Identity of Polynomials in positive charcteristic












2












$begingroup$


In positive charcteristic $p$, we know that for every field element $xinmathbb{F}_{p}$ we get $x^p = x$.



Then I think (and I might be wrong, but I don't see how) monomials of the form $t^{p^i}inmathbb{F}_p[t]$ for arbitrary $iin mathbb{N}$ are all the same, since the functions $p_0(t)=t$, $p_1(t)=t^p$, $p_2(t)=t^{p^2}$ and so on are all actually the same functions. Not? I mean certainly they are equal on all elements, that is



$$p_i(x)= p_j(x)$$ for all $xinmathbb{F}_p$ and $i,jinmathbb{N}_0$. But in textbooks on finite function field and such, they are treated as if they where different. But this seems to contradict the (poinwise) definition of a function in terms of evaluation on elements.



So would be great to get some clarification here.






polynomials finite-fields







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$endgroup$

















    2












    $begingroup$


    In positive charcteristic $p$, we know that for every field element $xinmathbb{F}_{p}$ we get $x^p = x$.



    Then I think (and I might be wrong, but I don't see how) monomials of the form $t^{p^i}inmathbb{F}_p[t]$ for arbitrary $iin mathbb{N}$ are all the same, since the functions $p_0(t)=t$, $p_1(t)=t^p$, $p_2(t)=t^{p^2}$ and so on are all actually the same functions. Not? I mean certainly they are equal on all elements, that is



    $$p_i(x)= p_j(x)$$ for all $xinmathbb{F}_p$ and $i,jinmathbb{N}_0$. But in textbooks on finite function field and such, they are treated as if they where different. But this seems to contradict the (poinwise) definition of a function in terms of evaluation on elements.



    So would be great to get some clarification here.






    polynomials finite-fields







    share|cite|improve this question









    $endgroup$















      2












      2








      2





      $begingroup$


      In positive charcteristic $p$, we know that for every field element $xinmathbb{F}_{p}$ we get $x^p = x$.



      Then I think (and I might be wrong, but I don't see how) monomials of the form $t^{p^i}inmathbb{F}_p[t]$ for arbitrary $iin mathbb{N}$ are all the same, since the functions $p_0(t)=t$, $p_1(t)=t^p$, $p_2(t)=t^{p^2}$ and so on are all actually the same functions. Not? I mean certainly they are equal on all elements, that is



      $$p_i(x)= p_j(x)$$ for all $xinmathbb{F}_p$ and $i,jinmathbb{N}_0$. But in textbooks on finite function field and such, they are treated as if they where different. But this seems to contradict the (poinwise) definition of a function in terms of evaluation on elements.



      So would be great to get some clarification here.






      polynomials finite-fields







      share|cite|improve this question









      $endgroup$




      In positive charcteristic $p$, we know that for every field element $xinmathbb{F}_{p}$ we get $x^p = x$.



      Then I think (and I might be wrong, but I don't see how) monomials of the form $t^{p^i}inmathbb{F}_p[t]$ for arbitrary $iin mathbb{N}$ are all the same, since the functions $p_0(t)=t$, $p_1(t)=t^p$, $p_2(t)=t^{p^2}$ and so on are all actually the same functions. Not? I mean certainly they are equal on all elements, that is



      $$p_i(x)= p_j(x)$$ for all $xinmathbb{F}_p$ and $i,jinmathbb{N}_0$. But in textbooks on finite function field and such, they are treated as if they where different. But this seems to contradict the (poinwise) definition of a function in terms of evaluation on elements.



