Irreducibility of a polynomial over Rationals with condition given on its coefficients.
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
asked Nov 22 at 14:21
Mittal G
1,188515
add a comment |
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
asked Nov 22 at 14:21
Mittal G
1,188515
add a comment |
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
asked Nov 22 at 14:21
Mittal G
1,188515
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
roots irreducible-polynomials rational-numbers
asked Nov 22 at 14:21
Mittal G
1,188515
asked Nov 22 at 14:21
Mittal G
1,188515
asked Nov 22 at 14:21
Mittal G
1,188515
asked Nov 22 at 14:21
Mittal G
1,188515
asked Nov 22 at 14:21
Mittal G
1,188515
1,188515
add a comment |
add a comment |
1 Answer
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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 $f(x)=x^2+x-6$
answered Nov 22 at 14:39
John_Wick
1,314111
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
add a comment |
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1 Answer
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1 Answer
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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 $f(x)=x^2+x-6$
answered Nov 22 at 14:39
John_Wick
1,314111
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
add a comment |
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 $f(x)=x^2+x-6$
answered Nov 22 at 14:39
John_Wick
1,314111
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
add a comment |
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 $f(x)=x^2+x-6$
answered Nov 22 at 14:39
John_Wick
1,314111
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 $f(x)=x^2+x-6$
answered Nov 22 at 14:39
John_Wick
1,314111
answered Nov 22 at 14:39
John_Wick
1,314111
answered Nov 22 at 14:39
John_Wick
1,314111
answered Nov 22 at 14:39
John_Wick
1,314111
1,314111
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
add a comment |
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
add a comment |
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