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Irreducibility of a simple polynomial

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3

1

$begingroup$

For an integer $a$, I'm trying to find a criterion to tell me if $x^4+a^2$ is irreducible over $mathbbQ$.

What I've done so far is shown that if $a$ is odd, then $a^2$ is congruent to $1 mod 4$, and so the Eisenstein criterion with $p = 2$ tells me that $(x+1)^4 + a^2$ is irreducible and hence $x^4+a^2$ is irreducible.

For $a$ even I have had no such luck. I know that the polynomial is not irreducible for all such $a$ because when $a = 2$, for example, we have the factorization $x^4+4 = (x^2-2x+2)(x^2+2x+2)$. It's easy to show that this polynomial has no linear factors, but I'm at a loss trying to decide when there are a pair of irreducible quadratic factors.

A thought I had was to try and consider when $mathbbQ(sqrtai)$ is a degree $4$ extension, but this didn't seem to help.

abstract-algebra field-theory irreducible-polynomials

share|cite|improve this question

asked 5 hours ago

JonHalesJonHales

520311

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  If $a=2n^2$, then $x^4+a^2=x^4+(2n^2)^2=(x^2-2nx+2n^2)(x^2+2nx+2n^2)$. In other cases it seems to be irreducible.
  $endgroup$
  – Sil
  5 hours ago
  
  

  
  
  
  

add a comment | 

3

1

$begingroup$

For an integer $a$, I'm trying to find a criterion to tell me if $x^4+a^2$ is irreducible over $mathbbQ$.

What I've done so far is shown that if $a$ is odd, then $a^2$ is congruent to $1 mod 4$, and so the Eisenstein criterion with $p = 2$ tells me that $(x+1)^4 + a^2$ is irreducible and hence $x^4+a^2$ is irreducible.

For $a$ even I have had no such luck. I know that the polynomial is not irreducible for all such $a$ because when $a = 2$, for example, we have the factorization $x^4+4 = (x^2-2x+2)(x^2+2x+2)$. It's easy to show that this polynomial has no linear factors, but I'm at a loss trying to decide when there are a pair of irreducible quadratic factors.

A thought I had was to try and consider when $mathbbQ(sqrtai)$ is a degree $4$ extension, but this didn't seem to help.

abstract-algebra field-theory irreducible-polynomials

share|cite|improve this question

asked 5 hours ago

JonHalesJonHales

520311

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  If $a=2n^2$, then $x^4+a^2=x^4+(2n^2)^2=(x^2-2nx+2n^2)(x^2+2nx+2n^2)$. In other cases it seems to be irreducible.
  $endgroup$
  – Sil
  5 hours ago
  
  

  
  
  
  

add a comment | 

$begingroup$

For an integer $a$, I'm trying to find a criterion to tell me if $x^4+a^2$ is irreducible over $mathbbQ$.

What I've done so far is shown that if $a$ is odd, then $a^2$ is congruent to $1 mod 4$, and so the Eisenstein criterion with $p = 2$ tells me that $(x+1)^4 + a^2$ is irreducible and hence $x^4+a^2$ is irreducible.

For $a$ even I have had no such luck. I know that the polynomial is not irreducible for all such $a$ because when $a = 2$, for example, we have the factorization $x^4+4 = (x^2-2x+2)(x^2+2x+2)$. It's easy to show that this polynomial has no linear factors, but I'm at a loss trying to decide when there are a pair of irreducible quadratic factors.

