Coordinate ring of a scheme in functorial algebraic geometry











up vote
4
down vote

favorite
1












I will preface this by saying that I am new to algebraic geometry, but I am somewhat experienced with category theory.



I'm just reading the introduction to Milne's notes "Basic Theory of Affine Group Schemes". He uses the functorial point of view here, so I am viewing an affine scheme over $K$ as a representable functor $X: Kmathsf{Alg} to mathsf{Sets}$, and a scheme is likewise defined as a functor satisfying appropriate gluing properties. We can think of some general functor as a generalized scheme.



In section I.3 he has a subsection titled "The canonical coordinate ring of an affine group" but I noticed that his construction seems to define a canonical "coordinate ring" for every sort of "generalized scheme", not just affine group schemes. Indeed, if $X: Kmathsf{Alg} to mathsf{Sets}$ is a functor then $mathrm{Nat}(X, mathbb{A}^1_K)$ is a $K$-algebra (with operations defined pointwise), since the affine line over $K$ is the forgetful functor $mathbb{A}^1_K: Kmathsf{Alg} to mathsf{Sets}$.



So we have a functor $mathsf{Sets}^{Kmathsf{Alg}} to Kmathsf{Alg}$ defined by $X mapsto mathrm{Nat}(X, mathbb{A}^1_K)$.



Moreover, we have an obvious natural transformation $alpha: X to mathrm{Spec_K}(mathrm{Nat}(X, mathbb{A}^1_K))$,
where $mathrm{Spec}_K$ here is just the contravariant Yoneda embedding (since I am thinking of affine schemes as functors rather than ringed spaces). This natural transformation has components $alpha_A: X(A) to mathrm{Hom}(mathrm{Nat}(X, mathbb{A}^1_K), A)$ given by $x mapsto (f mapsto f_A(x))$.



My question is:





  1. Is it reasonable to call $mathrm{Nat}(X, mathbb{A}^1_K), A)$ the coordinate ring for any "generalize scheme" given by a functor $X: Kmathsf{Alg} to mathsf{Sets}$? If not, what should we call this?


  2. Is the functor $mathsf{Sets}^{Kmathsf{Alg}} to Kmathsf{Alg}$ mapping $X$ to $mathrm{Nat}(X, mathbb{A}^1_K)$ adjoint (on the left or right) to $mathrm{Spec}_K: Kmathsf{Alg}^{mathrm{opp}} to mathsf{Sets}^{Kmathsf{Alg}}$? My guess is that it is the left adjoint to $mathrm{Spec}_K$.


  3. Is there a name and interpretation for this natural transformation $alpha_A: X(A) to mathrm{Hom}(mathrm{Nat}(X, mathbb{A}^1_K), A)$? I can see that $X$ is an affine scheme over $K$ if and only if this is an isomorphism. But what if $X$ is not affine? How do we interpret this?












share|cite|improve this question




























    up vote
    4
    down vote

    favorite
    1












    I will preface this by saying that I am new to algebraic geometry, but I am somewhat experienced with category theory.



    I'm just reading the introduction to Milne's notes "Basic Theory of Affine Group Schemes". He uses the functorial point of view here, so I am viewing an affine scheme over $K$ as a representable functor $X: Kmathsf{Alg} to mathsf{Sets}$, and a scheme is likewise defined as a functor satisfying appropriate gluing properties. We can think of some general functor as a generalized scheme.



    In section I.3 he has a subsection titled "The canonical coordinate ring of an affine group" but I noticed that his construction seems to define a canonical "coordinate ring" for every sort of "generalized scheme", not just affine group schemes. Indeed, if $X: Kmathsf{Alg} to mathsf{Sets}$ is a functor then $mathrm{Nat}(X, mathbb{A}^1_K)$ is a $K$-algebra (with operations defined pointwise), since the affine line over $K$ is the forgetful functor $mathbb{A}^1_K: Kmathsf{Alg} to mathsf{Sets}$.



    So we have a functor $mathsf{Sets}^{Kmathsf{Alg}} to Kmathsf{Alg}$ defined by $X mapsto mathrm{Nat}(X, mathbb{A}^1_K)$.



