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Is $x^4+2x^2-8m^2x+1$ irreducible over $mathbbQ$?

How to prove that $x^4+x^3+x^2+3x+3 $ is irreducible over ring $mathbbZ$ of integers?Irreducible over the rationalsWhy can't Eisenstein Criterion be used for certain polynomials (to show that it's irreducible over $mathbbQ$)?Showing a polynomial is irreducible over $mathbbC[x,y]$How to prove a polynomial irreducible over $mathbbC$Why is $x^3+2x^2+x-9$ irreducible over $mathbbQ$?Is $x^6 + 3x^3 -2$ irreducible over $mathbb Q$?Show that $2x^5-4x^2-5$ is irreducible over $mathbbQ$Is $f(y) = 5 - 4y^2 + y^4$ irreducible over $mathbbQ(sqrt5)$?How to prove that $x^5-5$ is irreducible over $mathbb Q(sqrt 2, sqrt[3] 3) $

4

$begingroup$

Is $x^4+2x^2-8m^2x+1$ irreducible over $mathbbQ$ for $m in mathbbZ$, $m>0$?

Wolfram alpha says it is but gives no proof why. Eisenstein's criterion doesn't apply. I have no idea how to proceed. Can anyone help?

abstract-algebra polynomials irreducible-polynomials

share|cite|improve this question

edited Sep 6 '15 at 15:16

user26857

39.4k124183

asked Sep 6 '15 at 12:37

ShevakShevak

1155

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Have you tried reducing modulo a prime? There is a theorem along the following lines; if you reduce f modulo a prime and the degree doesn't change, then irreducibility modulo the prime implies irreducibility over $mathbbQ$. Reducing modulo 3 shows that it is irreducible if $mneq 1$
  $endgroup$
  – User0112358
  Sep 6 '15 at 12:42
  
  

  
  
  
  

add a comment | 

4

$begingroup$

Is $x^4+2x^2-8m^2x+1$ irreducible over $mathbbQ$ for $m in mathbbZ$, $m>0$?

Wolfram alpha says it is but gives no proof why. Eisenstein's criterion doesn't apply. I have no idea how to proceed. Can anyone help?

abstract-algebra polynomials irreducible-polynomials

share|cite|improve this question

edited Sep 6 '15 at 15:16

user26857

39.4k124183

asked Sep 6 '15 at 12:37

ShevakShevak

1155

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Have you tried reducing modulo a prime? There is a theorem along the following lines; if you reduce f modulo a prime and the degree doesn't change, then irreducibility modulo the prime implies irreducibility over $mathbbQ$. Reducing modulo 3 shows that it is irreducible if $mneq 1$
  $endgroup$
  – User0112358
  Sep 6 '15 at 12:42
  
  

  
  
  
  

add a comment | 

$begingroup$

Is $x^4+2x^2-8m^2x+1$ irreducible over $mathbbQ$ for $m in mathbbZ$, $m>0$?

Wolfram alpha says it is but gives no proof why. Eisenstein's criterion doesn't apply. I have no idea how to proceed. Can anyone help?

abstract-algebra polynomials irreducible-polynomials

share|cite|improve this question

edited Sep 6 '15 at 15:16

user26857

39.4k124183

asked Sep 6 '15 at 12:37

ShevakShevak

1155

$endgroup$

share|cite|improve this question

edited Sep 6 '15 at 15:16

user26857

39.4k124183

asked Sep 6 '15 at 12:37

ShevakShevak

1155

share|cite|improve this question

edited Sep 6 '15 at 15:16

user26857

39.4k124183

asked Sep 6 '15 at 12:37

ShevakShevak

1155

edited Sep 6 '15 at 15:16

user26857

39.4k124183

edited Sep 6 '15 at 15:16

user26857

39.4k124183

asked Sep 6 '15 at 12:37

ShevakShevak

1155

asked Sep 6 '15 at 12:37

ShevakShevak

1155

2 Answers
2

active

oldest

votes

5

$begingroup$

Clearly there are no integer (hence no rational roots), so we need to only check for quadratic factors. So we need to only check for the form
$$x^4+2x^2-8m^2x+1=(x^2+axpm1)(x^2-axpm1)$$
which gives $2-a^2=2$ and $pm2a=-8m^2$ which is possible only when $a=m=0$.

share|cite|improve this answer

answered Sep 6 '15 at 12:51

MacavityMacavity

35.7k52554

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  That works. I was struggling for a while trying to show that it has no rational roots. But from the rational root theorem $x=pm 1$ if x is a rational root of the above polynomial which leads to a contradiction.
  $endgroup$
  – Shevak
  Sep 6 '15 at 13:46
  
  

  
  
  
  

add a comment | 

2

$begingroup$

Its irreducibility follows nicely from Perron's criterion when applied to reciprocal $g(x)=x^4f(1/x)=x^4-8m^2x^3+2x^2+1$:

Perron's criterion: Let $P(x) = x^n + a_n-1 x^n-1 + dots + a_0 in mathbbZ[x]$. If $|a_n-1| > 1 + |a_n-2| + dots + |a_0|$, then $P(x)$ is irreducible.

