The ladder of the real closure of an ordered field











up vote
2
down vote

favorite












$DeclareMathOperator{lad}{Lad}$Let $(K,<)$ be an ordered field, and $(R,<)$ be the real closure of $(K,<)$. Let $lad(K)$ denote the ladder of $(K,<)$ (for the definition of the ladder, see the link given below). Then the canonical injection $iota colon lad(K) to lad(R)$ is induced. My first question is the following.



Question 1. Is $iota colon lad(K) to lad(R)$ an isomorphism?



I came up with this question when I read https://mathoverflow.net/a/23529/131576. According to the answer, Question 1 is true in the following case.



Proposition. Let $L$ be a linear order. Write $F = bar{mathbb{Q}} cap mathbb{R}$. Suppose $(K,<)$ be the ordered field $F(x_l : l in L)$ as in the answer. Then $iota$ is an isomorphism.



In fact, $(R,<)$ is contained in the field of Hahn series $F[[t^G]]$, where $G$ is the group $mathbb{Q}^{{oplus} L}$ with a suitable order. However, this proof seems a little bit hard in that it uses Hahn series. Then my second question is the following.



Question 2. Are there any easier proofs of the Proposition?



Any help would be appreciated.










share|cite|improve this question




























    up vote
    2
    down vote

    favorite












    $DeclareMathOperator{lad}{Lad}$Let $(K,<)$ be an ordered field, and $(R,<)$ be the real closure of $(K,<)$. Let $lad(K)$ denote the ladder of $(K,<)$ (for the definition of the ladder, see the link given below). Then the canonical injection $iota colon lad(K) to lad(R)$ is induced. My first question is the following.



    Question 1. Is $iota colon lad(K) to lad(R)$ an isomorphism?



    I came up with this question when I read https://mathoverflow.net/a/23529/131576. According to the answer, Question 1 is true in the following case.



    Proposition. Let $L$ be a linear order. Write $F = bar{mathbb{Q}} cap mathbb{R}$. Suppose $(K,<)$ be the ordered field $F(x_l : l in L)$ as in the answer. Then $iota$ is an isomorphism.



    In fact, $(R,<)$ is contained in the field of Hahn series $F[[t^G]]$, where $G$ is the group $mathbb{Q}^{{oplus} L}$ with a suitable order. However, this proof seems a little bit hard in that it uses Hahn series. Then my second question is the following.



    Question 2. Are there any easier proofs of the Proposition?



    Any help would be appreciated.










    share|cite|improve this question


























      up vote
      2
      down vote

      favorite









      up vote
      2
      down vote

      favorite











      $DeclareMathOperator{lad}{Lad}$Let $(K,<)$ be an ordered field, and $(R,<)$ be the real closure of $(K,<)$. Let $lad(K)$ denote the ladder of $(K,<)$ (for the definition of the ladder, see the link given below). Then the canonical injection $iota colon lad(K) to lad(R)$ is induced. My first question is the following.



      Question 1. Is $iota colon lad(K) to lad(R)$ an isomorphism?



      I came up with this question when I read https://mathoverflow.net/a/23529/131576. According to the answer, Question 1 is true in the following case.



      Proposition. Let $L$ be a linear order. Write $F = bar{mathbb{Q}} cap mathbb{R}$. Suppose $(K,<)$ be the ordered field $F(x_l : l in L)$ as in the answer. Then $iota$ is an isomorphism.



      In fact, $(R,<)$ is contained in the field of Hahn series $F[[t^G]]$, where $G$ is the group $mathbb{Q}^{{oplus} L}$ with a suitable order. However, this proof seems a little bit hard in that it uses Hahn series. Then my second question is the following.



      Question 2. Are there any easier proofs of the Proposition?



      Any help would be appreciated.










      share|cite|improve this question















      $DeclareMathOperator{lad}{Lad}$Let $(K,<)$ be an ordered field, and $(R,<)$ be the real closure of $(K,<)$. Let $lad(K)$ denote the ladder of $(K,<)$ (for the definition of the ladder, see the link given below). Then the canonical injection $iota colon lad(K) to lad(R)$ is induced. My first question is the following.



      Question 1. Is $iota colon lad(K) to lad(R)$ an isomorphism?



      I came up with this question when I read https://mathoverflow.net/a/23529/131576. According to the answer, Question 1 is true in the following case.



      Proposition. Let $L$ be a linear order. Write $F = bar{mathbb{Q}} cap mathbb{R}$. Suppose $(K,<)$ be the ordered field $F(x_l : l in L)$ as in the answer. Then $iota$ is an isomorphism.