    Moreover, we have an obvious natural transformation $alpha: X to mathrm{Spec_K}(mathrm{Nat}(X, mathbb{A}^1_K))$,
    where $mathrm{Spec}_K$ here is just the contravariant Yoneda embedding (since I am thinking of affine schemes as functors rather than ringed spaces). This natural transformation has components $alpha_A: X(A) to mathrm{Hom}(mathrm{Nat}(X, mathbb{A}^1_K), A)$ given by $x mapsto (f mapsto f_A(x))$.



    My question is:





    1. Is it reasonable to call $mathrm{Nat}(X, mathbb{A}^1_K), A)$ the coordinate ring for any "generalize scheme" given by a functor $X: Kmathsf{Alg} to mathsf{Sets}$? If not, what should we call this?


    2. Is the functor $mathsf{Sets}^{Kmathsf{Alg}} to Kmathsf{Alg}$ mapping $X$ to $mathrm{Nat}(X, mathbb{A}^1_K)$ adjoint (on the left or right) to $mathrm{Spec}_K: Kmathsf{Alg}^{mathrm{opp}} to mathsf{Sets}^{Kmathsf{Alg}}$? My guess is that it is the left adjoint to $mathrm{Spec}_K$.


    3. Is there a name and interpretation for this natural transformation $alpha_A: X(A) to mathrm{Hom}(mathrm{Nat}(X, mathbb{A}^1_K), A)$? I can see that $X$ is an affine scheme over $K$ if and only if this is an isomorphism. But what if $X$ is not affine? How do we interpret this?












    share|cite|improve this question


























      up vote
      4
      down vote

      favorite
      1









      up vote
      4
      down vote

      favorite
      1






      1





      I will preface this by saying that I am new to algebraic geometry, but I am somewhat experienced with category theory.



      I'm just reading the introduction to Milne's notes "Basic Theory of Affine Group Schemes". He uses the functorial point of view here, so I am viewing an affine scheme over $K$ as a representable functor $X: Kmathsf{Alg} to mathsf{Sets}$, and a scheme is likewise defined as a functor satisfying appropriate gluing properties. We can think of some general functor as a generalized scheme.



      In section I.3 he has a subsection titled "The canonical coordinate ring of an affine group" but I noticed that his construction seems to define a canonical "coordinate ring" for every sort of "generalized scheme", not just affine group schemes. Indeed, if $X: Kmathsf{Alg} to mathsf{Sets}$ is a functor then $mathrm{Nat}(X, mathbb{A}^1_K)$ is a $K$-algebra (with operations defined pointwise), since the affine line over $K$ is the forgetful functor $mathbb{A}^1_K: Kmathsf{Alg} to mathsf{Sets}$.



      So we have a functor $mathsf{Sets}^{Kmathsf{Alg}} to Kmathsf{Alg}$ defined by $X mapsto mathrm{Nat}(X, mathbb{A}^1_K)$.



      Moreover, we have an obvious natural transformation $alpha: X to mathrm{Spec_K}(mathrm{Nat}(X, mathbb{A}^1_K))$,
      where $mathrm{Spec}_K$ here is just the contravariant Yoneda embedding (since I am thinking of affine schemes as functors rather than ringed spaces). This natural transformation has components $alpha_A: X(A) to mathrm{Hom}(mathrm{Nat}(X, mathbb{A}^1_K), A)$ given by $x mapsto (f mapsto f_A(x))$.



      My question is:





      1. Is it reasonable to call $mathrm{Nat}(X, mathbb{A}^1_K), A)$ the coordinate ring for any "generalize scheme" given by a functor $X: Kmathsf{Alg} to mathsf{Sets}$? If not, what should we call this?


      2. Is the functor $mathsf{Sets}^{Kmathsf{Alg}} to Kmathsf{Alg}$ mapping $X$ to $mathrm{Nat}(X, mathbb{A}^1_K)$ adjoint (on the left or right) to $mathrm{Spec}_K: Kmathsf{Alg}^{mathrm{opp}} to mathsf{Sets}^{Kmathsf{Alg}}$? My guess is that it is the left adjoint to $mathrm{Spec}_K$.


      3. Is there a name and interpretation for this natural transformation $alpha_A: X(A) to mathrm{Hom}(mathrm{Nat}(X, mathbb{A}^1_K), A)$? I can see that $X$ is an affine scheme over $K$ if and only if this is an isomorphism. But what if $X$ is not affine? How do we interpret this?












      share|cite|improve this question















      I will preface this by saying that I am new to algebraic geometry, but I am somewhat experienced with category theory.