      So would be great to get some clarification here.






      polynomials finite-fields




      polynomials finite-fields






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked 35 mins ago









      BobbyBobby

      1847




      1847






















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          $begingroup$

          There are indeed two different notions of a polynomial over a field $K$: the first is that of formal expressions of the form $sum_{i=0}^k a_iX^i$ where $a_i in K$, and the second is that of functions $f : K to K$ where there exist $a_i in K$ such that for all $x in K$ we have
          $$
          f(x) = sum_{i = 0}^k a_ix^k.
          $$
          If we write $K[X]$ for the formal expressions, and $mathrm{Pol}(K)$ for the functions, then there is a mapping $K[X] to mathrm{Pol}(K)$ given by interpreting the formal expression as a function. In fact, it is a homomorphism whose kernel is precisely those polynomial expressions which evaluate as the zero function.



          Note that if $K$ is infinite, then the mapping $K[X] to mathrm{Pol}(K)$ is a bijection, and we tend to identify the two sets without too much worry.



          When $K$ is finite, identifying the sets makes little sense: the set $mathrm{Pol}(K)$ is a finite set -- in fact the set of all functions $K to K$ -- that does not catch the structure we find interesting about polynomials over $K$. For a concrete example, $mathrm{Pol}(mathbb F_2)$ only has four elements, and they can be written (as polynomial expressions) as $0, 1, X, X - 1$. There are no polynomials of degree at least $2$ to form any splitting fields!






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Nice! That clears everything exactly. Unfortunately this is rarely mentioned. Likely because of the identification in char zero. Good answer.
            $endgroup$
            – Bobby
            5 mins ago












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          $begingroup$

          There are indeed two different notions of a polynomial over a field $K$: the first is that of formal expressions of the form $sum_{i=0}^k a_iX^i$ where $a_i in K$, and the second is that of functions $f : K to K$ where there exist $a_i in K$ such that for all $x in K$ we have
          $$
          f(x) = sum_{i = 0}^k a_ix^k.
          $$
          If we write $K[X]$ for the formal expressions, and $mathrm{Pol}(K)$ for the functions, then there is a mapping $K[X] to mathrm{Pol}(K)$ given by interpreting the formal expression as a function. In fact, it is a homomorphism whose kernel is precisely those polynomial expressions which evaluate as the zero function.



          Note that if $K$ is infinite, then the mapping $K[X] to mathrm{Pol}(K)$ is a bijection, and we tend to identify the two sets without too much worry.



          When $K$ is finite, identifying the sets makes little sense: the set $mathrm{Pol}(K)$ is a finite set -- in fact the set of all functions $K to K$ -- that does not catch the structure we find interesting about polynomials over $K$. For a concrete example, $mathrm{Pol}(mathbb F_2)$ only has four elements, and they can be written (as polynomial expressions) as $0, 1, X, X - 1$. There are no polynomials of degree at least $2$ to form any splitting fields!






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Nice! That clears everything exactly. Unfortunately this is rarely mentioned. Likely because of the identification in char zero. Good answer.
            $endgroup$
            – Bobby
            5 mins ago
















          4












          $begingroup$

          There are indeed two different notions of a polynomial over a field $K$: the first is that of formal expressions of the form $sum_{i=0}^k a_iX^i$ where $a_i in K$, and the second is that of functions $f : K to K$ where there exist $a_i in K$ such that for all $x in K$ we have
          $$
          f(x) = sum_{i = 0}^k a_ix^k.
          $$
          If we write $K[X]$ for the formal expressions, and $mathrm{Pol}(K)$ for the functions, then there is a mapping $K[X] to mathrm{Pol}(K)$ given by interpreting the formal expression as a function. In fact, it is a homomorphism whose kernel is precisely those polynomial expressions which evaluate as the zero function.



          Note that if $K$ is infinite, then the mapping $K[X] to mathrm{Pol}(K)$ is a bijection, and we tend to identify the two sets without too much worry.



          When $K$ is finite, identifying the sets makes little sense: the set $mathrm{Pol}(K)$ is a finite set -- in fact the set of all functions $K to K$ -- that does not catch the structure we find interesting about polynomials over $K$. For a concrete example, $mathrm{Pol}(mathbb F_2)$ only has four elements, and they can be written (as polynomial expressions) as $0, 1, X, X - 1$. There are no polynomials of degree at least $2$ to form any splitting fields!






