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Show that $2x^5-4x^2-5$ is irreducible over $mathbbQ$

How to show that $x^6-72$ is irreducible over the rationals?Check if polynomial is minimal over $mathbbQ$Prove particular quintic is irreducibleThe degree of $sqrt2 + sqrt[3]5$ over $mathbb Q$Factorising $x^7-7x^6+21x^5-35x^4+35x^3-21x^2+20x+14$ into $Bbb Q$-irreducible factors (Eisenstein's Criterion)Proving Irreducibility of $x^4-16x^3+20x^2+12$ in $mathbb Q[x]$Show that $p(x) = x^5+5x^4+10x^3+10x^2-x-2$ is irreducible over $mathbbQ$ and has exactly $2$ nonreal roots.Is this polynomial irreducible over the rationals?Why is $x^3+2x^2+x-9$ irreducible over $mathbbQ$?Is $x^3+y^2x^2+3yx-y$ irreducible in $mathbbQ[y][x]$?

3

1

$begingroup$

The excercise is to show that $2x^5-4x^2-5$ is irreducible over $mathbbQ$.

I actually managed to solve this, but using some luck and previous excercises, so I wonder if you can think of a better way.
My way is as follows:

• The given polynomial is irreducible iff the polynomial obtained by the substitution of $x$ by $x+1$ is irrdeucible (why $x+1$? a lucky guess).
  This is a previous excercise.
  The result is $$ 2x^5+10x^4+20x^3+16x^2+2x-5 $$


• Another excercise says that a polynomial is irreducible iff the polynomial obtained by "flipping" the coefficients is irreducible. That is:
  $$ -5x^5+2x^4+16x^3+20x^2+10x+2 $$


• Apply eisenstein's criterion for $p=2$ to deduce irreducibillity.

Do you have any other, maybe more natural solutions?

abstract-algebra irreducible-polynomials

share|cite|improve this question

edited Mar 25 '17 at 12:33

35T41

asked Mar 25 '17 at 12:22

35T4135T41

1,758417

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Can you have a look at cyclotomic polynomials ?
  $endgroup$
  – Hirak
  Mar 25 '17 at 12:26
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  We haven't got there yet...
  $endgroup$
  – 35T41
  Mar 25 '17 at 12:29
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Why can't you do the flipping in the original polynomial?
  $endgroup$
  – themaker
  Mar 25 '17 at 13:23
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You're right... I guess it's easy to miss
  $endgroup$
  – 35T41
  Mar 25 '17 at 14:15
  

  
  
  
  

add a comment | 

3

1

$begingroup$

The excercise is to show that $2x^5-4x^2-5$ is irreducible over $mathbbQ$.

I actually managed to solve this, but using some luck and previous excercises, so I wonder if you can think of a better way.
My way is as follows:

• The given polynomial is irreducible iff the polynomial obtained by the substitution of $x$ by $x+1$ is irrdeucible (why $x+1$? a lucky guess).
  This is a previous excercise.
  The result is $$ 2x^5+10x^4+20x^3+16x^2+2x-5 $$


• Another excercise says that a polynomial is irreducible iff the polynomial obtained by "flipping" the coefficients is irreducible. That is:
  $$ -5x^5+2x^4+16x^3+20x^2+10x+2 $$


• Apply eisenstein's criterion for $p=2$ to deduce irreducibillity.

Do you have any other, maybe more natural solutions?

abstract-algebra irreducible-polynomials

share|cite|improve this question

edited Mar 25 '17 at 12:33

35T41

asked Mar 25 '17 at 12:22

35T4135T41

1,758417

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Can you have a look at cyclotomic polynomials ?
  $endgroup$
  – Hirak
  Mar 25 '17 at 12:26
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  We haven't got there yet...
  $endgroup$
  – 35T41
  Mar 25 '17 at 12:29
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Why can't you do the flipping in the original polynomial?
  $endgroup$
  – themaker
  Mar 25 '17 at 13:23
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You're right... I guess it's easy to miss
  $endgroup$
  – 35T41
  Mar 25 '17 at 14:15
  

  
  
  
  

add a comment | 

$begingroup$

The excercise is to show that $2x^5-4x^2-5$ is irreducible over $mathbbQ$.