      I'm just reading the introduction to Milne's notes "Basic Theory of Affine Group Schemes". He uses the functorial point of view here, so I am viewing an affine scheme over $K$ as a representable functor $X: Kmathsf{Alg} to mathsf{Sets}$, and a scheme is likewise defined as a functor satisfying appropriate gluing properties. We can think of some general functor as a generalized scheme.



      In section I.3 he has a subsection titled "The canonical coordinate ring of an affine group" but I noticed that his construction seems to define a canonical "coordinate ring" for every sort of "generalized scheme", not just affine group schemes. Indeed, if $X: Kmathsf{Alg} to mathsf{Sets}$ is a functor then $mathrm{Nat}(X, mathbb{A}^1_K)$ is a $K$-algebra (with operations defined pointwise), since the affine line over $K$ is the forgetful functor $mathbb{A}^1_K: Kmathsf{Alg} to mathsf{Sets}$.



      So we have a functor $mathsf{Sets}^{Kmathsf{Alg}} to Kmathsf{Alg}$ defined by $X mapsto mathrm{Nat}(X, mathbb{A}^1_K)$.



      Moreover, we have an obvious natural transformation $alpha: X to mathrm{Spec_K}(mathrm{Nat}(X, mathbb{A}^1_K))$,
      where $mathrm{Spec}_K$ here is just the contravariant Yoneda embedding (since I am thinking of affine schemes as functors rather than ringed spaces). This natural transformation has components $alpha_A: X(A) to mathrm{Hom}(mathrm{Nat}(X, mathbb{A}^1_K), A)$ given by $x mapsto (f mapsto f_A(x))$.



      My question is:





      1. Is it reasonable to call $mathrm{Nat}(X, mathbb{A}^1_K), A)$ the coordinate ring for any "generalize scheme" given by a functor $X: Kmathsf{Alg} to mathsf{Sets}$? If not, what should we call this?


      2. Is the functor $mathsf{Sets}^{Kmathsf{Alg}} to Kmathsf{Alg}$ mapping $X$ to $mathrm{Nat}(X, mathbb{A}^1_K)$ adjoint (on the left or right) to $mathrm{Spec}_K: Kmathsf{Alg}^{mathrm{opp}} to mathsf{Sets}^{Kmathsf{Alg}}$? My guess is that it is the left adjoint to $mathrm{Spec}_K$.


      3. Is there a name and interpretation for this natural transformation $alpha_A: X(A) to mathrm{Hom}(mathrm{Nat}(X, mathbb{A}^1_K), A)$? I can see that $X$ is an affine scheme over $K$ if and only if this is an isomorphism. But what if $X$ is not affine? How do we interpret this?









      algebraic-geometry ring-theory category-theory schemes algebraic-groups






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited 1 hour ago

























      asked 3 hours ago









      ಠ_ಠ

      5,34221242




      5,34221242






















          2 Answers
          2






          active

          oldest

          votes

















          up vote
          3
          down vote



          accepted











          1. Yes.


          2. The opposites confuse me about which of "left" and "right" I'm supposed to say. It should be left with the correct choice of ops.


          3. $alpha$ deserves to be called "affinization." It's the universal map from a scheme or generalized scheme into an affine scheme; that is, it's the left adjoint of the inclusion of affine schemes into schemes / generalized schemes. As a simple example, the affinization of projective space is a point.







          share|cite|improve this answer





















          • Thank you very much for your answer!
            – ಠ_ಠ
            2 hours ago










          • I am also curious: it seems like this construction works if we replace $K$-algebras by any sort of algebraic category, like say groups or non-commutative algebras. Is this a reasonable approach to non-commutative geometry?
            – ಠ_ಠ
            1 hour ago










          • ಠ_ಠ: there's not really anything here to do geometry with.
            – Qiaochu Yuan
            1 hour ago


















          up vote
          1
          down vote













          "Indeed, if $X:KAlg→Sets$ is a functor then $Nat(X,mathbb{A}^1_K)$ is a $K$-algebra (with operations defined pointwise),'' Not quite --- it may be a proper class (i.e., the underlying "set" may not be a set).






          share|cite|improve this answer





















          • Hmm...good point. Maybe somehow it follows from the affine line $mathbb{A}^1_K$ being representable? Otherwise maybe we can use a Grothendieck Universe to fix it (I don't know much about this to be honest).
            – ಠ_ಠ
            1 hour ago






          • 1




            We can require $X$ to be small (a small colimit of representables); I think that should fix it.
            – Qiaochu Yuan
            1 hour ago











          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%2f3037533%2fcoordinate-ring-of-a-scheme-in-functorial-algebraic-geometry%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








          up vote
          3
          down vote



          accepted











          1. Yes.


          2. The opposites confuse me about which of "left" and "right" I'm supposed to say. It should be left with the correct choice of ops.