      In fact, $(R,<)$ is contained in the field of Hahn series $F[[t^G]]$, where $G$ is the group $mathbb{Q}^{{oplus} L}$ with a suitable order. However, this proof seems a little bit hard in that it uses Hahn series. Then my second question is the following.



      Question 2. Are there any easier proofs of the Proposition?



      Any help would be appreciated.







      abstract-algebra field-theory model-theory ordered-fields






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Nov 22 at 14:16

























      asked Nov 22 at 7:25









      unigermalto

      134




      134






















          1 Answer
          1






          active

          oldest

          votes

















          up vote
          1
          down vote



          accepted










          I don't know if there is an easy proof of $1$, but there is a somewhat elementary one using valuation theory and more specifically the natural valuation on an ordered field. The idea is that the value group of $R$ is the divisible hull of that of $K$, but here I only prove what's necessary for your result.



          So let $v$ denote the generic function of natural valuation on an ordered field, and let $operatorname{Lad}$ denote the generic function of comparability class on an ordered field. Notice that if $K$ is an ordered field and $x in K$ is positive infinite, then $operatorname{Lad}(y)=operatorname{Lad}(x)$ for all $y in v x$.
          (in fact, the Ladder functor is naturally isomorphic to the composition of the natural valuations functors on the categories of ordered fields and ordered groups).



          Claim: if $A$ is an algebraic (ordered field) extension of $K$ and $x in A$, we have $d_x:=[v K[x]: v K] <infty$.



          It is no secret that $R$ is an algebraic extension of $A$ so we may apply the claim. Let $x in R^{times}$ be positive infinite. We have $d_x.vK[x] subseteq v K$ so $d_x. v x =v x^{d_x}=v y$ for some element $y in K^{times}$ which must be infinite. We have $operatorname{Lad}(x)=operatorname{Lad}(x^{d_x})=operatorname{Lad}(y) in operatorname{Lad}(K)$, hence $operatorname{Lad}(R)=operatorname{Lad}(K)$.





          Proof of the claim: notice that every finite family $(gamma_i)_{1leq ileq n}$ in $Gamma_{K[x]}$ such that for $ineq j$ we have $gamma_i - gamma_j notin Gamma_K$, and for $x_1,...,x_n in K[x]^{times}$ with $forall 1 leq i leq n,gamma_i= v x_i$, the family $(x_1,...,x_n)$ is free in the $K$-vector space $K[x]$. Indeed for $(0,...,0)neq (a_1,...,a_n) in K^n$ and $1leq i neq j leq n$, we have $gamma_j - gamma_inotin v K ni v a_i - v a_j$ so $v a_i x_i =v a_i + gamma_ineq v a_j + gamma_j =v a_j x_j$. By standard properties of valuations, we have $v sum limits_{i=1}^n a_i x_i=min_{i=1}^n v a_i x_i$, and in particular the element $sum limits_{i=1}^n a_i x_i$ is non zero. The extension $K[x] / K$ is finite, so $[v K[x]: vK]$ must be finite.









          share|cite|improve this answer





















            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',
            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%2f3008838%2fthe-ladder-of-the-real-closure-of-an-ordered-field%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








            up vote
            1
            down vote



            accepted










            I don't know if there is an easy proof of $1$, but there is a somewhat elementary one using valuation theory and more specifically the natural valuation on an ordered field. The idea is that the value group of $R$ is the divisible hull of that of $K$, but here I only prove what's necessary for your result.



            So let $v$ denote the generic function of natural valuation on an ordered field, and let $operatorname{Lad}$ denote the generic function of comparability class on an ordered field. Notice that if $K$ is an ordered field and $x in K$ is positive infinite, then $operatorname{Lad}(y)=operatorname{Lad}(x)$ for all $y in v x$.
            (in fact, the Ladder functor is naturally isomorphic to the composition of the natural valuations functors on the categories of ordered fields and ordered groups).



            Claim: if $A$ is an algebraic (ordered field) extension of $K$ and $x in A$, we have $d_x:=[v K[x]: v K] <infty$.



            It is no secret that $R$ is an algebraic extension of $A$ so we may apply the claim. Let $x in R^{times}$ be positive infinite. We have $d_x.vK[x] subseteq v K$ so $d_x. v x =v x^{d_x}=v y$ for some element $y in K^{times}$ which must be infinite. We have $operatorname{Lad}(x)=operatorname{Lad}(x^{d_x})=operatorname{Lad}(y) in operatorname{Lad}(K)$, hence $operatorname{Lad}(R)=operatorname{Lad}(K)$.





