Showing the lemma $operatorname{ord}_p(1+ζ_p)=0$ if $p>2$












2












$begingroup$


Just to give some background regarding my motivation, I'm trying to prove a lemma to help me solve How do we prove p-order of $g_k$ is $frac {k} {p-1}$?



Let $Z_p$ denote the p-adic integers, and let us adjoin a primitive p-th root of unity $ζ_p$. Assume $p>2DeclareMathOperator{ord}{ord}$.



I want to show that $ord_p(1-ζ_p)=ord_p(1-ζ_p^2)$, but I see that it is enough to prove $ord_p( frac {1-ζ_p^2} {1-ζ_p})=0$ or, equivalently, $ord_p(1+ζ_p)=0$.



I have tried applying the properties from here. (Even though they are for rational numbers, I'm going to assume for now that they hold in $Z_p(ζ_p)$ also. If not, please correct me.)



I see that, if $ord_p(ζ_p) neq ord_p(1)=0$, we will have $ord_p(1+ζ_p)=inf {ord_p(1), ord_p(ζ_p)} leq 0$ and we will get a contradiction if $<0$
.



But if instead $ord_p(ζ_p)=0,$ then I don't know what to do. We will have $ord_p(1+ζ_p) geq inf {ord_p(1), ord_p(ζ_p)}=0$, but I don't see how $>0$ would give us a contradiction.






abstract-algebra field-theory prime-numbers p-adic-number-theory roots-of-unity







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    That $|zeta_p|_p = 1$ is obvious from $zeta_p^p = 1$
    $endgroup$
    – reuns
    Dec 7 '18 at 20:15
















2












$begingroup$


Just to give some background regarding my motivation, I'm trying to prove a lemma to help me solve How do we prove p-order of $g_k$ is $frac {k} {p-1}$?



Let $Z_p$ denote the p-adic integers, and let us adjoin a primitive p-th root of unity $ζ_p$. Assume $p>2DeclareMathOperator{ord}{ord}$.



I want to show that $ord_p(1-ζ_p)=ord_p(1-ζ_p^2)$, but I see that it is enough to prove $ord_p( frac {1-ζ_p^2} {1-ζ_p})=0$ or, equivalently, $ord_p(1+ζ_p)=0$.



I have tried applying the properties from here. (Even though they are for rational numbers, I'm going to assume for now that they hold in $Z_p(ζ_p)$ also. If not, please correct me.)



I see that, if $ord_p(ζ_p) neq ord_p(1)=0$, we will have $ord_p(1+ζ_p)=inf {ord_p(1), ord_p(ζ_p)} leq 0$ and we will get a contradiction if $<0$
.



But if instead $ord_p(ζ_p)=0,$ then I don't know what to do. We will have $ord_p(1+ζ_p) geq inf {ord_p(1), ord_p(ζ_p)}=0$, but I don't see how $>0$ would give us a contradiction.






abstract-algebra field-theory prime-numbers p-adic-number-theory roots-of-unity







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    That $|zeta_p|_p = 1$ is obvious from $zeta_p^p = 1$
    $endgroup$
    – reuns
    Dec 7 '18 at 20:15














2












2








2


1



$begingroup$


Just to give some background regarding my motivation, I'm trying to prove a lemma to help me solve How do we prove p-order of $g_k$ is $frac {k} {p-1}$?



Let $Z_p$ denote the p-adic integers, and let us adjoin a primitive p-th root of unity $ζ_p$. Assume $p>2DeclareMathOperator{ord}{ord}$.



I want to show that $ord_p(1-ζ_p)=ord_p(1-ζ_p^2)$, but I see that it is enough to prove $ord_p( frac {1-ζ_p^2} {1-ζ_p})=0$ or, equivalently, $ord_p(1+ζ_p)=0$.



I have tried applying the properties from here. (Even though they are for rational numbers, I'm going to assume for now that they hold in $Z_p(ζ_p)$ also. If not, please correct me.)



I see that, if $ord_p(ζ_p) neq ord_p(1)=0$, we will have $ord_p(1+ζ_p)=inf {ord_p(1), ord_p(ζ_p)} leq 0$ and we will get a contradiction if $<0$
.



