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












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












    $begingroup$



    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











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 13 '18 at 14:46









      AlessarAlessar

      313115




      313115






















          1 Answer
          1






          active

          oldest

          votes


















          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













          Your Answer





          StackExchange.ifUsing("editor", function () {
          return StackExchange.using("mathjaxEditing", function () {
          StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
          StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
          });
          });
          }, "mathjax-editing");

          StackExchange.ready(function() {
          var channelOptions = {
          tags: "".split(" "),
          id: "69"
          };
          initTagRenderer("".split(" "), "".split(" "), channelOptions);

          StackExchange.using("externalEditor", function() {
          // Have to fire editor after snippets, if snippets enabled
          if (StackExchange.settings.snippets.snippetsEnabled) {
          StackExchange.using("snippets", function() {
          createEditor();
          });
          }
          else {
          createEditor();
          }
          });

          function createEditor() {
          StackExchange.prepareEditor({
          heartbeatType: 'answer',
          autoActivateHeartbeat: false,
          convertImagesToLinks: true,
          noModals: true,
          showLowRepImageUploadWarning: true,
          reputationToPostImages: 10,
          bindNavPrevention: true,
          postfix: "",
          imageUploader: {
          brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
          contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
          allowUrls: true
          },
          noCode: true, onDemand: true,
          discardSelector: ".discard-answer"
          ,immediatelyShowMarkdownHelp:true
          });


          }
          });














          draft saved

          draft discarded


















          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3038096%2fshow-that-for-p-prime-mathbbz-p-is-the-only-finite-field-with-cardinalit%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown

























          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          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












          $begingroup$

          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




















          draft saved

          draft discarded




















































          Thanks for contributing an answer to Mathematics Stack Exchange!


          • Please be sure to answer the question. Provide details and share your research!

          But avoid …



          • Asking for help, clarification, or responding to other answers.

          • Making statements based on opinion; back them up with references or personal experience.


          Use MathJax to format equations. MathJax reference.


          To learn more, see our tips on writing great answers.




          draft saved


          draft discarded














          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3038096%2fshow-that-for-p-prime-mathbbz-p-is-the-only-finite-field-with-cardinalit%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown





















































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown

































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown







          -

          Popular posts from this blog

          Probability when a professor distributes a quiz and homework assignment to a class of n students.

          Image
          0 0 $begingroup$ Need help with this problem. Suppose our lazy professor collects a quiz and a homework assignment from a class of n students one day, then distributes both the quizzes and the homework assignments back to the class in a random fashion for grading. Each student receives one quiz and one homework assignment to grade. (a) What is the probability that every student receives someone else's quiz to grade, and someone else's homework to grade? (b) What is the probability that no student receives both their own quiz and their own homework assignment to grade? In this case, some students may receive their own quiz, and others may receive their own homework assignment. (c) Compute the limiting probability as n approaches infinity in each case. ...
          Read more

          Aardman Animations

          Image
          Aardman Animations, Ltd. Aardman Animations Tipo privada Atividade Animação em stop-motion , Animação em CGI Fundação 1972 Fundador(es) Peter Lord David Sproxton Sede Bristol, Inglaterra  Reino Unido Proprietário(s) Dreamworks Pessoas-chave Peter Lord David Sproxton Nick Park Divisões Aardman Features Aardman Digital Aardman Commercials Aardman Broadcast Aardman International Aardman Rights Aardman Effects Aardman 3-D Systems Aardman Nathan Love AardBoiled Website oficial aardman.com ...
          Read more

          Are they similar matrix

          Image
          0 $begingroup$ Do $begin{bmatrix} 0&i&0\0&0&1\0&0&0 end{bmatrix} $ and $begin{bmatrix} 0&0&0\-i&0&0\0&1&0 end{bmatrix} $ are similar.Is this True/false Clearly both are nilpotent and one is conjucate transpose of other but how to know if they are similar.i'm stuck. Please help me linear-algebra vector-spaces eigenvalues-eigenvectors generalizedeigenvector share | cite | improve this question edited Dec 10 '18 at 6:38 Vasanth Kris asked Dec 10 '18 at 6:19 ...
          Read more