Does the ring of regular expressions exist?












1














It is well known that the set of Regular Expressions R over some alphabet form a semiring with:




  • Concatenation as multiplication

  • The empty string as the multiplicative identity

  • 'Or' as addition

  • The empty set of strings as the additive identity


Q. Does there exist an extension of R to a ring?






abstract-algebra formal-languages







share|cite|improve this question






















  • Yes, but it's the zero ring. Union is idempotent, so every element satisfies $a + a = a$. In a ring this implies $a = 0$.
    – Qiaochu Yuan
    Nov 25 '18 at 9:25










  • Thanks. So that extension is therefore unique, right (i.e. it implies that no other extensions can exist)?
    – NietzscheanAI
    Nov 25 '18 at 9:41










  • I don't know what exactly you mean by "extension." There's a universal map from any semiring to a ring, namely the ring given by adjoining additive inverses to every element. In this case that ring and that map are zero.
    – Qiaochu Yuan
    Nov 25 '18 at 10:54










  • 'Extension' = 'alternative mapping'. In other words, is the zero ring the only ring that can be constructed from REs (given the above definition)?
    – NietzscheanAI
    Nov 25 '18 at 11:18










  • I don't know what you mean by "constructed from." In this case, the only semiring homomorphism from the semiring above to a ring is the homomorphism to the zero ring, because the zero ring doesn't admit homomorphisms to any nonzero rings. In general there will be many such homomorphisms, for example quotients of the universal one.
    – Qiaochu Yuan
    Nov 25 '18 at 19:02
















1














It is well known that the set of Regular Expressions R over some alphabet form a semiring with:




  • Concatenation as multiplication

  • The empty string as the multiplicative identity

  • 'Or' as addition

  • The empty set of strings as the additive identity


Q. Does there exist an extension of R to a ring?






abstract-algebra formal-languages







share|cite|improve this question






















  • Yes, but it's the zero ring. Union is idempotent, so every element satisfies $a + a = a$. In a ring this implies $a = 0$.
    – Qiaochu Yuan
    Nov 25 '18 at 9:25










  • Thanks. So that extension is therefore unique, right (i.e. it implies that no other extensions can exist)?
    – NietzscheanAI
    Nov 25 '18 at 9:41










  • I don't know what exactly you mean by "extension." There's a universal map from any semiring to a ring, namely the ring given by adjoining additive inverses to every element. In this case that ring and that map are zero.
    – Qiaochu Yuan
    Nov 25 '18 at 10:54










  • 'Extension' = 'alternative mapping'. In other words, is the zero ring the only ring that can be constructed from REs (given the above definition)?
    – NietzscheanAI
    Nov 25 '18 at 11:18










  • I don't know what you mean by "constructed from." In this case, the only semiring homomorphism from the semiring above to a ring is the homomorphism to the zero ring, because the zero ring doesn't admit homomorphisms to any nonzero rings. In general there will be many such homomorphisms, for example quotients of the universal one.
    – Qiaochu Yuan
    Nov 25 '18 at 19:02














1












1








1







It is well known that the set of Regular Expressions R over some alphabet form a semiring with:




  • Concatenation as multiplication

  • The empty string as the multiplicative identity

  • 'Or' as addition

  • The empty set of strings as the additive identity


Q. Does there exist an extension of R to a ring?






abstract-algebra formal-languages







share|cite|improve this question













It is well known that the set of Regular Expressions R over some alphabet form a semiring with:




  • Concatenation as multiplication

  • The empty string as the multiplicative identity

  • 'Or' as addition

  • The empty set of strings as the additive identity


Q. Does there exist an extension of R to a ring?






abstract-algebra formal-languages




abstract-algebra formal-languages






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 25 '18 at 9:12









NietzscheanAI

19510




19510












  • Yes, but it's the zero ring. Union is idempotent, so every element satisfies $a + a = a$. In a ring this implies $a = 0$.
    – Qiaochu Yuan
    Nov 25 '18 at 9:25










  • Thanks. So that extension is therefore unique, right (i.e. it implies that no other extensions can exist)?
    – NietzscheanAI
    Nov 25 '18 at 9:41










  • I don't know what exactly you mean by "extension." There's a universal map from any semiring to a ring, namely the ring given by adjoining additive inverses to every element. In this case that ring and that map are zero.
    – Qiaochu Yuan
    Nov 25 '18 at 10:54










