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Prove $x^n+x^{n-1}+2$ is irreducible in $mathbb{Q}[x]$ for $ngeq 2$

Let $f (x) in mathbb{Z}[x]$ be irreducible. Prove that $f(x)$ is primitive.Is $x^8+1$ irreducible in $mathbb{R}[x]$$f(Y),g(Y)in K[Y]$ coprime $Rightarrow f(Y)-g(Y)X$ irreducible in $K(X)[Y]$condition for $f(x)$ to be irreducible in $mathbb{Q}$ when it is reducible in $mathbb{Z}$If $x^p−x−c$ is irreducible in $F[x]$ then it has no root in the field.Question on splitting field and irreducible polynomials.$x^6+2x^3-3x^2+1$, irreducible over $mathbb{Q}$Calculating the number of irreducible polynomials over a finite fieldCan we find an irreducible polynomial of any degree in $mathbb C[x,y]$?If $degleft(fright) = minleft(left{d in mathbb{N}^times;; fvert X^{q^d} - Xright}right),$, then $f$ is irreducible.

1

$begingroup$

I want to prove $f(x)=x^n+x^{n-1}+2$ is irreducible in $mathbb{Q}[x]$ for $ngeq 2$. I think I'm very close to doing this, and I have used the following theorem:

Generalized Eisenstein criterion: Let $f(x)inmathbb{Z}[x]$ be
primitive and let $p$ be a prime number. Write $f(x)=f_nx^n+ ... +
> f_mx^m + ... + f_1x + f_0$ with $mleq n$. Suppose that $p$ does not
divide $f_nf_m$, and $p|f_i$ for all $i<m$, and $p^2$ does not divide
$f_0=f(0)$. Then $f(x)$ has an irreducible factor in $mathbb{Q}[x]$
of degree $geq m$.

This is what I got so far:

Let $f(x)=x^n+x^{n-1}+2$. We see that $f(x)inmathbb{Z}[x]$ and that
it is primitive ($gcd(1,1,2)=1$, and $1$ a unit in $mathbb{Z}$). Let
$m=n-1$ and let $p=2$ a prime number. Then $f_nf_m=1$ and $2$ does not
divide $1$. $2$ does divide all coefficients $f_i$ with $i<m$, since
only $f_0neq 0$, and $2|0$, and $2|2$ and $f_0=2$. We also see that
$p^2=2^2=4$ does not divide $f_0$. By theorem 2 we have that $f(x)$
has an irreducible factor $g(x)$ in $mathbb{Q}[x]$ of degree $geq
> m$. So $deg(g(x))$ is either $n-1$ or $n$. Suppose $f(x)$ is
reducible in $mathbb{Q}[x]$ and can be written as $f(x)=g(x)h(x)$,
with $h(x)inmathbb{Q}[x]$.

If $deg(g(x))=n$, then $h(x)$ is a constant term and in $mathbb{Q}$.
Every non-zero element of $mathbb{Q}$ is a unit, and $h(x)$ cannot be
zero, because then $f(x)equiv 0$ which is not the case. So if
$deg(g(x))=n$, we have that $f(x)$ irreducible.

If $deg(g(x))=n-1$, then $h(x)$ is a linear term, not constant, and
thus there is an element $ainmathbb{Q}$ such that $h(a)equiv 0$.

Assuming this is all correct, how do I show that either $h(x)$ with degree $1$ is a unit (which I think cannot happen in $mathbb{Q}[x]$), or $h(x)$ cannot exist for some reason, (maybe because $f(x)$ cannot have roots)?

And a question about my proof so far: $h(x)$ is linear, so it has a root (zero point). In what number field does this root lie? As you can see in my prove I thought it would be $mathbb{Q}$ since we are talking about $h(x)inmathbb{Q}[x]$

ring-theory irreducible-polynomials

share|cite|improve this question

edited Mar 14 at 13:42

The Coding Wombat

asked Mar 14 at 13:22

The Coding WombatThe Coding Wombat

23919

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, the root if it existed would need to be in $mathbb Q$
  $endgroup$
  – Ewan Delanoy
  Mar 14 at 13:46
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You can also prove this by showing that this polynomial has no complex roots inside or on the unit circle, which together with constant coefficient $p=2$ is enough to prove it is irreducible.
  $endgroup$
  – Sil
  Mar 15 at 23:50
  

  
  
  
  

add a comment | 

1

$begingroup$

I want to prove $f(x)=x^n+x^{n-1}+2$ is irreducible in $mathbb{Q}[x]$ for $ngeq 2$. I think I'm very close to doing this, and I have used the following theorem:

Generalized Eisenstein criterion: Let $f(x)inmathbb{Z}[x]$ be
primitive and let $p$ be a prime number. Write $f(x)=f_nx^n+ ... +
> f_mx^m + ... + f_1x + f_0$ with $mleq n$. Suppose that $p$ does not
divide $f_nf_m$, and $p|f_i$ for all $i<m$, and $p^2$ does not divide
$f_0=f(0)$. Then $f(x)$ has an irreducible factor in $mathbb{Q}[x]$
of degree $geq m$.