A thought I had was to try and consider when $mathbbQ(sqrtai)$ is a degree $4$ extension, but this didn't seem to help.

abstract-algebra field-theory irreducible-polynomials

share|cite|improve this question

asked 5 hours ago

JonHalesJonHales

520311

$endgroup$

share|cite|improve this question

asked 5 hours ago

JonHalesJonHales

520311

share|cite|improve this question

asked 5 hours ago

JonHalesJonHales

520311

asked 5 hours ago

JonHalesJonHales

520311

asked 5 hours ago

JonHalesJonHales

520311

2 Answers
2

active

oldest

votes

4

$begingroup$

Notice that we are trying to reduce that polynomial by this way:

$$x^4+a^2=(x^2-bx+a)(x^2+bx+a)=x^4+(2a-b^2)x^2+a^2$$

We need:

$$2a-b^2=0$$
$$b=sqrt2a$$

But since we are working on integers then $$a=2k^2$$ .So your polynomial is reducible if and only if it can be written i nthis form:

$$x^4+4k^4=(x^2-2kx+2k^2)(x^2+2kx+2k^2)$$

Which is also known as Sophie Germain Identity.

share|cite|improve this answer

answered 4 hours ago

EurekaEureka

25611

New contributor

Eureka is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I think you should justify why $(x^2-bx+a)(x^2+bx+a)$ is the only possible form, and not generic $(x^2+mx+n)(x^2+px+q)$
  $endgroup$
  – Sil
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Sil It's pretty easy with complex factorization. And it can also be done with a system of equations.
  $endgroup$
  – Eureka
  4 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @Sil Try expanding out $(x^2+mx+n)(x^2+px+q)$ and look at the $x^3$ term to why $m=-p$. Then expand $(x^2-bx+n)(x^2+bx+q)$ and look at the $x$ term to see why $n=q$.
  $endgroup$
  – Ethan MacBrough
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @EthanMacBrough I understand, my point though was that things like that should be in answer itself.
  $endgroup$
  – Sil
  4 hours ago
  

  
  
  
  

add a comment | 

2

$begingroup$

Assuming, without loss of generality, $a>0$, the polynomial can be rewritten as
$$
x^4+2ax^2+a^2-2ax^2=(x^2+a)^2-(sqrt2ax)^2=
(x^2-sqrt2ax+a)(x^2+sqrt2ax+a)
$$
and it's obvious that the two polynomials are irreducible over $mathbbR$. By uniqueness of factorization, this is a factorization in $mathbbQ[x]$ if and only if $2a$ is a square in $mathbbQ$.

share|cite|improve this answer

answered 4 hours ago

egregegreg

185k1486206

$endgroup$

add a comment | 

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4

$begingroup$

Notice that we are trying to reduce that polynomial by this way:

$$x^4+a^2=(x^2-bx+a)(x^2+bx+a)=x^4+(2a-b^2)x^2+a^2$$

We need:

$$2a-b^2=0$$
$$b=sqrt2a$$

But since we are working on integers then $$a=2k^2$$ .So your polynomial is reducible if and only if it can be written i nthis form:

$$x^4+4k^4=(x^2-2kx+2k^2)(x^2+2kx+2k^2)$$

Which is also known as Sophie Germain Identity.

share|cite|improve this answer

answered 4 hours ago

EurekaEureka

25611

New contributor

Eureka is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I think you should justify why $(x^2-bx+a)(x^2+bx+a)$ is the only possible form, and not generic $(x^2+mx+n)(x^2+px+q)$
  $endgroup$
  – Sil
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Sil It's pretty easy with complex factorization. And it can also be done with a system of equations.
  $endgroup$
  – Eureka
  4 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @Sil Try expanding out $(x^2+mx+n)(x^2+px+q)$ and look at the $x^3$ term to why $m=-p$. Then expand $(x^2-bx+n)(x^2+bx+q)$ and look at the $x$ term to see why $n=q$.
  $endgroup$
  – Ethan MacBrough
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @EthanMacBrough I understand, my point though was that things like that should be in answer itself.
  $endgroup$
  – Sil
  4 hours ago
  

  
  
  
  

add a comment | 

4

$begingroup$

Notice that we are trying to reduce that polynomial by this way:

$$x^4+a^2=(x^2-bx+a)(x^2+bx+a)=x^4+(2a-b^2)x^2+a^2$$

We need:

$$2a-b^2=0$$
$$b=sqrt2a$$

But since we are working on integers then $$a=2k^2$$ .So your polynomial is reducible if and only if it can be written i nthis form:

$$x^4+4k^4=(x^2-2kx+2k^2)(x^2+2kx+2k^2)$$

Which is also known as Sophie Germain Identity.

