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Isomorphism of fields via the forgetful functor


Why is $text{Aut}(F)$ of the forgetful functor $F$ on $G$-sets isomorphic to $G$?Trying to find a left adjointadjoint of forgetful functor related to localizationAlgebraic theories and forgetful functorWhat categorical property do these forgetful functors have in common?For what $k$ does the forgetful functor $mathsf{Fld}_k to mathsf{Set}$ have a left adjoint?Fields on uncountable setsExistence of a cofree functor (right adjoint)Legitimate functor metacategory.Trying to find a left adjoint to the forgetful functor from [C,Set] to Set^{Ob(C)}













2












$begingroup$


The following article https://tinyurl.com/yydxzxe3 says that



"For the case of fields, given a field F and an isomorphism of sets U(F) → S, there is a unique field whose underlying set is S and which is isomorphic to F as a field via the given function"



The U above is the forgetful functor which we know for fields doesn't have a left adjoint but this could still be possible as we need a weaker condition (i.e. all bijections give us isomorphisms not all morphisms ).



But I don't get how. Say $mathbb{F}_{2^n} = mathbb{F}_{2}[x]/p(x) $ is a fixed finite field and I have a permutation $pi$ on $2^n$ elements. How do I construct $f(x)$ such that $ mathbb{F}_{2}[x]/f(x) cong mathbb{F}_{2^n}$ via $pi$?






abstract-algebra category-theory finite-fields







share|cite|improve this question









$endgroup$

















    2












    $begingroup$


    The following article https://tinyurl.com/yydxzxe3 says that



    "For the case of fields, given a field F and an isomorphism of sets U(F) → S, there is a unique field whose underlying set is S and which is isomorphic to F as a field via the given function"



    The U above is the forgetful functor which we know for fields doesn't have a left adjoint but this could still be possible as we need a weaker condition (i.e. all bijections give us isomorphisms not all morphisms ).



    But I don't get how. Say $mathbb{F}_{2^n} = mathbb{F}_{2}[x]/p(x) $ is a fixed finite field and I have a permutation $pi$ on $2^n$ elements. How do I construct $f(x)$ such that $ mathbb{F}_{2}[x]/f(x) cong mathbb{F}_{2^n}$ via $pi$?






    abstract-algebra category-theory finite-fields







    share|cite|improve this question









    $endgroup$















      2












      2








      2





      $begingroup$


      The following article https://tinyurl.com/yydxzxe3 says that



      "For the case of fields, given a field F and an isomorphism of sets U(F) → S, there is a unique field whose underlying set is S and which is isomorphic to F as a field via the given function"



      The U above is the forgetful functor which we know for fields doesn't have a left adjoint but this could still be possible as we need a weaker condition (i.e. all bijections give us isomorphisms not all morphisms ).



      But I don't get how. Say $mathbb{F}_{2^n} = mathbb{F}_{2}[x]/p(x) $ is a fixed finite field and I have a permutation $pi$ on $2^n$ elements. How do I construct $f(x)$ such that $ mathbb{F}_{2}[x]/f(x) cong mathbb{F}_{2^n}$ via $pi$?






      abstract-algebra category-theory finite-fields







      share|cite|improve this question









      $endgroup$




      The following article https://tinyurl.com/yydxzxe3 says that



      "For the case of fields, given a field F and an isomorphism of sets U(F) → S, there is a unique field whose underlying set is S and which is isomorphic to F as a field via the given function"



      The U above is the forgetful functor which we know for fields doesn't have a left adjoint but this could still be possible as we need a weaker condition (i.e. all bijections give us isomorphisms not all morphisms ).



      But I don't get how. Say $mathbb{F}_{2^n} = mathbb{F}_{2}[x]/p(x) $ is a fixed finite field and I have a permutation $pi$ on $2^n$ elements. How do I construct $f(x)$ such that $ mathbb{F}_{2}[x]/f(x) cong mathbb{F}_{2^n}$ via $pi$?






      abstract-algebra category-theory finite-fields




      abstract-algebra category-theory finite-fields






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked 7 hours ago









      Tushant MittalTushant Mittal

      752412




      752412






















          1 Answer
          1






          active

          oldest

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          7












          $begingroup$

          The observation here is just that we can "transport structure" along any isomorphism.