This is what I got so far:

Let $f(x)=x^n+x^{n-1}+2$. We see that $f(x)inmathbb{Z}[x]$ and that
it is primitive ($gcd(1,1,2)=1$, and $1$ a unit in $mathbb{Z}$). Let
$m=n-1$ and let $p=2$ a prime number. Then $f_nf_m=1$ and $2$ does not
divide $1$. $2$ does divide all coefficients $f_i$ with $i<m$, since
only $f_0neq 0$, and $2|0$, and $2|2$ and $f_0=2$. We also see that
$p^2=2^2=4$ does not divide $f_0$. By theorem 2 we have that $f(x)$
has an irreducible factor $g(x)$ in $mathbb{Q}[x]$ of degree $geq
> m$. So $deg(g(x))$ is either $n-1$ or $n$. Suppose $f(x)$ is
reducible in $mathbb{Q}[x]$ and can be written as $f(x)=g(x)h(x)$,
with $h(x)inmathbb{Q}[x]$.

If $deg(g(x))=n$, then $h(x)$ is a constant term and in $mathbb{Q}$.
Every non-zero element of $mathbb{Q}$ is a unit, and $h(x)$ cannot be
zero, because then $f(x)equiv 0$ which is not the case. So if
$deg(g(x))=n$, we have that $f(x)$ irreducible.

If $deg(g(x))=n-1$, then $h(x)$ is a linear term, not constant, and
thus there is an element $ainmathbb{Q}$ such that $h(a)equiv 0$.

Assuming this is all correct, how do I show that either $h(x)$ with degree $1$ is a unit (which I think cannot happen in $mathbb{Q}[x]$), or $h(x)$ cannot exist for some reason, (maybe because $f(x)$ cannot have roots)?

And a question about my proof so far: $h(x)$ is linear, so it has a root (zero point). In what number field does this root lie? As you can see in my prove I thought it would be $mathbb{Q}$ since we are talking about $h(x)inmathbb{Q}[x]$

ring-theory irreducible-polynomials

share|cite|improve this question

edited Mar 14 at 13:42

The Coding Wombat

asked Mar 14 at 13:22

The Coding WombatThe Coding Wombat

23919

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, the root if it existed would need to be in $mathbb Q$
  $endgroup$
  – Ewan Delanoy
  Mar 14 at 13:46
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You can also prove this by showing that this polynomial has no complex roots inside or on the unit circle, which together with constant coefficient $p=2$ is enough to prove it is irreducible.
  $endgroup$
  – Sil
  Mar 15 at 23:50
  

  
  
  
  

add a comment | 

$begingroup$

I want to prove $f(x)=x^n+x^{n-1}+2$ is irreducible in $mathbb{Q}[x]$ for $ngeq 2$. I think I'm very close to doing this, and I have used the following theorem:

Generalized Eisenstein criterion: Let $f(x)inmathbb{Z}[x]$ be
primitive and let $p$ be a prime number. Write $f(x)=f_nx^n+ ... +
> f_mx^m + ... + f_1x + f_0$ with $mleq n$. Suppose that $p$ does not
divide $f_nf_m$, and $p|f_i$ for all $i<m$, and $p^2$ does not divide
$f_0=f(0)$. Then $f(x)$ has an irreducible factor in $mathbb{Q}[x]$
of degree $geq m$.

This is what I got so far:

Let $f(x)=x^n+x^{n-1}+2$. We see that $f(x)inmathbb{Z}[x]$ and that
it is primitive ($gcd(1,1,2)=1$, and $1$ a unit in $mathbb{Z}$). Let
$m=n-1$ and let $p=2$ a prime number. Then $f_nf_m=1$ and $2$ does not
divide $1$. $2$ does divide all coefficients $f_i$ with $i<m$, since
only $f_0neq 0$, and $2|0$, and $2|2$ and $f_0=2$. We also see that
$p^2=2^2=4$ does not divide $f_0$. By theorem 2 we have that $f(x)$
has an irreducible factor $g(x)$ in $mathbb{Q}[x]$ of degree $geq
> m$. So $deg(g(x))$ is either $n-1$ or $n$. Suppose $f(x)$ is
reducible in $mathbb{Q}[x]$ and can be written as $f(x)=g(x)h(x)$,
with $h(x)inmathbb{Q}[x]$.