          3. $alpha$ deserves to be called "affinization." It's the universal map from a scheme or generalized scheme into an affine scheme; that is, it's the left adjoint of the inclusion of affine schemes into schemes / generalized schemes. As a simple example, the affinization of projective space is a point.







          share|cite|improve this answer





















          • Thank you very much for your answer!
            – ಠ_ಠ
            2 hours ago










          • I am also curious: it seems like this construction works if we replace $K$-algebras by any sort of algebraic category, like say groups or non-commutative algebras. Is this a reasonable approach to non-commutative geometry?
            – ಠ_ಠ
            1 hour ago










          • ಠ_ಠ: there's not really anything here to do geometry with.
            – Qiaochu Yuan
            1 hour ago















          up vote
          3
          down vote



          accepted











          1. Yes.


          2. The opposites confuse me about which of "left" and "right" I'm supposed to say. It should be left with the correct choice of ops.


          3. $alpha$ deserves to be called "affinization." It's the universal map from a scheme or generalized scheme into an affine scheme; that is, it's the left adjoint of the inclusion of affine schemes into schemes / generalized schemes. As a simple example, the affinization of projective space is a point.







          share|cite|improve this answer





















          • Thank you very much for your answer!
            – ಠ_ಠ
            2 hours ago










          • I am also curious: it seems like this construction works if we replace $K$-algebras by any sort of algebraic category, like say groups or non-commutative algebras. Is this a reasonable approach to non-commutative geometry?
            – ಠ_ಠ
            1 hour ago










          • ಠ_ಠ: there's not really anything here to do geometry with.
            – Qiaochu Yuan
            1 hour ago













          up vote
          3
          down vote



          accepted







          up vote
          3
          down vote



          accepted







          1. Yes.


          2. The opposites confuse me about which of "left" and "right" I'm supposed to say. It should be left with the correct choice of ops.


          3. $alpha$ deserves to be called "affinization." It's the universal map from a scheme or generalized scheme into an affine scheme; that is, it's the left adjoint of the inclusion of affine schemes into schemes / generalized schemes. As a simple example, the affinization of projective space is a point.







          share|cite|improve this answer













          1. Yes.


          2. The opposites confuse me about which of "left" and "right" I'm supposed to say. It should be left with the correct choice of ops.


          3. $alpha$ deserves to be called "affinization." It's the universal map from a scheme or generalized scheme into an affine scheme; that is, it's the left adjoint of the inclusion of affine schemes into schemes / generalized schemes. As a simple example, the affinization of projective space is a point.








          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 2 hours ago









          Qiaochu Yuan

          276k32579916




          276k32579916












          • Thank you very much for your answer!
            – ಠ_ಠ
            2 hours ago










          • I am also curious: it seems like this construction works if we replace $K$-algebras by any sort of algebraic category, like say groups or non-commutative algebras. Is this a reasonable approach to non-commutative geometry?
            – ಠ_ಠ
            1 hour ago










          • ಠ_ಠ: there's not really anything here to do geometry with.
            – Qiaochu Yuan
            1 hour ago


















          • Thank you very much for your answer!
            – ಠ_ಠ
            2 hours ago










          • I am also curious: it seems like this construction works if we replace $K$-algebras by any sort of algebraic category, like say groups or non-commutative algebras. Is this a reasonable approach to non-commutative geometry?
            – ಠ_ಠ
            1 hour ago










          • ಠ_ಠ: there's not really anything here to do geometry with.
            – Qiaochu Yuan
            1 hour ago
















          Thank you very much for your answer!
          – ಠ_ಠ
          2 hours ago




          Thank you very much for your answer!
          – ಠ_ಠ
          2 hours ago












          I am also curious: it seems like this construction works if we replace $K$-algebras by any sort of algebraic category, like say groups or non-commutative algebras. Is this a reasonable approach to non-commutative geometry?
          – ಠ_ಠ
          1 hour ago