          Suppose $F$ is a field and $picolon U(F)to S$ is a bijection of sets. We define operations $+$ and $times$ on $S$ in such a way that $(S;+,times)$ is a field and $picolon Fto (S;+,times)$ is a field isomorphism. For any $a,bin S$, define: begin{align*} a+b &= pi(pi^{-1}(a) + pi^{-1}(b))\ atimes b &= pi(pi^{-1}(a)times pi^{-1}(b)).end{align*}
          Note that on the right hand sides of the above equations, $+$ and $times$ are the field operations in $F$.



          In your example of a permutation $pi$ of the set $U(mathbb{F}_{2^n}) = mathbb{F}_2[x]/p(x)$, this doesn't amount to finding a new quotient of $mathbb{F}_2[x]$, we just get totally new field structure on the set $mathbb{F}_2[x]/p(x)$ which has nothing to do with the original one, or with the ring structure on $mathbb{F}_2[x]$ (e.g. in most cases it will have a different $0$ and a different $1$).






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Exactly, but since we know that all finite fields are of the above type we should be able to find a new quotient, right? The motivation for the question is that a bijection in F_2^n is given by 2^n images but by converting it to a field isomorphism I just need to specify images of the n generators. This would be helpful in creating smaller computational circuits (given that I somehow precompute the f)
            $endgroup$
            – Tushant Mittal
            6 hours ago






          • 1




            $begingroup$
            All finite fields are of the above type up to isomorphism. The new field is isomorphic to $mathbb{F}_{2^n}$ by construction, so it's isomorphic to the same quotient of $mathbb{F}_2[x]$, not a new one!
            $endgroup$
            – Alex Kruckman
            6 hours ago












          • $begingroup$
            Ah. I get it now. The isomorphism, equality problem again. That's exactly what the article is about and I still made that error!
            $endgroup$
            – Tushant Mittal
            6 hours ago














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          1 Answer
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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          7












          $begingroup$

          The observation here is just that we can "transport structure" along any isomorphism.



          Suppose $F$ is a field and $picolon U(F)to S$ is a bijection of sets. We define operations $+$ and $times$ on $S$ in such a way that $(S;+,times)$ is a field and $picolon Fto (S;+,times)$ is a field isomorphism. For any $a,bin S$, define: begin{align*} a+b &= pi(pi^{-1}(a) + pi^{-1}(b))\ atimes b &= pi(pi^{-1}(a)times pi^{-1}(b)).end{align*}
          Note that on the right hand sides of the above equations, $+$ and $times$ are the field operations in $F$.



          In your example of a permutation $pi$ of the set $U(mathbb{F}_{2^n}) = mathbb{F}_2[x]/p(x)$, this doesn't amount to finding a new quotient of $mathbb{F}_2[x]$, we just get totally new field structure on the set $mathbb{F}_2[x]/p(x)$ which has nothing to do with the original one, or with the ring structure on $mathbb{F}_2[x]$ (e.g. in most cases it will have a different $0$ and a different $1$).






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Exactly, but since we know that all finite fields are of the above type we should be able to find a new quotient, right? The motivation for the question is that a bijection in F_2^n is given by 2^n images but by converting it to a field isomorphism I just need to specify images of the n generators. This would be helpful in creating smaller computational circuits (given that I somehow precompute the f)
            $endgroup$
            – Tushant Mittal
            6 hours ago






          • 1




            $begingroup$
            All finite fields are of the above type up to isomorphism. The new field is isomorphic to $mathbb{F}_{2^n}$ by construction, so it's isomorphic to the same quotient of $mathbb{F}_2[x]$, not a new one!
            $endgroup$
            – Alex Kruckman
            6 hours ago












          • $begingroup$
            Ah. I get it now. The isomorphism, equality problem again. That's exactly what the article is about and I still made that error!
            $endgroup$
            – Tushant Mittal
            6 hours ago


















          7












          $begingroup$

          The observation here is just that we can "transport structure" along any isomorphism.