If $deg(g(x))=n$, then $h(x)$ is a constant term and in $mathbb{Q}$.
Every non-zero element of $mathbb{Q}$ is a unit, and $h(x)$ cannot be
zero, because then $f(x)equiv 0$ which is not the case. So if
$deg(g(x))=n$, we have that $f(x)$ irreducible.

If $deg(g(x))=n-1$, then $h(x)$ is a linear term, not constant, and
thus there is an element $ainmathbb{Q}$ such that $h(a)equiv 0$.

Assuming this is all correct, how do I show that either $h(x)$ with degree $1$ is a unit (which I think cannot happen in $mathbb{Q}[x]$), or $h(x)$ cannot exist for some reason, (maybe because $f(x)$ cannot have roots)?

And a question about my proof so far: $h(x)$ is linear, so it has a root (zero point). In what number field does this root lie? As you can see in my prove I thought it would be $mathbb{Q}$ since we are talking about $h(x)inmathbb{Q}[x]$

ring-theory irreducible-polynomials

share|cite|improve this question

edited Mar 14 at 13:42

The Coding Wombat

asked Mar 14 at 13:22

The Coding WombatThe Coding Wombat

23919

$endgroup$

share|cite|improve this question

edited Mar 14 at 13:42

The Coding Wombat

asked Mar 14 at 13:22

The Coding WombatThe Coding Wombat

23919

share|cite|improve this question

edited Mar 14 at 13:42

The Coding Wombat

asked Mar 14 at 13:22

The Coding WombatThe Coding Wombat

23919

asked Mar 14 at 13:22

The Coding WombatThe Coding Wombat

23919

asked Mar 14 at 13:22

The Coding WombatThe Coding Wombat

23919

2 Answers
2

active

oldest

votes

1

$begingroup$

It is indeed true that $f$ has no roots in ${mathbb Q}$. If $frac{p}{q}$ (with $p,q$ coprime integers and $q>0$) were a root of $f$, then $p^n+p^{n-1}q+2q^n=0$, so that $q$ divides $p^n$. By Gauss' lemma, $q$ divides $1$, so $q=1$. Then $p^n+p^{n-1}=(-2)$, so $p$ divides $2$, and is therefore one of $1,2,-1,-2$. It is easy to see that none of those is a solution.

share|cite|improve this answer

answered Mar 14 at 13:45

Ewan DelanoyEwan Delanoy

42k443104

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What do you mean by "so that $q$ divides $p^n$? You got the equation before that by filling in the fraction and multiplying everything by $q^n$ right?
  $endgroup$
  – The Coding Wombat
  Mar 14 at 13:59
  
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I mean that, from $p^n+p^{n-1}q+2q^n=0$ you deduce $p^n=q(-p^{n-1}-2q^{n-1})$, so $p^n$ is $q$ times something, so $q$ divides $p^n$
  $endgroup$
  – Ewan Delanoy
  Mar 14 at 16:53
  

  
  
  
  

add a comment | 

0

$begingroup$

Hint: For $n>2$ apply Eisenstein to $f(x+2)$ and study $n=1,2$.

share|cite|improve this answer

answered Mar 14 at 13:28

Tsemo AristideTsemo Aristide

59.8k11446

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Does this continue on the partial proof I've given?
  $endgroup$
  – The Coding Wombat
  Mar 14 at 13:36
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  And I forgot to add a criterion: $ngeq 2$.
  $endgroup$
  – The Coding Wombat
  Mar 14 at 13:37
  

  
  
  
  

add a comment | 

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1

$begingroup$

It is indeed true that $f$ has no roots in ${mathbb Q}$. If $frac{p}{q}$ (with $p,q$ coprime integers and $q>0$) were a root of $f$, then $p^n+p^{n-1}q+2q^n=0$, so that $q$ divides $p^n$. By Gauss' lemma, $q$ divides $1$, so $q=1$. Then $p^n+p^{n-1}=(-2)$, so $p$ divides $2$, and is therefore one of $1,2,-1,-2$. It is easy to see that none of those is a solution.