  • 'Extension' = 'alternative mapping'. In other words, is the zero ring the only ring that can be constructed from REs (given the above definition)?
    – NietzscheanAI
    Nov 25 '18 at 11:18










  • I don't know what you mean by "constructed from." In this case, the only semiring homomorphism from the semiring above to a ring is the homomorphism to the zero ring, because the zero ring doesn't admit homomorphisms to any nonzero rings. In general there will be many such homomorphisms, for example quotients of the universal one.
    – Qiaochu Yuan
    Nov 25 '18 at 19:02


















  • Yes, but it's the zero ring. Union is idempotent, so every element satisfies $a + a = a$. In a ring this implies $a = 0$.
    – Qiaochu Yuan
    Nov 25 '18 at 9:25










  • Thanks. So that extension is therefore unique, right (i.e. it implies that no other extensions can exist)?
    – NietzscheanAI
    Nov 25 '18 at 9:41










  • I don't know what exactly you mean by "extension." There's a universal map from any semiring to a ring, namely the ring given by adjoining additive inverses to every element. In this case that ring and that map are zero.
    – Qiaochu Yuan
    Nov 25 '18 at 10:54










  • 'Extension' = 'alternative mapping'. In other words, is the zero ring the only ring that can be constructed from REs (given the above definition)?
    – NietzscheanAI
    Nov 25 '18 at 11:18










  • I don't know what you mean by "constructed from." In this case, the only semiring homomorphism from the semiring above to a ring is the homomorphism to the zero ring, because the zero ring doesn't admit homomorphisms to any nonzero rings. In general there will be many such homomorphisms, for example quotients of the universal one.
    – Qiaochu Yuan
    Nov 25 '18 at 19:02
















Yes, but it's the zero ring. Union is idempotent, so every element satisfies $a + a = a$. In a ring this implies $a = 0$.
– Qiaochu Yuan
Nov 25 '18 at 9:25




Yes, but it's the zero ring. Union is idempotent, so every element satisfies $a + a = a$. In a ring this implies $a = 0$.
– Qiaochu Yuan
Nov 25 '18 at 9:25












Thanks. So that extension is therefore unique, right (i.e. it implies that no other extensions can exist)?
– NietzscheanAI
Nov 25 '18 at 9:41




Thanks. So that extension is therefore unique, right (i.e. it implies that no other extensions can exist)?
– NietzscheanAI
Nov 25 '18 at 9:41












I don't know what exactly you mean by "extension." There's a universal map from any semiring to a ring, namely the ring given by adjoining additive inverses to every element. In this case that ring and that map are zero.
– Qiaochu Yuan
Nov 25 '18 at 10:54




I don't know what exactly you mean by "extension." There's a universal map from any semiring to a ring, namely the ring given by adjoining additive inverses to every element. In this case that ring and that map are zero.
– Qiaochu Yuan
Nov 25 '18 at 10:54












'Extension' = 'alternative mapping'. In other words, is the zero ring the only ring that can be constructed from REs (given the above definition)?
– NietzscheanAI
Nov 25 '18 at 11:18




'Extension' = 'alternative mapping'. In other words, is the zero ring the only ring that can be constructed from REs (given the above definition)?
– NietzscheanAI
Nov 25 '18 at 11:18












I don't know what you mean by "constructed from." In this case, the only semiring homomorphism from the semiring above to a ring is the homomorphism to the zero ring, because the zero ring doesn't admit homomorphisms to any nonzero rings. In general there will be many such homomorphisms, for example quotients of the universal one.
– Qiaochu Yuan
Nov 25 '18 at 19:02




I don't know what you mean by "constructed from." In this case, the only semiring homomorphism from the semiring above to a ring is the homomorphism to the zero ring, because the zero ring doesn't admit homomorphisms to any nonzero rings. In general there will be many such homomorphisms, for example quotients of the universal one.
– Qiaochu Yuan
Nov 25 '18 at 19:02










2 Answers
2






active

oldest

votes


















2














The zero element is the empty set (i.e., empty language) and the unit element contains as its sole element the empty string.
A ring requires to have an additive inverse for each element. But this is not the case when the operation is union of languages (i.e., union as sets).






share|cite|improve this answer





























    2














    Regular languages on an alphabet $A$ form a semiring with union as addition (and the empty set as $0$) and concatenation product as product (and the language reduced to the empty word as $1$). This semiring is noncommutative if and only if $|A| > 1$.