          I am also curious: it seems like this construction works if we replace $K$-algebras by any sort of algebraic category, like say groups or non-commutative algebras. Is this a reasonable approach to non-commutative geometry?
          – ಠ_ಠ
          1 hour ago












          ಠ_ಠ: there's not really anything here to do geometry with.
          – Qiaochu Yuan
          1 hour ago




          ಠ_ಠ: there's not really anything here to do geometry with.
          – Qiaochu Yuan
          1 hour ago










          up vote
          1
          down vote













          "Indeed, if $X:KAlg→Sets$ is a functor then $Nat(X,mathbb{A}^1_K)$ is a $K$-algebra (with operations defined pointwise),'' Not quite --- it may be a proper class (i.e., the underlying "set" may not be a set).






          share|cite|improve this answer





















          • Hmm...good point. Maybe somehow it follows from the affine line $mathbb{A}^1_K$ being representable? Otherwise maybe we can use a Grothendieck Universe to fix it (I don't know much about this to be honest).
            – ಠ_ಠ
            1 hour ago






          • 1




            We can require $X$ to be small (a small colimit of representables); I think that should fix it.
            – Qiaochu Yuan
            1 hour ago















          up vote
          1
          down vote













          "Indeed, if $X:KAlg→Sets$ is a functor then $Nat(X,mathbb{A}^1_K)$ is a $K$-algebra (with operations defined pointwise),'' Not quite --- it may be a proper class (i.e., the underlying "set" may not be a set).






          share|cite|improve this answer





















          • Hmm...good point. Maybe somehow it follows from the affine line $mathbb{A}^1_K$ being representable? Otherwise maybe we can use a Grothendieck Universe to fix it (I don't know much about this to be honest).
            – ಠ_ಠ
            1 hour ago






          • 1




            We can require $X$ to be small (a small colimit of representables); I think that should fix it.
            – Qiaochu Yuan
            1 hour ago













          up vote
          1
          down vote










          up vote
          1
          down vote









          "Indeed, if $X:KAlg→Sets$ is a functor then $Nat(X,mathbb{A}^1_K)$ is a $K$-algebra (with operations defined pointwise),'' Not quite --- it may be a proper class (i.e., the underlying "set" may not be a set).






          share|cite|improve this answer












          "Indeed, if $X:KAlg→Sets$ is a functor then $Nat(X,mathbb{A}^1_K)$ is a $K$-algebra (with operations defined pointwise),'' Not quite --- it may be a proper class (i.e., the underlying "set" may not be a set).







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 2 hours ago









          anon

          462




          462












          • Hmm...good point. Maybe somehow it follows from the affine line $mathbb{A}^1_K$ being representable? Otherwise maybe we can use a Grothendieck Universe to fix it (I don't know much about this to be honest).
            – ಠ_ಠ
            1 hour ago






          • 1




            We can require $X$ to be small (a small colimit of representables); I think that should fix it.
            – Qiaochu Yuan
            1 hour ago


















          • Hmm...good point. Maybe somehow it follows from the affine line $mathbb{A}^1_K$ being representable? Otherwise maybe we can use a Grothendieck Universe to fix it (I don't know much about this to be honest).
            – ಠ_ಠ
            1 hour ago






          • 1




            We can require $X$ to be small (a small colimit of representables); I think that should fix it.
            – Qiaochu Yuan
            1 hour ago
















          Hmm...good point. Maybe somehow it follows from the affine line $mathbb{A}^1_K$ being representable? Otherwise maybe we can use a Grothendieck Universe to fix it (I don't know much about this to be honest).
          – ಠ_ಠ
          1 hour ago




          Hmm...good point. Maybe somehow it follows from the affine line $mathbb{A}^1_K$ being representable? Otherwise maybe we can use a Grothendieck Universe to fix it (I don't know much about this to be honest).
          – ಠ_ಠ
          1 hour ago




          1




          1




          We can require $X$ to be small (a small colimit of representables); I think that should fix it.
          – Qiaochu Yuan
          1 hour ago




          We can require $X$ to be small (a small colimit of representables); I think that should fix it.
          – Qiaochu Yuan
          1 hour ago


















          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%2f3037533%2fcoordinate-ring-of-a-scheme-in-functorial-algebraic-geometry%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

          404 Error Contact Form 7 ajax form submitting

          How to know if a Active Directory user can login interactively

          TypeError: fit_transform() missing 1 required positional argument: 'X'