But if instead $ord_p(ζ_p)=0,$ then I don't know what to do. We will have $ord_p(1+ζ_p) geq inf {ord_p(1), ord_p(ζ_p)}=0$, but I don't see how $>0$ would give us a contradiction.






abstract-algebra field-theory prime-numbers p-adic-number-theory roots-of-unity







share|cite|improve this question











$endgroup$




Just to give some background regarding my motivation, I'm trying to prove a lemma to help me solve How do we prove p-order of $g_k$ is $frac {k} {p-1}$?



Let $Z_p$ denote the p-adic integers, and let us adjoin a primitive p-th root of unity $ζ_p$. Assume $p>2DeclareMathOperator{ord}{ord}$.



I want to show that $ord_p(1-ζ_p)=ord_p(1-ζ_p^2)$, but I see that it is enough to prove $ord_p( frac {1-ζ_p^2} {1-ζ_p})=0$ or, equivalently, $ord_p(1+ζ_p)=0$.



I have tried applying the properties from here. (Even though they are for rational numbers, I'm going to assume for now that they hold in $Z_p(ζ_p)$ also. If not, please correct me.)



I see that, if $ord_p(ζ_p) neq ord_p(1)=0$, we will have $ord_p(1+ζ_p)=inf {ord_p(1), ord_p(ζ_p)} leq 0$ and we will get a contradiction if $<0$
.



But if instead $ord_p(ζ_p)=0,$ then I don't know what to do. We will have $ord_p(1+ζ_p) geq inf {ord_p(1), ord_p(ζ_p)}=0$, but I don't see how $>0$ would give us a contradiction.






abstract-algebra field-theory prime-numbers p-adic-number-theory roots-of-unity




abstract-algebra field-theory prime-numbers p-adic-number-theory roots-of-unity






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 7 '18 at 19:41









Bernard

119k740113




119k740113










asked Dec 7 '18 at 19:28









Pascal's WagerPascal's Wager

357315




357315








  • 1




    $begingroup$
    That $|zeta_p|_p = 1$ is obvious from $zeta_p^p = 1$
    $endgroup$
    – reuns
    Dec 7 '18 at 20:15














  • 1




    $begingroup$
    That $|zeta_p|_p = 1$ is obvious from $zeta_p^p = 1$
    $endgroup$
    – reuns
    Dec 7 '18 at 20:15








1




1




$begingroup$
That $|zeta_p|_p = 1$ is obvious from $zeta_p^p = 1$
$endgroup$
– reuns
Dec 7 '18 at 20:15




$begingroup$
That $|zeta_p|_p = 1$ is obvious from $zeta_p^p = 1$
$endgroup$
– reuns
Dec 7 '18 at 20:15










3 Answers
3






active

oldest

votes


















0












$begingroup$

Let $$(x-1)^p-1 = f(x)g(x) in mathbf{Z}_p[x]$$ where $f(x)$ is the minimal polynomial of $1+zeta_p$ so $$f(x) = prod_{alpha in Gal(overline{mathbf{Q}_p}/mathbf{Q}_p)cdot (1+zeta_p)} (x-alpha)$$



by definition $|.|_p$ is $Gal(overline{mathbf{Q}_p}/mathbf{Q}_p)$ invariant and$$|1+zeta_p|_p=|f(0)|_p^{1/deg(f)}$$



and since $g(0) in mathbf{Z}_p$
$$1 ge |f(0)|_p ge |f(0)g(0)|_p = |(-1)^p-1|_p = 1$$



whence $|f(0)|_p= 1$ and $|1+zeta_p|_p = 1$.






share|cite|improve this answer











$endgroup$





















    0












    $begingroup$

    Note that $frac {ζ^2-1} {ζ-1}=1+ζ$ and $frac {ζ-1} {ζ^2-1} =1+ζ^2+ζ^4+...+ζ^{p-1}$.



    In the case that $ord_p(ζ)=0$, we will have $ord_p(ζ^2)=ord_p(ζ^3)=...=0$.



    This would imply $ord_p(ζ^2-1)-οrd_p(ζ-1)=ord_p(1+ζ) geq inf {ord_p(1),ord_p(ζ)}=0 $ and also $οrd_p(ζ-1)-ord_p(ζ^2-1)=ord_p(1+ζ^2+ζ^4+...+ζ^{p-1})geq inf{ord_p(1),...,ord_p(ζ^{p-1})}=0$.