          Suppose $F$ is a field and $picolon U(F)to S$ is a bijection of sets. We define operations $+$ and $times$ on $S$ in such a way that $(S;+,times)$ is a field and $picolon Fto (S;+,times)$ is a field isomorphism. For any $a,bin S$, define: begin{align*} a+b &= pi(pi^{-1}(a) + pi^{-1}(b))\ atimes b &= pi(pi^{-1}(a)times pi^{-1}(b)).end{align*}
          Note that on the right hand sides of the above equations, $+$ and $times$ are the field operations in $F$.



          In your example of a permutation $pi$ of the set $U(mathbb{F}_{2^n}) = mathbb{F}_2[x]/p(x)$, this doesn't amount to finding a new quotient of $mathbb{F}_2[x]$, we just get totally new field structure on the set $mathbb{F}_2[x]/p(x)$ which has nothing to do with the original one, or with the ring structure on $mathbb{F}_2[x]$ (e.g. in most cases it will have a different $0$ and a different $1$).






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Exactly, but since we know that all finite fields are of the above type we should be able to find a new quotient, right? The motivation for the question is that a bijection in F_2^n is given by 2^n images but by converting it to a field isomorphism I just need to specify images of the n generators. This would be helpful in creating smaller computational circuits (given that I somehow precompute the f)
            $endgroup$
            – Tushant Mittal
            6 hours ago






          • 1




            $begingroup$
            All finite fields are of the above type up to isomorphism. The new field is isomorphic to $mathbb{F}_{2^n}$ by construction, so it's isomorphic to the same quotient of $mathbb{F}_2[x]$, not a new one!
            $endgroup$
            – Alex Kruckman
            6 hours ago












          • $begingroup$
            Ah. I get it now. The isomorphism, equality problem again. That's exactly what the article is about and I still made that error!
            $endgroup$
            – Tushant Mittal
            6 hours ago
















          7












          7








          7





          $begingroup$

          The observation here is just that we can "transport structure" along any isomorphism.



          Suppose $F$ is a field and $picolon U(F)to S$ is a bijection of sets. We define operations $+$ and $times$ on $S$ in such a way that $(S;+,times)$ is a field and $picolon Fto (S;+,times)$ is a field isomorphism. For any $a,bin S$, define: begin{align*} a+b &= pi(pi^{-1}(a) + pi^{-1}(b))\ atimes b &= pi(pi^{-1}(a)times pi^{-1}(b)).end{align*}
          Note that on the right hand sides of the above equations, $+$ and $times$ are the field operations in $F$.



          In your example of a permutation $pi$ of the set $U(mathbb{F}_{2^n}) = mathbb{F}_2[x]/p(x)$, this doesn't amount to finding a new quotient of $mathbb{F}_2[x]$, we just get totally new field structure on the set $mathbb{F}_2[x]/p(x)$ which has nothing to do with the original one, or with the ring structure on $mathbb{F}_2[x]$ (e.g. in most cases it will have a different $0$ and a different $1$).






          share|cite|improve this answer











          $endgroup$



          The observation here is just that we can "transport structure" along any isomorphism.



          Suppose $F$ is a field and $picolon U(F)to S$ is a bijection of sets. We define operations $+$ and $times$ on $S$ in such a way that $(S;+,times)$ is a field and $picolon Fto (S;+,times)$ is a field isomorphism. For any $a,bin S$, define: begin{align*} a+b &= pi(pi^{-1}(a) + pi^{-1}(b))\ atimes b &= pi(pi^{-1}(a)times pi^{-1}(b)).end{align*}
          Note that on the right hand sides of the above equations, $+$ and $times$ are the field operations in $F$.