            Proof of the claim: notice that every finite family $(gamma_i)_{1leq ileq n}$ in $Gamma_{K[x]}$ such that for $ineq j$ we have $gamma_i - gamma_j notin Gamma_K$, and for $x_1,...,x_n in K[x]^{times}$ with $forall 1 leq i leq n,gamma_i= v x_i$, the family $(x_1,...,x_n)$ is free in the $K$-vector space $K[x]$. Indeed for $(0,...,0)neq (a_1,...,a_n) in K^n$ and $1leq i neq j leq n$, we have $gamma_j - gamma_inotin v K ni v a_i - v a_j$ so $v a_i x_i =v a_i + gamma_ineq v a_j + gamma_j =v a_j x_j$. By standard properties of valuations, we have $v sum limits_{i=1}^n a_i x_i=min_{i=1}^n v a_i x_i$, and in particular the element $sum limits_{i=1}^n a_i x_i$ is non zero. The extension $K[x] / K$ is finite, so $[v K[x]: vK]$ must be finite.









            share|cite|improve this answer

























              up vote
              1
              down vote



              accepted










              I don't know if there is an easy proof of $1$, but there is a somewhat elementary one using valuation theory and more specifically the natural valuation on an ordered field. The idea is that the value group of $R$ is the divisible hull of that of $K$, but here I only prove what's necessary for your result.



              So let $v$ denote the generic function of natural valuation on an ordered field, and let $operatorname{Lad}$ denote the generic function of comparability class on an ordered field. Notice that if $K$ is an ordered field and $x in K$ is positive infinite, then $operatorname{Lad}(y)=operatorname{Lad}(x)$ for all $y in v x$.
              (in fact, the Ladder functor is naturally isomorphic to the composition of the natural valuations functors on the categories of ordered fields and ordered groups).



              Claim: if $A$ is an algebraic (ordered field) extension of $K$ and $x in A$, we have $d_x:=[v K[x]: v K] <infty$.



              It is no secret that $R$ is an algebraic extension of $A$ so we may apply the claim. Let $x in R^{times}$ be positive infinite. We have $d_x.vK[x] subseteq v K$ so $d_x. v x =v x^{d_x}=v y$ for some element $y in K^{times}$ which must be infinite. We have $operatorname{Lad}(x)=operatorname{Lad}(x^{d_x})=operatorname{Lad}(y) in operatorname{Lad}(K)$, hence $operatorname{Lad}(R)=operatorname{Lad}(K)$.





              Proof of the claim: notice that every finite family $(gamma_i)_{1leq ileq n}$ in $Gamma_{K[x]}$ such that for $ineq j$ we have $gamma_i - gamma_j notin Gamma_K$, and for $x_1,...,x_n in K[x]^{times}$ with $forall 1 leq i leq n,gamma_i= v x_i$, the family $(x_1,...,x_n)$ is free in the $K$-vector space $K[x]$. Indeed for $(0,...,0)neq (a_1,...,a_n) in K^n$ and $1leq i neq j leq n$, we have $gamma_j - gamma_inotin v K ni v a_i - v a_j$ so $v a_i x_i =v a_i + gamma_ineq v a_j + gamma_j =v a_j x_j$. By standard properties of valuations, we have $v sum limits_{i=1}^n a_i x_i=min_{i=1}^n v a_i x_i$, and in particular the element $sum limits_{i=1}^n a_i x_i$ is non zero. The extension $K[x] / K$ is finite, so $[v K[x]: vK]$ must be finite.









              share|cite|improve this answer























                up vote
                1
                down vote



                accepted







                up vote
                1
                down vote



                accepted






                I don't know if there is an easy proof of $1$, but there is a somewhat elementary one using valuation theory and more specifically the natural valuation on an ordered field. The idea is that the value group of $R$ is the divisible hull of that of $K$, but here I only prove what's necessary for your result.



                So let $v$ denote the generic function of natural valuation on an ordered field, and let $operatorname{Lad}$ denote the generic function of comparability class on an ordered field. Notice that if $K$ is an ordered field and $x in K$ is positive infinite, then $operatorname{Lad}(y)=operatorname{Lad}(x)$ for all $y in v x$.
                (in fact, the Ladder functor is naturally isomorphic to the composition of the natural valuations functors on the categories of ordered fields and ordered groups).