    From these two inequalities, it should be easy to see $ord_p(ζ^2-1)=οrd_p(ζ-1)$ as needed.






    share|cite|improve this answer









    $endgroup$





















      0












      $begingroup$

      Here’s yet another argument:

      Start with the minimal polynomial for $zeta_p$ ,
      begin{align}
      text{Irr}(zeta_p,Bbb Q_p)&=frac{X^p-1}{X-1}\
      f=text{Irr}(zeta_p-1,Bbb Q_p)&=frac{(X+1)^p-1}X&text{($p$-Eisenstein, root $pi$)}\
      v_p(pi)=v_p(zeta_p-1)&=frac1{p-1}\
      v_p(pi+2)=v_p(zeta_p+1)&=v_p(2)=0&text{(’cause $pne2$)},.
      end{align}






      share|cite|improve this answer









      $endgroup$













        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%2f3030282%2fshowing-the-lemma-operatornameord-p1%25ce%25b6-p-0-if-p2%23new-answer', 'question_page');
        }
        );

        Post as a guest















        Required, but never shown

























        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        0












        $begingroup$

        Let $$(x-1)^p-1 = f(x)g(x) in mathbf{Z}_p[x]$$ where $f(x)$ is the minimal polynomial of $1+zeta_p$ so $$f(x) = prod_{alpha in Gal(overline{mathbf{Q}_p}/mathbf{Q}_p)cdot (1+zeta_p)} (x-alpha)$$



        by definition $|.|_p$ is $Gal(overline{mathbf{Q}_p}/mathbf{Q}_p)$ invariant and$$|1+zeta_p|_p=|f(0)|_p^{1/deg(f)}$$



        and since $g(0) in mathbf{Z}_p$
        $$1 ge |f(0)|_p ge |f(0)g(0)|_p = |(-1)^p-1|_p = 1$$



        whence $|f(0)|_p= 1$ and $|1+zeta_p|_p = 1$.






        share|cite|improve this answer











        $endgroup$


















          0












          $begingroup$

          Let $$(x-1)^p-1 = f(x)g(x) in mathbf{Z}_p[x]$$ where $f(x)$ is the minimal polynomial of $1+zeta_p$ so $$f(x) = prod_{alpha in Gal(overline{mathbf{Q}_p}/mathbf{Q}_p)cdot (1+zeta_p)} (x-alpha)$$



          by definition $|.|_p$ is $Gal(overline{mathbf{Q}_p}/mathbf{Q}_p)$ invariant and$$|1+zeta_p|_p=|f(0)|_p^{1/deg(f)}$$



          and since $g(0) in mathbf{Z}_p$
          $$1 ge |f(0)|_p ge |f(0)g(0)|_p = |(-1)^p-1|_p = 1$$



          whence $|f(0)|_p= 1$ and $|1+zeta_p|_p = 1$.






          share|cite|improve this answer











          $endgroup$
















            0












            0








            0





            $begingroup$

            Let $$(x-1)^p-1 = f(x)g(x) in mathbf{Z}_p[x]$$ where $f(x)$ is the minimal polynomial of $1+zeta_p$ so $$f(x) = prod_{alpha in Gal(overline{mathbf{Q}_p}/mathbf{Q}_p)cdot (1+zeta_p)} (x-alpha)$$



            by definition $|.|_p$ is $Gal(overline{mathbf{Q}_p}/mathbf{Q}_p)$ invariant and$$|1+zeta_p|_p=|f(0)|_p^{1/deg(f)}$$



            and since $g(0) in mathbf{Z}_p$
            $$1 ge |f(0)|_p ge |f(0)g(0)|_p = |(-1)^p-1|_p = 1$$



            whence $|f(0)|_p= 1$ and $|1+zeta_p|_p = 1$.