          In your example of a permutation $pi$ of the set $U(mathbb{F}_{2^n}) = mathbb{F}_2[x]/p(x)$, this doesn't amount to finding a new quotient of $mathbb{F}_2[x]$, we just get totally new field structure on the set $mathbb{F}_2[x]/p(x)$ which has nothing to do with the original one, or with the ring structure on $mathbb{F}_2[x]$ (e.g. in most cases it will have a different $0$ and a different $1$).







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited 6 hours ago

























          answered 6 hours ago









          Alex KruckmanAlex Kruckman

          29.2k32758




          29.2k32758












          • $begingroup$
            Exactly, but since we know that all finite fields are of the above type we should be able to find a new quotient, right? The motivation for the question is that a bijection in F_2^n is given by 2^n images but by converting it to a field isomorphism I just need to specify images of the n generators. This would be helpful in creating smaller computational circuits (given that I somehow precompute the f)
            $endgroup$
            – Tushant Mittal
            6 hours ago






          • 1




            $begingroup$
            All finite fields are of the above type up to isomorphism. The new field is isomorphic to $mathbb{F}_{2^n}$ by construction, so it's isomorphic to the same quotient of $mathbb{F}_2[x]$, not a new one!
            $endgroup$
            – Alex Kruckman
            6 hours ago












          • $begingroup$
            Ah. I get it now. The isomorphism, equality problem again. That's exactly what the article is about and I still made that error!
            $endgroup$
            – Tushant Mittal
            6 hours ago




















          • $begingroup$
            Exactly, but since we know that all finite fields are of the above type we should be able to find a new quotient, right? The motivation for the question is that a bijection in F_2^n is given by 2^n images but by converting it to a field isomorphism I just need to specify images of the n generators. This would be helpful in creating smaller computational circuits (given that I somehow precompute the f)
            $endgroup$
            – Tushant Mittal
            6 hours ago






          • 1




            $begingroup$
            All finite fields are of the above type up to isomorphism. The new field is isomorphic to $mathbb{F}_{2^n}$ by construction, so it's isomorphic to the same quotient of $mathbb{F}_2[x]$, not a new one!
            $endgroup$
            – Alex Kruckman
            6 hours ago












          • $begingroup$
            Ah. I get it now. The isomorphism, equality problem again. That's exactly what the article is about and I still made that error!
            $endgroup$
            – Tushant Mittal
            6 hours ago


















          $begingroup$
          Exactly, but since we know that all finite fields are of the above type we should be able to find a new quotient, right? The motivation for the question is that a bijection in F_2^n is given by 2^n images but by converting it to a field isomorphism I just need to specify images of the n generators. This would be helpful in creating smaller computational circuits (given that I somehow precompute the f)
          $endgroup$
          – Tushant Mittal
          6 hours ago




          $begingroup$
          Exactly, but since we know that all finite fields are of the above type we should be able to find a new quotient, right? The motivation for the question is that a bijection in F_2^n is given by 2^n images but by converting it to a field isomorphism I just need to specify images of the n generators. This would be helpful in creating smaller computational circuits (given that I somehow precompute the f)
          $endgroup$
          – Tushant Mittal
          6 hours ago




          1




          1




          $begingroup$
          All finite fields are of the above type up to isomorphism. The new field is isomorphic to $mathbb{F}_{2^n}$ by construction, so it's isomorphic to the same quotient of $mathbb{F}_2[x]$, not a new one!
          $endgroup$
          – Alex Kruckman
          6 hours ago






          $begingroup$
          All finite fields are of the above type up to isomorphism. The new field is isomorphic to $mathbb{F}_{2^n}$ by construction, so it's isomorphic to the same quotient of $mathbb{F}_2[x]$, not a new one!
          $endgroup$
          – Alex Kruckman
          6 hours ago














          $begingroup$
          Ah. I get it now. The isomorphism, equality problem again. That's exactly what the article is about and I still made that error!
          $endgroup$
          – Tushant Mittal
          6 hours ago






          $begingroup$
          Ah. I get it now. The isomorphism, equality problem again. That's exactly what the article is about and I still made that error!
          $endgroup$
          – Tushant Mittal
          6 hours ago




















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