    This semiring can be identified with the semiring of rational series over the Boolean semiring $mathbb{B} = {0, 1}$. As a set, this semiring can be embedded into the ring of rational series over the ring $mathbb{Z}$, but this embedding is not a semiring embedding since in $mathbb{B}$, $1 + 1 = 1$.
    A very good reference on this topic is [1].



    [1] J. Berstel and C. Reutenauer, Noncommutative rational series with applications, Encyclopedia of Mathematics and its Applications, 137. Cambridge University Press, Cambridge, 2011. xiv+248 pp. ISBN 978-0-521-19022-0






    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',
      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%2f3012608%2fdoes-the-ring-of-regular-expressions-exist%23new-answer', 'question_page');
      }
      );

      Post as a guest















      Required, but never shown

























      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      2














      The zero element is the empty set (i.e., empty language) and the unit element contains as its sole element the empty string.
      A ring requires to have an additive inverse for each element. But this is not the case when the operation is union of languages (i.e., union as sets).






      share|cite|improve this answer


























        2














        The zero element is the empty set (i.e., empty language) and the unit element contains as its sole element the empty string.
        A ring requires to have an additive inverse for each element. But this is not the case when the operation is union of languages (i.e., union as sets).






        share|cite|improve this answer
























          2












          2








          2






          The zero element is the empty set (i.e., empty language) and the unit element contains as its sole element the empty string.
          A ring requires to have an additive inverse for each element. But this is not the case when the operation is union of languages (i.e., union as sets).






          share|cite|improve this answer












          The zero element is the empty set (i.e., empty language) and the unit element contains as its sole element the empty string.
          A ring requires to have an additive inverse for each element. But this is not the case when the operation is union of languages (i.e., union as sets).







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 25 '18 at 9:21









          Wuestenfux

          3,5531411




          3,5531411























              2














              Regular languages on an alphabet $A$ form a semiring with union as addition (and the empty set as $0$) and concatenation product as product (and the language reduced to the empty word as $1$). This semiring is noncommutative if and only if $|A| > 1$.



              This semiring can be identified with the semiring of rational series over the Boolean semiring $mathbb{B} = {0, 1}$. As a set, this semiring can be embedded into the ring of rational series over the ring $mathbb{Z}$, but this embedding is not a semiring embedding since in $mathbb{B}$, $1 + 1 = 1$.
              A very good reference on this topic is [1].



              [1] J. Berstel and C. Reutenauer, Noncommutative rational series with applications, Encyclopedia of Mathematics and its Applications, 137. Cambridge University Press, Cambridge, 2011. xiv+248 pp. ISBN 978-0-521-19022-0






              share|cite|improve this answer


























                2














                Regular languages on an alphabet $A$ form a semiring with union as addition (and the empty set as $0$) and concatenation product as product (and the language reduced to the empty word as $1$). This semiring is noncommutative if and only if $|A| > 1$.



                This semiring can be identified with the semiring of rational series over the Boolean semiring $mathbb{B} = {0, 1}$. As a set, this semiring can be embedded into the ring of rational series over the ring $mathbb{Z}$, but this embedding is not a semiring embedding since in $mathbb{B}$, $1 + 1 = 1$.
                A very good reference on this topic is [1].



                [1] J. Berstel and C. Reutenauer, Noncommutative rational series with applications, Encyclopedia of Mathematics and its Applications, 137. Cambridge University Press, Cambridge, 2011. xiv+248 pp. ISBN 978-0-521-19022-0






                share|cite|improve this answer
























                  2












                  2








                  2






                  Regular languages on an alphabet $A$ form a semiring with union as addition (and the empty set as $0$) and concatenation product as product (and the language reduced to the empty word as $1$). This semiring is noncommutative if and only if $|A| > 1$.



                  This semiring can be identified with the semiring of rational series over the Boolean semiring $mathbb{B} = {0, 1}$. As a set, this semiring can be embedded into the ring of rational series over the ring $mathbb{Z}$, but this embedding is not a semiring embedding since in $mathbb{B}$, $1 + 1 = 1$.
                  A very good reference on this topic is [1].



                  [1] J. Berstel and C. Reutenauer, Noncommutative rational series with applications, Encyclopedia of Mathematics and its Applications, 137. Cambridge University Press, Cambridge, 2011. xiv+248 pp. ISBN 978-0-521-19022-0






                  share|cite|improve this answer












                  Regular languages on an alphabet $A$ form a semiring with union as addition (and the empty set as $0$) and concatenation product as product (and the language reduced to the empty word as $1$). This semiring is noncommutative if and only if $|A| > 1$.