                Claim: if $A$ is an algebraic (ordered field) extension of $K$ and $x in A$, we have $d_x:=[v K[x]: v K] <infty$.



                It is no secret that $R$ is an algebraic extension of $A$ so we may apply the claim. Let $x in R^{times}$ be positive infinite. We have $d_x.vK[x] subseteq v K$ so $d_x. v x =v x^{d_x}=v y$ for some element $y in K^{times}$ which must be infinite. We have $operatorname{Lad}(x)=operatorname{Lad}(x^{d_x})=operatorname{Lad}(y) in operatorname{Lad}(K)$, hence $operatorname{Lad}(R)=operatorname{Lad}(K)$.





                Proof of the claim: notice that every finite family $(gamma_i)_{1leq ileq n}$ in $Gamma_{K[x]}$ such that for $ineq j$ we have $gamma_i - gamma_j notin Gamma_K$, and for $x_1,...,x_n in K[x]^{times}$ with $forall 1 leq i leq n,gamma_i= v x_i$, the family $(x_1,...,x_n)$ is free in the $K$-vector space $K[x]$. Indeed for $(0,...,0)neq (a_1,...,a_n) in K^n$ and $1leq i neq j leq n$, we have $gamma_j - gamma_inotin v K ni v a_i - v a_j$ so $v a_i x_i =v a_i + gamma_ineq v a_j + gamma_j =v a_j x_j$. By standard properties of valuations, we have $v sum limits_{i=1}^n a_i x_i=min_{i=1}^n v a_i x_i$, and in particular the element $sum limits_{i=1}^n a_i x_i$ is non zero. The extension $K[x] / K$ is finite, so $[v K[x]: vK]$ must be finite.









                share|cite|improve this answer












                I don't know if there is an easy proof of $1$, but there is a somewhat elementary one using valuation theory and more specifically the natural valuation on an ordered field. The idea is that the value group of $R$ is the divisible hull of that of $K$, but here I only prove what's necessary for your result.



                So let $v$ denote the generic function of natural valuation on an ordered field, and let $operatorname{Lad}$ denote the generic function of comparability class on an ordered field. Notice that if $K$ is an ordered field and $x in K$ is positive infinite, then $operatorname{Lad}(y)=operatorname{Lad}(x)$ for all $y in v x$.
                (in fact, the Ladder functor is naturally isomorphic to the composition of the natural valuations functors on the categories of ordered fields and ordered groups).



                Claim: if $A$ is an algebraic (ordered field) extension of $K$ and $x in A$, we have $d_x:=[v K[x]: v K] <infty$.



                It is no secret that $R$ is an algebraic extension of $A$ so we may apply the claim. Let $x in R^{times}$ be positive infinite. We have $d_x.vK[x] subseteq v K$ so $d_x. v x =v x^{d_x}=v y$ for some element $y in K^{times}$ which must be infinite. We have $operatorname{Lad}(x)=operatorname{Lad}(x^{d_x})=operatorname{Lad}(y) in operatorname{Lad}(K)$, hence $operatorname{Lad}(R)=operatorname{Lad}(K)$.





                Proof of the claim: notice that every finite family $(gamma_i)_{1leq ileq n}$ in $Gamma_{K[x]}$ such that for $ineq j$ we have $gamma_i - gamma_j notin Gamma_K$, and for $x_1,...,x_n in K[x]^{times}$ with $forall 1 leq i leq n,gamma_i= v x_i$, the family $(x_1,...,x_n)$ is free in the $K$-vector space $K[x]$. Indeed for $(0,...,0)neq (a_1,...,a_n) in K^n$ and $1leq i neq j leq n$, we have $gamma_j - gamma_inotin v K ni v a_i - v a_j$ so $v a_i x_i =v a_i + gamma_ineq v a_j + gamma_j =v a_j x_j$. By standard properties of valuations, we have $v sum limits_{i=1}^n a_i x_i=min_{i=1}^n v a_i x_i$, and in particular the element $sum limits_{i=1}^n a_i x_i$ is non zero. The extension $K[x] / K$ is finite, so $[v K[x]: vK]$ must be finite.










                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 22 at 17:33









                nombre

                2,559913




                2,559913






























                    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.





                    Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


                    Please pay close attention to the following guidance:


                    • 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.


                    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%2f3008838%2fthe-ladder-of-the-real-closure-of-an-ordered-field%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

                    Ellipse (mathématiques)

                    Quarter-circle Tiles

                    Mont Emei