_i = k_i$ so the mapping should be complete, and the proof too. Do I miss something yet? I have great difficulties to use the formalism, sorry
          $endgroup$
          – Alessar
          Dec 13 '18 at 15:18




          $begingroup$
          $phi : mathbb{Z}_p rightarrow K$ is an homomorphism because $phi(1+z_i) = phi(1)+phi(z_i) = 1_k+k_i$. For the product $phi(1 times z_i) = phi(1) times phi(z_i) = 1_k times k_i = k_i$ so the mapping should be complete, and the proof too. Do I miss something yet? I have great difficulties to use the formalism, sorry
          $endgroup$
          – Alessar
          Dec 13 '18 at 15:18




          1




          1




          $begingroup$
          Think about the kernel of $phi$ and what we can say about it, knowing that $mathbb{Z}/(p)$ has prime order.
          $endgroup$
          – ODF
          Dec 13 '18 at 15:23




          $begingroup$
          Think about the kernel of $phi$ and what we can say about it, knowing that $mathbb{Z}/(p)$ has prime order.
          $endgroup$
          – ODF
          Dec 13 '18 at 15:23












          $begingroup$
          We know that if $mathbb{Z_p}$ has order $p$ with $p$ prime, then $Ker phi not = 0$; this reminds me of the algebraic condition or the minimal polynomial. For the algebraic condition, it also mean that is a finite field...?
          $endgroup$
          – Alessar
          Dec 13 '18 at 15:33






          $begingroup$
          We know that if $mathbb{Z_p}$ has order $p$ with $p$ prime, then $Ker phi not = 0$; this reminds me of the algebraic condition or the minimal polynomial. For the algebraic condition, it also mean that is a finite field...?
          $endgroup$
          – Alessar
          Dec 13 '18 at 15:33




















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