Let $F$ be a field and $f(x) in F[x]$ be a polynomial of degree $> 1$. If $f(a) = 0$ for some $a in F$,...












0














My Attempt at this is as follows:



Let $f(x) = a_0+a_1x+a_2x^2+...+a_mx^m$ $forall$ $a_i$ $in$ $F[X]$.



Given $f(a)=0$



$Rightarrow$ $$f(x) = (x-a)t(x) label{a}tag{1}$$ such that $deg(t(x))geq1$,



and since $deg(x-alpha)=1$, we can express $f(x)$ as a product of two functions such that their degree is not zero, hence $f(x)$ is reducible.



So for proving the above result is this approach allowed? Or should I use some other way or do I have to prove the equation which I used in $ref{a}$?










share|cite|improve this question





























    0














    My Attempt at this is as follows:



    Let $f(x) = a_0+a_1x+a_2x^2+...+a_mx^m$ $forall$ $a_i$ $in$ $F[X]$.



    Given $f(a)=0$



    $Rightarrow$ $$f(x) = (x-a)t(x) label{a}tag{1}$$ such that $deg(t(x))geq1$,



    and since $deg(x-alpha)=1$, we can express $f(x)$ as a product of two functions such that their degree is not zero, hence $f(x)$ is reducible.



    So for proving the above result is this approach allowed? Or should I use some other way or do I have to prove the equation which I used in $ref{a}$?










    share|cite|improve this question



























      0












      0








      0







      My Attempt at this is as follows:



      Let $f(x) = a_0+a_1x+a_2x^2+...+a_mx^m$ $forall$ $a_i$ $in$ $F[X]$.



      Given $f(a)=0$



      $Rightarrow$ $$f(x) = (x-a)t(x) label{a}tag{1}$$ such that $deg(t(x))geq1$,



      and since $deg(x-alpha)=1$, we can express $f(x)$ as a product of two functions such that their degree is not zero, hence $f(x)$ is reducible.



      So for proving the above result is this approach allowed? Or should I use some other way or do I have to prove the equation which I used in $ref{a}$?










      share|cite|improve this question















      My Attempt at this is as follows:



      Let $f(x) = a_0+a_1x+a_2x^2+...+a_mx^m$ $forall$ $a_i$ $in$ $F[X]$.



      Given $f(a)=0$



      $Rightarrow$ $$f(x) = (x-a)t(x) label{a}tag{1}$$ such that $deg(t(x))geq1$,



      and since $deg(x-alpha)=1$, we can express $f(x)$ as a product of two functions such that their degree is not zero, hence $f(x)$ is reducible.



      So for proving the above result is this approach allowed? Or should I use some other way or do I have to prove the equation which I used in $ref{a}$?







      abstract-algebra ring-theory






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      edited Nov 25 '18 at 11:05









      the_fox

      2,43411431




      2,43411431










      asked Nov 25 '18 at 10:06









      Rohit Bharadwaj

      518




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          2 Answers
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          You should use polynomial division, otherwise you're simply stating what you want to prove. The division algorithm states that if $f(x)$ and $g(x)$ are polynomials, with $gne0$, then there exist unique polynomials $t(x)$ and $r(x)$ such that





          1. $f(x)=g(x)t(x)+r(x)$;


          2. $deg r(x)<deg g(x)$ (or $r=0$).


          In this case we can take $g(x)=x-alpha$, so
          $$
          f(x)=(x-alpha)t(x)+r(x)
          $$

          where $r=0$ or has degree less than $deg(x-alpha)=1$. Therefore $r(x)=c$ is constant.