            share|cite|improve this answer











            $endgroup$



            Let $$(x-1)^p-1 = f(x)g(x) in mathbf{Z}_p[x]$$ where $f(x)$ is the minimal polynomial of $1+zeta_p$ so $$f(x) = prod_{alpha in Gal(overline{mathbf{Q}_p}/mathbf{Q}_p)cdot (1+zeta_p)} (x-alpha)$$



            by definition $|.|_p$ is $Gal(overline{mathbf{Q}_p}/mathbf{Q}_p)$ invariant and$$|1+zeta_p|_p=|f(0)|_p^{1/deg(f)}$$



            and since $g(0) in mathbf{Z}_p$
            $$1 ge |f(0)|_p ge |f(0)g(0)|_p = |(-1)^p-1|_p = 1$$



            whence $|f(0)|_p= 1$ and $|1+zeta_p|_p = 1$.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 7 '18 at 20:16

























            answered Dec 7 '18 at 20:10









            reunsreuns

            19.9k21148




            19.9k21148























                0












                $begingroup$

                Note that $frac {ζ^2-1} {ζ-1}=1+ζ$ and $frac {ζ-1} {ζ^2-1} =1+ζ^2+ζ^4+...+ζ^{p-1}$.



                In the case that $ord_p(ζ)=0$, we will have $ord_p(ζ^2)=ord_p(ζ^3)=...=0$.



                This would imply $ord_p(ζ^2-1)-οrd_p(ζ-1)=ord_p(1+ζ) geq inf {ord_p(1),ord_p(ζ)}=0 $ and also $οrd_p(ζ-1)-ord_p(ζ^2-1)=ord_p(1+ζ^2+ζ^4+...+ζ^{p-1})geq inf{ord_p(1),...,ord_p(ζ^{p-1})}=0$.



                From these two inequalities, it should be easy to see $ord_p(ζ^2-1)=οrd_p(ζ-1)$ as needed.






                share|cite|improve this answer









                $endgroup$


















                  0












                  $begingroup$

                  Note that $frac {ζ^2-1} {ζ-1}=1+ζ$ and $frac {ζ-1} {ζ^2-1} =1+ζ^2+ζ^4+...+ζ^{p-1}$.



                  In the case that $ord_p(ζ)=0$, we will have $ord_p(ζ^2)=ord_p(ζ^3)=...=0$.



                  This would imply $ord_p(ζ^2-1)-οrd_p(ζ-1)=ord_p(1+ζ) geq inf {ord_p(1),ord_p(ζ)}=0 $ and also $οrd_p(ζ-1)-ord_p(ζ^2-1)=ord_p(1+ζ^2+ζ^4+...+ζ^{p-1})geq inf{ord_p(1),...,ord_p(ζ^{p-1})}=0$.



                  From these two inequalities, it should be easy to see $ord_p(ζ^2-1)=οrd_p(ζ-1)$ as needed.






                  share|cite|improve this answer









                  $endgroup$
















                    0












                    0








                    0





                    $begingroup$

                    Note that $frac {ζ^2-1} {ζ-1}=1+ζ$ and $frac {ζ-1} {ζ^2-1} =1+ζ^2+ζ^4+...+ζ^{p-1}$.



                    In the case that $ord_p(ζ)=0$, we will have $ord_p(ζ^2)=ord_p(ζ^3)=...=0$.



                    This would imply $ord_p(ζ^2-1)-οrd_p(ζ-1)=ord_p(1+ζ) geq inf {ord_p(1),ord_p(ζ)}=0 $ and also $οrd_p(ζ-1)-ord_p(ζ^2-1)=ord_p(1+ζ^2+ζ^4+...+ζ^{p-1})geq inf{ord_p(1),...,ord_p(ζ^{p-1})}=0$.



                    From these two inequalities, it should be easy to see $ord_p(ζ^2-1)=οrd_p(ζ-1)$ as needed.






                    share|cite|improve this answer









                    $endgroup$



                    Note that $frac {ζ^2-1} {ζ-1}=1+ζ$ and $frac {ζ-1} {ζ^2-1} =1+ζ^2+ζ^4+...+ζ^{p-1}$.



                    In the case that $ord_p(ζ)=0$, we will have $ord_p(ζ^2)=ord_p(ζ^3)=...=0$.



                    This would imply $ord_p(ζ^2-1)-οrd_p(ζ-1)=ord_p(1+ζ) geq inf {ord_p(1),ord_p(ζ)}=0 $ and also $οrd_p(ζ-1)-ord_p(ζ^2-1)=ord_p(1+ζ^2+ζ^4+...+ζ^{p-1})geq inf{ord_p(1),...,ord_p(ζ^{p-1})}=0$.