share|cite|improve this answer

answered 4 hours ago

EurekaEureka

25611

New contributor

Eureka is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I think you should justify why $(x^2-bx+a)(x^2+bx+a)$ is the only possible form, and not generic $(x^2+mx+n)(x^2+px+q)$
  $endgroup$
  – Sil
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Sil It's pretty easy with complex factorization. And it can also be done with a system of equations.
  $endgroup$
  – Eureka
  4 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @Sil Try expanding out $(x^2+mx+n)(x^2+px+q)$ and look at the $x^3$ term to why $m=-p$. Then expand $(x^2-bx+n)(x^2+bx+q)$ and look at the $x$ term to see why $n=q$.
  $endgroup$
  – Ethan MacBrough
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @EthanMacBrough I understand, my point though was that things like that should be in answer itself.
  $endgroup$
  – Sil
  4 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

Notice that we are trying to reduce that polynomial by this way:

$$x^4+a^2=(x^2-bx+a)(x^2+bx+a)=x^4+(2a-b^2)x^2+a^2$$

We need:

$$2a-b^2=0$$
$$b=sqrt2a$$

But since we are working on integers then $$a=2k^2$$ .So your polynomial is reducible if and only if it can be written i nthis form:

$$x^4+4k^4=(x^2-2kx+2k^2)(x^2+2kx+2k^2)$$

Which is also known as Sophie Germain Identity.

share|cite|improve this answer

answered 4 hours ago

EurekaEureka

25611

New contributor

Eureka is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this answer

answered 4 hours ago

EurekaEureka

25611

New contributor

Eureka is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

answered 4 hours ago

EurekaEureka

25611

New contributor

Eureka is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

answered 4 hours ago

EurekaEureka

25611

2

$begingroup$

Assuming, without loss of generality, $a>0$, the polynomial can be rewritten as
$$
x^4+2ax^2+a^2-2ax^2=(x^2+a)^2-(sqrt2ax)^2=
(x^2-sqrt2ax+a)(x^2+sqrt2ax+a)
$$
and it's obvious that the two polynomials are irreducible over $mathbbR$. By uniqueness of factorization, this is a factorization in $mathbbQ[x]$ if and only if $2a$ is a square in $mathbbQ$.

share|cite|improve this answer

answered 4 hours ago

egregegreg

185k1486206

$endgroup$

add a comment | 

2

$begingroup$

Assuming, without loss of generality, $a>0$, the polynomial can be rewritten as
$$
x^4+2ax^2+a^2-2ax^2=(x^2+a)^2-(sqrt2ax)^2=
(x^2-sqrt2ax+a)(x^2+sqrt2ax+a)
$$
and it's obvious that the two polynomials are irreducible over $mathbbR$. By uniqueness of factorization, this is a factorization in $mathbbQ[x]$ if and only if $2a$ is a square in $mathbbQ$.

share|cite|improve this answer

answered 4 hours ago

egregegreg

185k1486206

$endgroup$

add a comment | 

$begingroup$

Assuming, without loss of generality, $a>0$, the polynomial can be rewritten as
$$
x^4+2ax^2+a^2-2ax^2=(x^2+a)^2-(sqrt2ax)^2=
(x^2-sqrt2ax+a)(x^2+sqrt2ax+a)
$$
and it's obvious that the two polynomials are irreducible over $mathbbR$. By uniqueness of factorization, this is a factorization in $mathbbQ[x]$ if and only if $2a$ is a square in $mathbbQ$.

share|cite|improve this answer

answered 4 hours ago

egregegreg

185k1486206

$endgroup$

share|cite|improve this answer

answered 4 hours ago

egregegreg

185k1486206

answered 4 hours ago

egregegreg

185k1486206

answered 4 hours ago

egregegreg

185k1486206