I actually managed to solve this, but using some luck and previous excercises, so I wonder if you can think of a better way.
My way is as follows:

• The given polynomial is irreducible iff the polynomial obtained by the substitution of $x$ by $x+1$ is irrdeucible (why $x+1$? a lucky guess).
  This is a previous excercise.
  The result is $$ 2x^5+10x^4+20x^3+16x^2+2x-5 $$


• Another excercise says that a polynomial is irreducible iff the polynomial obtained by "flipping" the coefficients is irreducible. That is:
  $$ -5x^5+2x^4+16x^3+20x^2+10x+2 $$


• Apply eisenstein's criterion for $p=2$ to deduce irreducibillity.

Do you have any other, maybe more natural solutions?

abstract-algebra irreducible-polynomials

share|cite|improve this question

edited Mar 25 '17 at 12:33

35T41

asked Mar 25 '17 at 12:22

35T4135T41

1,758417

$endgroup$

share|cite|improve this question

edited Mar 25 '17 at 12:33

35T41

asked Mar 25 '17 at 12:22

35T4135T41

1,758417

share|cite|improve this question

edited Mar 25 '17 at 12:33

35T41

asked Mar 25 '17 at 12:22

35T4135T41

1,758417

asked Mar 25 '17 at 12:22

35T4135T41

1,758417

asked Mar 25 '17 at 12:22

35T4135T41

1,758417

2 Answers
2

active

oldest

votes

2

$begingroup$

The polynomial is irreducible modulo $13$, and hence irreducible over $mathbbZ$, and hence over $mathbbQ$. If the polynomial had a non-trivial factorization over $mathbbZ$, then it had also one modulo $p$.

Irreducibility over a finite field $mathbbF_p$ is much easier than over $mathbbZ$, either by using the Berlekamp algorithm, or by assuming
$$
f(x)=(2x^3+ax^2+bx+c)(x^2+dx+e)
$$
and solving the equations over the finite field (which is easy).

share|cite|improve this answer

answered Mar 25 '17 at 13:04

Dietrich BurdeDietrich Burde

81.2k648106

$endgroup$

add a comment | 

1

$begingroup$

One way to do it would be using Cohn's criterion, which states that a polynomial $p$ is irreducible over the integers if its coefficients are non-negative, and if $p(b)$ is prime for some integer $b > maxa_n,cdots,a_0,2$.

Note that $p(x)$ is irreducible if and only if $p(-x)$ is. So let us look at the irreducibility of
$$p(x) = 2x^5 + 4x^2 + 5.$$
This is a form to which we can (try to) apply Cohn' criterion.

To that end, we can just quickly try some integers $b>5$ to see if $p(b)$ is prime. After a few trials, we find that $p(11)=322591$, which is indeed a prime number, so the polynomial is irreducible over $mathbbZ$.

share|cite|improve this answer

answered Mar 25 '17 at 13:10

EuYuEuYu

30.7k754102

$endgroup$

add a comment | 

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2

$begingroup$

The polynomial is irreducible modulo $13$, and hence irreducible over $mathbbZ$, and hence over $mathbbQ$. If the polynomial had a non-trivial factorization over $mathbbZ$, then it had also one modulo $p$.

Irreducibility over a finite field $mathbbF_p$ is much easier than over $mathbbZ$, either by using the Berlekamp algorithm, or by assuming
$$
f(x)=(2x^3+ax^2+bx+c)(x^2+dx+e)
$$
and solving the equations over the finite field (which is easy).

share|cite|improve this answer

answered Mar 25 '17 at 13:04

Dietrich BurdeDietrich Burde

81.2k648106

$endgroup$

add a comment | 

2

$begingroup$

The polynomial is irreducible modulo $13$, and hence irreducible over $mathbbZ$, and hence over $mathbbQ$. If the polynomial had a non-trivial factorization over $mathbbZ$, then it had also one modulo $p$.