                    From these two inequalities, it should be easy to see $ord_p(ζ^2-1)=οrd_p(ζ-1)$ as needed.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Dec 7 '18 at 21:11









                    Pascal's WagerPascal's Wager

                    357315




                    357315























                        0












                        $begingroup$

                        Here’s yet another argument:

                        Start with the minimal polynomial for $zeta_p$ ,
                        begin{align}
                        text{Irr}(zeta_p,Bbb Q_p)&=frac{X^p-1}{X-1}\
                        f=text{Irr}(zeta_p-1,Bbb Q_p)&=frac{(X+1)^p-1}X&text{($p$-Eisenstein, root $pi$)}\
                        v_p(pi)=v_p(zeta_p-1)&=frac1{p-1}\
                        v_p(pi+2)=v_p(zeta_p+1)&=v_p(2)=0&text{(’cause $pne2$)},.
                        end{align}






                        share|cite|improve this answer









                        $endgroup$


















                          0












                          $begingroup$

                          Here’s yet another argument:

                          Start with the minimal polynomial for $zeta_p$ ,
                          begin{align}
                          text{Irr}(zeta_p,Bbb Q_p)&=frac{X^p-1}{X-1}\
                          f=text{Irr}(zeta_p-1,Bbb Q_p)&=frac{(X+1)^p-1}X&text{($p$-Eisenstein, root $pi$)}\
                          v_p(pi)=v_p(zeta_p-1)&=frac1{p-1}\
                          v_p(pi+2)=v_p(zeta_p+1)&=v_p(2)=0&text{(’cause $pne2$)},.
                          end{align}






                          share|cite|improve this answer









                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$

                            Here’s yet another argument:

                            Start with the minimal polynomial for $zeta_p$ ,
                            begin{align}
                            text{Irr}(zeta_p,Bbb Q_p)&=frac{X^p-1}{X-1}\
                            f=text{Irr}(zeta_p-1,Bbb Q_p)&=frac{(X+1)^p-1}X&text{($p$-Eisenstein, root $pi$)}\
                            v_p(pi)=v_p(zeta_p-1)&=frac1{p-1}\
                            v_p(pi+2)=v_p(zeta_p+1)&=v_p(2)=0&text{(’cause $pne2$)},.
                            end{align}






                            share|cite|improve this answer









                            $endgroup$



                            Here’s yet another argument:

                            Start with the minimal polynomial for $zeta_p$ ,
                            begin{align}
                            text{Irr}(zeta_p,Bbb Q_p)&=frac{X^p-1}{X-1}\
                            f=text{Irr}(zeta_p-1,Bbb Q_p)&=frac{(X+1)^p-1}X&text{($p$-Eisenstein, root $pi$)}\
                            v_p(pi)=v_p(zeta_p-1)&=frac1{p-1}\
                            v_p(pi+2)=v_p(zeta_p+1)&=v_p(2)=0&text{(’cause $pne2$)},.
                            end{align}







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Dec 8 '18 at 5:45









                            LubinLubin

                            44.2k44585




                            44.2k44585






























                                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%2f3030282%2fshowing-the-lemma-operatornameord-p1%25ce%25b6-p-0-if-p2%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

                                Quarter-circle Tiles

                                Image
                                up vote 0 down vote favorite I have $99$ identical square tiles, each with a quarter-circle drawn on it like this: [asy] size(1.5cm); draw(Arc((2,0),1,90,180),red+1); draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)); [/asy] When I arrange the tiles in a $9times 11$ rectangular grid, each with a random orientation, what is the expected value of the number of full circles I form? I think this problem has to do with finding the chance any given 2x2 square has a circle, but I can't find it. expected-value share | cite | improve this question asked Nov 20 at 15:03 6minecraftninja 1 2
                                Read more

                                build a pushdown automaton that recognizes the reverse language of a given pushdown automaton?

                                Image
                                0 I think it's similiar to NFAs. I replace the finite states of the given automaton for start-states for my new automaton. I do it with $epsilon$ -transition from the start state to the actual finite states of the given automaton. The start state of the given automaton will be my accepting(finite) state of my new one. All transitions will be reversed of course to recognize the reversed word. But how do I have to change the stack of the pushdown automaton? That's tricky. formal-languages automata context-free-grammar share | cite | improve this question edited Nov 29 '18 at 14:20 joee
                                Read more

                                Mont Emei

                                Image
                                Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog
                                Read more