                  This semiring can be identified with the semiring of rational series over the Boolean semiring $mathbb{B} = {0, 1}$. As a set, this semiring can be embedded into the ring of rational series over the ring $mathbb{Z}$, but this embedding is not a semiring embedding since in $mathbb{B}$, $1 + 1 = 1$.
                  A very good reference on this topic is [1].



                  [1] J. Berstel and C. Reutenauer, Noncommutative rational series with applications, Encyclopedia of Mathematics and its Applications, 137. Cambridge University Press, Cambridge, 2011. xiv+248 pp. ISBN 978-0-521-19022-0







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 28 '18 at 16:57









                  J.-E. Pin

                  18.3k21754




                  18.3k21754






























                      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%2f3012608%2fdoes-the-ring-of-regular-expressions-exist%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

                      Plaza Victoria

                      Image
                      Plaza Victoria Plaza Victoria en 2017 Otros nombres Plaza de La Victoria Localización El Almendral, Valparaíso, Chile Vías adyacentes Chacabuco, Molina, Plaza Victoria y Edwards Creación 1841 Metro Bellavista Ubicación 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 Coordenadas: 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 [editar datos en Wikidata] La Plaza Victoria , también llamada Plaza de la Victoria , [ 1 ] ​ es una plaza ubicada en el sector El Almendral del plan de la ciudad de Valparaíso, Chile, de importancia patrimonial, que incluye valiosas estatuas y especies vegetales. [ 2 ] ​ La plaza se creó como tal en la primera mitad del siglo XIX, tomando previamente los nombres de Plaza Nueva y Plaza de Orrego . Si bien desde sus inicios tuvo relevancia histórica, no fue sino a partir de fines de los años 1860 cuando comenzó a cobrar mayor importancia y se convirtió en uno de los principales centros sociales de l...
                      Read more

                      Brian Clough

                      Image
                      Brian Clough Datos personales Nombre completo Brian Howard Clough Apodo(s) Old Big 'Ead, Cloughie [ 1 ] ​ Nacimiento Middlesbrough, Inglaterra, Reino Unido 21 de marzo de 1935 Nacionalidad(es) Fallecimiento Derby, Reino Unido 20 de septiembre de 2004 (69 años) Altura 1,78 [ 1 ] ​ metros Carrera Deporte Fútbol Debut deportivo 1955 (Como jugador) 1965 (Como entrenador) (Middlesbrough F. C. (Como jugador) Hartlepool United F. C. (Como entrenador) ) Posición Delantero Goles en clubes 269 [ 2 ] ​ Retirada deportiva 1964 (Como jugador) 1993 (Como entrenador) (Sunderland A. F. C. (Como jugador) Nottingham Forest (Como entrenador) ) Carrera internacional Part. 2 [editar datos en Wikidata] Brian Howard Clough , OBE [ 3 ] ​ (Middlesbrough, Inglaterra, Reino Unido, 21 de marzo de 1935 – Derby, Inglaterra, Reino Unido, 20 de septiembre de 2004) fue un jugador y entrenador británico de fútbol. Es conocido por...
                      Read more

                      Cáceres

                      Image
                      Para otros usos de este término, véase Cáceres (desambiguación). Cáceres Municipio y ciudad Bandera Escudo Collage de Cáceres Cáceres Ubicación de Cáceres en España. Cáceres Ubicación de Cáceres en la provincia de Cáceres. País  España • Com. autónoma  Extremadura • Provincia  Cáceres • Partido judicial Cáceres Ubicación 39°28′23″N 6°22′16″O  /  39.473055555556, -6.3711111111111 Coordenadas: 39°28′23″N 6°22′16″O  /  39.473055555556, -6.3711111111111 • Altitud 457 [ 1 ] ​ msnm (mín.:211 [ 2 ] ​, máx.:708 [ 2 ] ​) Superficie 1750,33 km² Núcleos de población Cáceres Estación Arroyo-Malpartida Rincón de Ballesteros Valdesalor Fundación 34 a. C. Población 96 068 hab. (2018) • Densidad 54,8 hab./km² Gentilicio Cacereño/a [ 3 ] ​ Código postal 10001-10005, 10195 Alcaldesa (2015) María Elena Nevado del Campo [ 4 ] ​ Presupuesto 69.045.004€ [ 5 ] ​  (año ...
                      Read more