Irreducibility over a finite field $mathbbF_p$ is much easier than over $mathbbZ$, either by using the Berlekamp algorithm, or by assuming
$$
f(x)=(2x^3+ax^2+bx+c)(x^2+dx+e)
$$
and solving the equations over the finite field (which is easy).

share|cite|improve this answer

answered Mar 25 '17 at 13:04

Dietrich BurdeDietrich Burde

81.2k648106

$endgroup$

add a comment | 

$begingroup$

The polynomial is irreducible modulo $13$, and hence irreducible over $mathbbZ$, and hence over $mathbbQ$. If the polynomial had a non-trivial factorization over $mathbbZ$, then it had also one modulo $p$.

Irreducibility over a finite field $mathbbF_p$ is much easier than over $mathbbZ$, either by using the Berlekamp algorithm, or by assuming
$$
f(x)=(2x^3+ax^2+bx+c)(x^2+dx+e)
$$
and solving the equations over the finite field (which is easy).

share|cite|improve this answer

answered Mar 25 '17 at 13:04

Dietrich BurdeDietrich Burde

81.2k648106

$endgroup$

share|cite|improve this answer

answered Mar 25 '17 at 13:04

Dietrich BurdeDietrich Burde

81.2k648106

answered Mar 25 '17 at 13:04

Dietrich BurdeDietrich Burde

81.2k648106

answered Mar 25 '17 at 13:04

Dietrich BurdeDietrich Burde

81.2k648106

1

$begingroup$

One way to do it would be using Cohn's criterion, which states that a polynomial $p$ is irreducible over the integers if its coefficients are non-negative, and if $p(b)$ is prime for some integer $b > maxa_n,cdots,a_0,2$.

Note that $p(x)$ is irreducible if and only if $p(-x)$ is. So let us look at the irreducibility of
$$p(x) = 2x^5 + 4x^2 + 5.$$
This is a form to which we can (try to) apply Cohn' criterion.

To that end, we can just quickly try some integers $b>5$ to see if $p(b)$ is prime. After a few trials, we find that $p(11)=322591$, which is indeed a prime number, so the polynomial is irreducible over $mathbbZ$.

share|cite|improve this answer

answered Mar 25 '17 at 13:10

EuYuEuYu

30.7k754102

$endgroup$

add a comment | 

1

$begingroup$

One way to do it would be using Cohn's criterion, which states that a polynomial $p$ is irreducible over the integers if its coefficients are non-negative, and if $p(b)$ is prime for some integer $b > maxa_n,cdots,a_0,2$.

Note that $p(x)$ is irreducible if and only if $p(-x)$ is. So let us look at the irreducibility of
$$p(x) = 2x^5 + 4x^2 + 5.$$
This is a form to which we can (try to) apply Cohn' criterion.

To that end, we can just quickly try some integers $b>5$ to see if $p(b)$ is prime. After a few trials, we find that $p(11)=322591$, which is indeed a prime number, so the polynomial is irreducible over $mathbbZ$.

share|cite|improve this answer

answered Mar 25 '17 at 13:10

EuYuEuYu

30.7k754102

$endgroup$

add a comment | 

$begingroup$

One way to do it would be using Cohn's criterion, which states that a polynomial $p$ is irreducible over the integers if its coefficients are non-negative, and if $p(b)$ is prime for some integer $b > maxa_n,cdots,a_0,2$.

Note that $p(x)$ is irreducible if and only if $p(-x)$ is. So let us look at the irreducibility of
$$p(x) = 2x^5 + 4x^2 + 5.$$
This is a form to which we can (try to) apply Cohn' criterion.

To that end, we can just quickly try some integers $b>5$ to see if $p(b)$ is prime. After a few trials, we find that $p(11)=322591$, which is indeed a prime number, so the polynomial is irreducible over $mathbbZ$.

share|cite|improve this answer

answered Mar 25 '17 at 13:10

EuYuEuYu

30.7k754102

$endgroup$

share|cite|improve this answer

answered Mar 25 '17 at 13:10

EuYuEuYu

30.7k754102

answered Mar 25 '17 at 13:10

EuYuEuYu

30.7k754102

answered Mar 25 '17 at 13:10

EuYuEuYu

30.7k754102