          Evaluate at $alpha$:
          $$
          0=f(alpha)=(alpha-alpha)t(alpha)+c=c
          $$

          Hence $f(x)=(x-alpha)t(x)$. Since $deg f(x)>1$, $deg t(x)>0$, so $f$ is reducible.






          share|cite|improve this answer























          • I think this is a more correct or better proof
            – Rohit Bharadwaj
            Nov 25 '18 at 11:01










          • one question, aren't you being specific in this case by taking $f(x)=(x-alpha)t(x)$ I mean you could have took any general polynomial right then why that one..... could we have any more general approach?
            – Rohit Bharadwaj
            Nov 26 '18 at 4:10










          • @RohitBharadwaj I stated the general theorem
            – egreg
            Nov 26 '18 at 8:52










          • i mean why did you take $(x-alpha)$, in the division algorithm it is $f(x)=g(x)t(x)+r(x)$
            – Rohit Bharadwaj
            Nov 26 '18 at 17:05










          • @RohitBharadwaj Where's the problem? You want to show divisibility by $x-alpha$, don't you?
            – egreg
            Nov 26 '18 at 17:10



















          0














          Your proof is correct. I would not write “s.t. $deg t(x)geqslant1$”. It's more specific than that (although what you wrote is correct): $deg t(x)=m-1$. And $f(x)$ doesn't became reducible; it is reducible.



          And the equality $(1)$ doesn't have to be proved (as part of this proof), since it is a standard theorem about polynomials.






          share|cite|improve this answer





















            Your Answer





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            2 Answers
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            active

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            2 Answers
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            active

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            active

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            active

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            0














            You should use polynomial division, otherwise you're simply stating what you want to prove. The division algorithm states that if $f(x)$ and $g(x)$ are polynomials, with $gne0$, then there exist unique polynomials $t(x)$ and $r(x)$ such that





            1. $f(x)=g(x)t(x)+r(x)$;


            2. $deg r(x)<deg g(x)$ (or $r=0$).


            In this case we can take $g(x)=x-alpha$, so
            $$
            f(x)=(x-alpha)t(x)+r(x)
            $$

            where $r=0$ or has degree less than $deg(x-alpha)=1$. Therefore $r(x)=c$ is constant.



            Evaluate at $alpha$:
            $$
            0=f(alpha)=(alpha-alpha)t(alpha)+c=c
            $$

            Hence $f(x)=(x-alpha)t(x)$. Since $deg f(x)>1$, $deg t(x)>0$, so $f$ is reducible.






            share|cite|improve this answer























            • I think this is a more correct or better proof
              – Rohit Bharadwaj
              Nov 25 '18 at 11:01










            • one question, aren't you being specific in this case by taking $f(x)=(x-alpha)t(x)$ I mean you could have took any general polynomial right then why that one..... could we have any more general approach?
              – Rohit Bharadwaj
              Nov 26 '18 at 4:10










            • @RohitBharadwaj I stated the general theorem
              – egreg
              Nov 26 '18 at 8:52










            • i mean why did you take $(x-alpha)$, in the division algorithm it is $f(x)=g(x)t(x)+r(x)$
              – Rohit Bharadwaj
              Nov 26 '18 at 17:05










            • @RohitBharadwaj Where's the problem? You want to show divisibility by $x-alpha$, don't you?
              – egreg
              Nov 26 '18 at 17:10
















            0














            You should use polynomial division, otherwise you're simply stating what you want to prove. The division algorithm states that if $f(x)$ and $g(x)$ are polynomials, with $gne0$, then there exist unique polynomials $t(x)$ and $r(x)$ such that





            1. $f(x)=g(x)t(x)+r(x)$;


            2. $deg r(x)<deg g(x)$ (or $r=0$).


            In this case we can take $g(x)=x-alpha$, so
            $$
            f(x)=(x-alpha)t(x)+r(x)
            $$

            where $r=0$ or has degree less than $deg(x-alpha)=1$. Therefore $r(x)=c$ is constant.