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Nice! That clears everything exactly. Unfortunately this is rarely mentioned. Likely because of the identification in char zero. Good answer.
            $endgroup$
            – Bobby
            5 mins ago














          4












          4








          4





          $begingroup$

          There are indeed two different notions of a polynomial over a field $K$: the first is that of formal expressions of the form $sum_{i=0}^k a_iX^i$ where $a_i in K$, and the second is that of functions $f : K to K$ where there exist $a_i in K$ such that for all $x in K$ we have
          $$
          f(x) = sum_{i = 0}^k a_ix^k.
          $$
          If we write $K[X]$ for the formal expressions, and $mathrm{Pol}(K)$ for the functions, then there is a mapping $K[X] to mathrm{Pol}(K)$ given by interpreting the formal expression as a function. In fact, it is a homomorphism whose kernel is precisely those polynomial expressions which evaluate as the zero function.



          Note that if $K$ is infinite, then the mapping $K[X] to mathrm{Pol}(K)$ is a bijection, and we tend to identify the two sets without too much worry.



          When $K$ is finite, identifying the sets makes little sense: the set $mathrm{Pol}(K)$ is a finite set -- in fact the set of all functions $K to K$ -- that does not catch the structure we find interesting about polynomials over $K$. For a concrete example, $mathrm{Pol}(mathbb F_2)$ only has four elements, and they can be written (as polynomial expressions) as $0, 1, X, X - 1$. There are no polynomials of degree at least $2$ to form any splitting fields!






          share|cite|improve this answer









          $endgroup$



          There are indeed two different notions of a polynomial over a field $K$: the first is that of formal expressions of the form $sum_{i=0}^k a_iX^i$ where $a_i in K$, and the second is that of functions $f : K to K$ where there exist $a_i in K$ such that for all $x in K$ we have
          $$
          f(x) = sum_{i = 0}^k a_ix^k.
          $$
          If we write $K[X]$ for the formal expressions, and $mathrm{Pol}(K)$ for the functions, then there is a mapping $K[X] to mathrm{Pol}(K)$ given by interpreting the formal expression as a function. In fact, it is a homomorphism whose kernel is precisely those polynomial expressions which evaluate as the zero function.



          Note that if $K$ is infinite, then the mapping $K[X] to mathrm{Pol}(K)$ is a bijection, and we tend to identify the two sets without too much worry.



          When $K$ is finite, identifying the sets makes little sense: the set $mathrm{Pol}(K)$ is a finite set -- in fact the set of all functions $K to K$ -- that does not catch the structure we find interesting about polynomials over $K$. For a concrete example, $mathrm{Pol}(mathbb F_2)$ only has four elements, and they can be written (as polynomial expressions) as $0, 1, X, X - 1$. There are no polynomials of degree at least $2$ to form any splitting fields!







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 25 mins ago









          Mees de VriesMees de Vries

          17.8k13061




          17.8k13061












          • $begingroup$
            Nice! That clears everything exactly. Unfortunately this is rarely mentioned. Likely because of the identification in char zero. Good answer.
            $endgroup$
            – Bobby
            5 mins ago


















          • $begingroup$
            Nice! That clears everything exactly. Unfortunately this is rarely mentioned. Likely because of the identification in char zero. Good answer.
            $endgroup$
            – Bobby
            5 mins ago
















          $begingroup$
          Nice! That clears everything exactly. Unfortunately this is rarely mentioned. Likely because of the identification in char zero. Good answer.
          $endgroup$
          – Bobby
          5 mins ago




          $begingroup$
          Nice! That clears everything exactly. Unfortunately this is rarely mentioned. Likely because of the identification in char zero. Good answer.
          $endgroup$
          – Bobby
          5 mins ago


















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