Since $8m^2> 1+2+1 = 4$ clearly holds for $m>0$, it follows $g(x)$ is irreducible, and particularly $f(x)$ is irreducible as well.

share|cite|improve this answer

answered Mar 16 at 10:53

SilSil

5,31221643

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  One can find Perron's criterion at yufeizhao.com/olympiad/intpoly.pdf
  $endgroup$
  – egreg
  Mar 16 at 11:33
  

  
  
  
  

add a comment | 

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5

$begingroup$

Clearly there are no integer (hence no rational roots), so we need to only check for quadratic factors. So we need to only check for the form
$$x^4+2x^2-8m^2x+1=(x^2+axpm1)(x^2-axpm1)$$
which gives $2-a^2=2$ and $pm2a=-8m^2$ which is possible only when $a=m=0$.

share|cite|improve this answer

answered Sep 6 '15 at 12:51

MacavityMacavity

35.7k52554

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  That works. I was struggling for a while trying to show that it has no rational roots. But from the rational root theorem $x=pm 1$ if x is a rational root of the above polynomial which leads to a contradiction.
  $endgroup$
  – Shevak
  Sep 6 '15 at 13:46
  
  

  
  
  
  

add a comment | 

5

$begingroup$

Clearly there are no integer (hence no rational roots), so we need to only check for quadratic factors. So we need to only check for the form
$$x^4+2x^2-8m^2x+1=(x^2+axpm1)(x^2-axpm1)$$
which gives $2-a^2=2$ and $pm2a=-8m^2$ which is possible only when $a=m=0$.

share|cite|improve this answer

answered Sep 6 '15 at 12:51

MacavityMacavity

35.7k52554

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  That works. I was struggling for a while trying to show that it has no rational roots. But from the rational root theorem $x=pm 1$ if x is a rational root of the above polynomial which leads to a contradiction.
  $endgroup$
  – Shevak
  Sep 6 '15 at 13:46
  
  

  
  
  
  

add a comment | 

$begingroup$

Clearly there are no integer (hence no rational roots), so we need to only check for quadratic factors. So we need to only check for the form
$$x^4+2x^2-8m^2x+1=(x^2+axpm1)(x^2-axpm1)$$
which gives $2-a^2=2$ and $pm2a=-8m^2$ which is possible only when $a=m=0$.

share|cite|improve this answer

answered Sep 6 '15 at 12:51

MacavityMacavity

35.7k52554

$endgroup$

share|cite|improve this answer

answered Sep 6 '15 at 12:51

MacavityMacavity

35.7k52554

answered Sep 6 '15 at 12:51

MacavityMacavity

35.7k52554

answered Sep 6 '15 at 12:51

MacavityMacavity

35.7k52554

2

$begingroup$

Its irreducibility follows nicely from Perron's criterion when applied to reciprocal $g(x)=x^4f(1/x)=x^4-8m^2x^3+2x^2+1$:

Perron's criterion: Let $P(x) = x^n + a_n-1 x^n-1 + dots + a_0 in mathbbZ[x]$. If $|a_n-1| > 1 + |a_n-2| + dots + |a_0|$, then $P(x)$ is irreducible.

Since $8m^2> 1+2+1 = 4$ clearly holds for $m>0$, it follows $g(x)$ is irreducible, and particularly $f(x)$ is irreducible as well.

share|cite|improve this answer

answered Mar 16 at 10:53

SilSil

5,31221643

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  One can find Perron's criterion at yufeizhao.com/olympiad/intpoly.pdf
  $endgroup$
  – egreg
  Mar 16 at 11:33
  

  
  
  
  

add a comment | 

2

$begingroup$

Its irreducibility follows nicely from Perron's criterion when applied to reciprocal $g(x)=x^4f(1/x)=x^4-8m^2x^3+2x^2+1$:

Perron's criterion: Let $P(x) = x^n + a_n-1 x^n-1 + dots + a_0 in mathbbZ[x]$. If $|a_n-1| > 1 + |a_n-2| + dots + |a_0|$, then $P(x)$ is irreducible.

Since $8m^2> 1+2+1 = 4$ clearly holds for $m>0$, it follows $g(x)$ is irreducible, and particularly $f(x)$ is irreducible as well.

share|cite|improve this answer

answered Mar 16 at 10:53

SilSil

5,31221643

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  One can find Perron's criterion at yufeizhao.com/olympiad/intpoly.pdf
  $endgroup$
  – egreg
  Mar 16 at 11:33
  

  
  
  
  

add a comment | 

$begingroup$

Its irreducibility follows nicely from Perron's criterion when applied to reciprocal $g(x)=x^4f(1/x)=x^4-8m^2x^3+2x^2+1$:

Perron's criterion: Let $P(x) = x^n + a_n-1 x^n-1 + dots + a_0 in mathbbZ[x]$. If $|a_n-1| > 1 + |a_n-2| + dots + |a_0|$, then $P(x)$ is irreducible.

Since $8m^2> 1+2+1 = 4$ clearly holds for $m>0$, it follows $g(x)$ is irreducible, and particularly $f(x)$ is irreducible as well.

share|cite|improve this answer

answered Mar 16 at 10:53

SilSil

5,31221643

$endgroup$

share|cite|improve this answer

answered Mar 16 at 10:53

SilSil

5,31221643

answered Mar 16 at 10:53

SilSil

5,31221643

answered Mar 16 at 10:53

SilSil

5,31221643