            Evaluate at $alpha$:
            $$
            0=f(alpha)=(alpha-alpha)t(alpha)+c=c
            $$

            Hence $f(x)=(x-alpha)t(x)$. Since $deg f(x)>1$, $deg t(x)>0$, so $f$ is reducible.






            share|cite|improve this answer























            • I think this is a more correct or better proof
              – Rohit Bharadwaj
              Nov 25 '18 at 11:01










            • one question, aren't you being specific in this case by taking $f(x)=(x-alpha)t(x)$ I mean you could have took any general polynomial right then why that one..... could we have any more general approach?
              – Rohit Bharadwaj
              Nov 26 '18 at 4:10










            • @RohitBharadwaj I stated the general theorem
              – egreg
              Nov 26 '18 at 8:52










            • i mean why did you take $(x-alpha)$, in the division algorithm it is $f(x)=g(x)t(x)+r(x)$
              – Rohit Bharadwaj
              Nov 26 '18 at 17:05










            • @RohitBharadwaj Where's the problem? You want to show divisibility by $x-alpha$, don't you?
              – egreg
              Nov 26 '18 at 17:10














            0












            0








            0






            You should use polynomial division, otherwise you're simply stating what you want to prove. The division algorithm states that if $f(x)$ and $g(x)$ are polynomials, with $gne0$, then there exist unique polynomials $t(x)$ and $r(x)$ such that





            1. $f(x)=g(x)t(x)+r(x)$;


            2. $deg r(x)<deg g(x)$ (or $r=0$).


            In this case we can take $g(x)=x-alpha$, so
            $$
            f(x)=(x-alpha)t(x)+r(x)
            $$

            where $r=0$ or has degree less than $deg(x-alpha)=1$. Therefore $r(x)=c$ is constant.



            Evaluate at $alpha$:
            $$
            0=f(alpha)=(alpha-alpha)t(alpha)+c=c
            $$

            Hence $f(x)=(x-alpha)t(x)$. Since $deg f(x)>1$, $deg t(x)>0$, so $f$ is reducible.






            share|cite|improve this answer














            You should use polynomial division, otherwise you're simply stating what you want to prove. The division algorithm states that if $f(x)$ and $g(x)$ are polynomials, with $gne0$, then there exist unique polynomials $t(x)$ and $r(x)$ such that





            1. $f(x)=g(x)t(x)+r(x)$;


            2. $deg r(x)<deg g(x)$ (or $r=0$).


            In this case we can take $g(x)=x-alpha$, so
            $$
            f(x)=(x-alpha)t(x)+r(x)
            $$

            where $r=0$ or has degree less than $deg(x-alpha)=1$. Therefore $r(x)=c$ is constant.



            Evaluate at $alpha$:
            $$
            0=f(alpha)=(alpha-alpha)t(alpha)+c=c
            $$

            Hence $f(x)=(x-alpha)t(x)$. Since $deg f(x)>1$, $deg t(x)>0$, so $f$ is reducible.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Nov 26 '18 at 8:52

























            answered Nov 25 '18 at 10:56









            egreg

            178k1484201




            178k1484201












            • I think this is a more correct or better proof
              – Rohit Bharadwaj
              Nov 25 '18 at 11:01










            • one question, aren't you being specific in this case by taking $f(x)=(x-alpha)t(x)$ I mean you could have took any general polynomial right then why that one..... could we have any more general approach?
              – Rohit Bharadwaj
              Nov 26 '18 at 4:10










            • @RohitBharadwaj I stated the general theorem
              – egreg
              Nov 26 '18 at 8:52










            • i mean why did you take $(x-alpha)$, in the division algorithm it is $f(x)=g(x)t(x)+r(x)$
              – Rohit Bharadwaj
              Nov 26 '18 at 17:05










            • @RohitBharadwaj Where's the problem? You want to show divisibility by $x-alpha$, don't you?
              – egreg
              Nov 26 '18 at 17:10


















            • I think this is a more correct or better proof
              – Rohit Bharadwaj
              Nov 25 '18 at 11:01