share|cite|improve this answer

answered Mar 14 at 13:45

Ewan DelanoyEwan Delanoy

42k443104

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What do you mean by "so that $q$ divides $p^n$? You got the equation before that by filling in the fraction and multiplying everything by $q^n$ right?
  $endgroup$
  – The Coding Wombat
  Mar 14 at 13:59
  
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I mean that, from $p^n+p^{n-1}q+2q^n=0$ you deduce $p^n=q(-p^{n-1}-2q^{n-1})$, so $p^n$ is $q$ times something, so $q$ divides $p^n$
  $endgroup$
  – Ewan Delanoy
  Mar 14 at 16:53
  

  
  
  
  

add a comment | 

1

$begingroup$

It is indeed true that $f$ has no roots in ${mathbb Q}$. If $frac{p}{q}$ (with $p,q$ coprime integers and $q>0$) were a root of $f$, then $p^n+p^{n-1}q+2q^n=0$, so that $q$ divides $p^n$. By Gauss' lemma, $q$ divides $1$, so $q=1$. Then $p^n+p^{n-1}=(-2)$, so $p$ divides $2$, and is therefore one of $1,2,-1,-2$. It is easy to see that none of those is a solution.

share|cite|improve this answer

answered Mar 14 at 13:45

Ewan DelanoyEwan Delanoy

42k443104

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What do you mean by "so that $q$ divides $p^n$? You got the equation before that by filling in the fraction and multiplying everything by $q^n$ right?
  $endgroup$
  – The Coding Wombat
  Mar 14 at 13:59
  
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I mean that, from $p^n+p^{n-1}q+2q^n=0$ you deduce $p^n=q(-p^{n-1}-2q^{n-1})$, so $p^n$ is $q$ times something, so $q$ divides $p^n$
  $endgroup$
  – Ewan Delanoy
  Mar 14 at 16:53
  

  
  
  
  

add a comment | 

$begingroup$

It is indeed true that $f$ has no roots in ${mathbb Q}$. If $frac{p}{q}$ (with $p,q$ coprime integers and $q>0$) were a root of $f$, then $p^n+p^{n-1}q+2q^n=0$, so that $q$ divides $p^n$. By Gauss' lemma, $q$ divides $1$, so $q=1$. Then $p^n+p^{n-1}=(-2)$, so $p$ divides $2$, and is therefore one of $1,2,-1,-2$. It is easy to see that none of those is a solution.

share|cite|improve this answer

answered Mar 14 at 13:45

Ewan DelanoyEwan Delanoy

42k443104

$endgroup$

share|cite|improve this answer

answered Mar 14 at 13:45

Ewan DelanoyEwan Delanoy

42k443104

answered Mar 14 at 13:45

Ewan DelanoyEwan Delanoy

42k443104

answered Mar 14 at 13:45

Ewan DelanoyEwan Delanoy

42k443104

0

$begingroup$

Hint: For $n>2$ apply Eisenstein to $f(x+2)$ and study $n=1,2$.

share|cite|improve this answer

answered Mar 14 at 13:28

Tsemo AristideTsemo Aristide

59.8k11446

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Does this continue on the partial proof I've given?
  $endgroup$
  – The Coding Wombat
  Mar 14 at 13:36
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  And I forgot to add a criterion: $ngeq 2$.
  $endgroup$
  – The Coding Wombat
  Mar 14 at 13:37
  

  
  
  
  

add a comment | 

0

$begingroup$

Hint: For $n>2$ apply Eisenstein to $f(x+2)$ and study $n=1,2$.

share|cite|improve this answer

answered Mar 14 at 13:28

Tsemo AristideTsemo Aristide

59.8k11446

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Does this continue on the partial proof I've given?
  $endgroup$
  – The Coding Wombat
  Mar 14 at 13:36
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  And I forgot to add a criterion: $ngeq 2$.
  $endgroup$
  – The Coding Wombat
  Mar 14 at 13:37
  

  
  
  
  

add a comment | 

$begingroup$

Hint: For $n>2$ apply Eisenstein to $f(x+2)$ and study $n=1,2$.

share|cite|improve this answer

answered Mar 14 at 13:28

Tsemo AristideTsemo Aristide

59.8k11446

$endgroup$

share|cite|improve this answer

answered Mar 14 at 13:28

Tsemo AristideTsemo Aristide

59.8k11446

answered Mar 14 at 13:28

Tsemo AristideTsemo Aristide

59.8k11446

answered Mar 14 at 13:28

Tsemo AristideTsemo Aristide

59.8k11446