Irreducibility of a polynomial over Rationals with condition given on its coefficients.











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Let $f = a_nX^n+cdots+a_1Xpm p in mathbb{Z}[X]$ with $sum_{i=1}^n |a_i| < p$. Show that $f$ is irreducible in $mathbb{Q}[X]$.



Hint: Show that every root of $f in mathbb{C}$ has modulus greater than $1$ and consider leading and constant terms of a factor of $f$.



I have been able to show that every roots has modulus greater than $1$. But I am not able to go any further? Please help.










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    Let $f = a_nX^n+cdots+a_1Xpm p in mathbb{Z}[X]$ with $sum_{i=1}^n |a_i| < p$. Show that $f$ is irreducible in $mathbb{Q}[X]$.



    Hint: Show that every root of $f in mathbb{C}$ has modulus greater than $1$ and consider leading and constant terms of a factor of $f$.



    I have been able to show that every roots has modulus greater than $1$. But I am not able to go any further? Please help.










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      up vote
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      favorite
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      Let $f = a_nX^n+cdots+a_1Xpm p in mathbb{Z}[X]$ with $sum_{i=1}^n |a_i| < p$. Show that $f$ is irreducible in $mathbb{Q}[X]$.



      Hint: Show that every root of $f in mathbb{C}$ has modulus greater than $1$ and consider leading and constant terms of a factor of $f$.



      I have been able to show that every roots has modulus greater than $1$. But I am not able to go any further? Please help.










      share|cite|improve this question













      Let $f = a_nX^n+cdots+a_1Xpm p in mathbb{Z}[X]$ with $sum_{i=1}^n |a_i| < p$. Show that $f$ is irreducible in $mathbb{Q}[X]$.



      Hint: Show that every root of $f in mathbb{C}$ has modulus greater than $1$ and consider leading and constant terms of a factor of $f$.



      I have been able to show that every roots has modulus greater than $1$. But I am not able to go any further? Please help.







      roots irreducible-polynomials rational-numbers






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      asked Nov 22 at 14:21









      Mittal G

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          If $f(x)=a_nx^n+a_{n-1}x^{n-1}+ldots+a_1x+(-1)^up$ has a solution in $mathbb{Q}$ which is $r/s$ with $gcd(r,s)=1$, then $a_nr^n+a_{n-1}r^{n-1}s+ldots+a_1rs^{n-1}+(-1)^ups^n=0\Rightarrow r(a_nr^{n-1}+a_{n-2}r^{n-1}s+ldots+a_1s^{n-1})=-(-1)^ups^n\Rightarrow r|p.$



          If $p$ is prime then we have a contradiction.



          If $p$ is not prime then we have a counterexample $f(x)=x^2+x-6$






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            How to show that $f(x)$ has no irreducible factor? With no roots doesn't means irreducibility.
            – Mittal G
            Nov 22 at 14:47













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          If $f(x)=a_nx^n+a_{n-1}x^{n-1}+ldots+a_1x+(-1)^up$ has a solution in $mathbb{Q}$ which is $r/s$ with $gcd(r,s)=1$, then $a_nr^n+a_{n-1}r^{n-1}s+ldots+a_1rs^{n-1}+(-1)^ups^n=0\Rightarrow r(a_nr^{n-1}+a_{n-2}r^{n-1}s+ldots+a_1s^{n-1})=-(-1)^ups^n\Rightarrow r|p.$



          If $p$ is prime then we have a contradiction.



          If $p$ is not prime then we have a counterexample $f(x)=x^2+x-6$






          share|cite|improve this answer

















          • 2




            How to show that $f(x)$ has no irreducible factor? With no roots doesn't means irreducibility.
            – Mittal G
            Nov 22 at 14:47

















          up vote
          0
          down vote













          If $f(x)=a_nx^n+a_{n-1}x^{n-1}+ldots+a_1x+(-1)^up$ has a solution in $mathbb{Q}$ which is $r/s$ with $gcd(r,s)=1$, then $a_nr^n+a_{n-1}r^{n-1}s+ldots+a_1rs^{n-1}+(-1)^ups^n=0\Rightarrow r(a_nr^{n-1}+a_{n-2}r^{n-1}s+ldots+a_1s^{n-1})=-(-1)^ups^n\Rightarrow r|p.$



          If $p$ is prime then we have a contradiction.



          If $p$ is not prime then we have a counterexample $f(x)=x^2+x-6$






          share|cite|improve this answer

















          • 2




            How to show that $f(x)$ has no irreducible factor? With no roots doesn't means irreducibility.
            – Mittal G
            Nov 22 at 14:47















          up vote
          0
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          up vote
          0
          down vote









          If $f(x)=a_nx^n+a_{n-1}x^{n-1}+ldots+a_1x+(-1)^up$ has a solution in $mathbb{Q}$ which is $r/s$ with $gcd(r,s)=1$, then $a_nr^n+a_{n-1}r^{n-1}s+ldots+a_1rs^{n-1}+(-1)^ups^n=0\Rightarrow r(a_nr^{n-1}+a_{n-2}r^{n-1}s+ldots+a_1s^{n-1})=-(-1)^ups^n\Rightarrow r|p.$



          If $p$ is prime then we have a contradiction.



          If $p$ is not prime then we have a counterexample $f(x)=x^2+x-6$






          share|cite|improve this answer












          If $f(x)=a_nx^n+a_{n-1}x^{n-1}+ldots+a_1x+(-1)^up$ has a solution in $mathbb{Q}$ which is $r/s$ with $gcd(r,s)=1$, then $a_nr^n+a_{n-1}r^{n-1}s+ldots+a_1rs^{n-1}+(-1)^ups^n=0\Rightarrow r(a_nr^{n-1}+a_{n-2}r^{n-1}s+ldots+a_1s^{n-1})=-(-1)^ups^n\Rightarrow r|p.$



          If $p$ is prime then we have a contradiction.



          If $p$ is not prime then we have a counterexample $f(x)=x^2+x-6$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 22 at 14:39









          John_Wick

          84919




          84919








          • 2




            How to show that $f(x)$ has no irreducible factor? With no roots doesn't means irreducibility.
            – Mittal G
            Nov 22 at 14:47
















          • 2




            How to show that $f(x)$ has no irreducible factor? With no roots doesn't means irreducibility.
            – Mittal G
            Nov 22 at 14:47










          2




          2




          How to show that $f(x)$ has no irreducible factor? With no roots doesn't means irreducibility.
          – Mittal G
          Nov 22 at 14:47






          How to show that $f(x)$ has no irreducible factor? With no roots doesn't means irreducibility.
          – Mittal G
          Nov 22 at 14:47




















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