            • one question, aren't you being specific in this case by taking $f(x)=(x-alpha)t(x)$ I mean you could have took any general polynomial right then why that one..... could we have any more general approach?
              – Rohit Bharadwaj
              Nov 26 '18 at 4:10










            • @RohitBharadwaj I stated the general theorem
              – egreg
              Nov 26 '18 at 8:52










            • i mean why did you take $(x-alpha)$, in the division algorithm it is $f(x)=g(x)t(x)+r(x)$
              – Rohit Bharadwaj
              Nov 26 '18 at 17:05










            • @RohitBharadwaj Where's the problem? You want to show divisibility by $x-alpha$, don't you?
              – egreg
              Nov 26 '18 at 17:10
















            I think this is a more correct or better proof
            – Rohit Bharadwaj
            Nov 25 '18 at 11:01




            I think this is a more correct or better proof
            – Rohit Bharadwaj
            Nov 25 '18 at 11:01












            one question, aren't you being specific in this case by taking $f(x)=(x-alpha)t(x)$ I mean you could have took any general polynomial right then why that one..... could we have any more general approach?
            – Rohit Bharadwaj
            Nov 26 '18 at 4:10




            one question, aren't you being specific in this case by taking $f(x)=(x-alpha)t(x)$ I mean you could have took any general polynomial right then why that one..... could we have any more general approach?
            – Rohit Bharadwaj
            Nov 26 '18 at 4:10












            @RohitBharadwaj I stated the general theorem
            – egreg
            Nov 26 '18 at 8:52




            @RohitBharadwaj I stated the general theorem
            – egreg
            Nov 26 '18 at 8:52












            i mean why did you take $(x-alpha)$, in the division algorithm it is $f(x)=g(x)t(x)+r(x)$
            – Rohit Bharadwaj
            Nov 26 '18 at 17:05




            i mean why did you take $(x-alpha)$, in the division algorithm it is $f(x)=g(x)t(x)+r(x)$
            – Rohit Bharadwaj
            Nov 26 '18 at 17:05












            @RohitBharadwaj Where's the problem? You want to show divisibility by $x-alpha$, don't you?
            – egreg
            Nov 26 '18 at 17:10




            @RohitBharadwaj Where's the problem? You want to show divisibility by $x-alpha$, don't you?
            – egreg
            Nov 26 '18 at 17:10











            0














            Your proof is correct. I would not write “s.t. $deg t(x)geqslant1$”. It's more specific than that (although what you wrote is correct): $deg t(x)=m-1$. And $f(x)$ doesn't became reducible; it is reducible.



            And the equality $(1)$ doesn't have to be proved (as part of this proof), since it is a standard theorem about polynomials.






            share|cite|improve this answer


























              0














              Your proof is correct. I would not write “s.t. $deg t(x)geqslant1$”. It's more specific than that (although what you wrote is correct): $deg t(x)=m-1$. And $f(x)$ doesn't became reducible; it is reducible.



              And the equality $(1)$ doesn't have to be proved (as part of this proof), since it is a standard theorem about polynomials.






              share|cite|improve this answer
























                0












                0








                0






                Your proof is correct. I would not write “s.t. $deg t(x)geqslant1$”. It's more specific than that (although what you wrote is correct): $deg t(x)=m-1$. And $f(x)$ doesn't became reducible; it is reducible.



                And the equality $(1)$ doesn't have to be proved (as part of this proof), since it is a standard theorem about polynomials.






                share|cite|improve this answer












                Your proof is correct. I would not write “s.t. $deg t(x)geqslant1$”. It's more specific than that (although what you wrote is correct): $deg t(x)=m-1$. And $f(x)$ doesn't became reducible; it is reducible.



                And the equality $(1)$ doesn't have to be proved (as part of this proof), since it is a standard theorem about polynomials.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 25 '18 at 10:20









                José Carlos Santos

                150k22121221




                150k22121221






























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