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Is there a cubic irreducible polynomial over $Bbb Q$ with all of its three roots are irrational?

proving that this family of angles, cannot be trisectedProving that either $x $ or $2x$ is a generator of the cyclic group $(mathbb Z_3[x]/langle f(x) rangle)^*$Proving this polynomial is irreducible over any algebraically closed field.Radical extension with root of cubic polynomialForming $K(alpha) = K[x]/langle frangle = L$ and $L(beta) = L[x]/langle grangle$ how do elements appearIs it possible for an irreducible polynomial with rational coefficients to have three zeros in an arithmetic progression?Totally real Galois extension of given degreeHow to determine the Galois Group of a polynomial with “ugly” roots?Proof of $inotinmathbbQ(alpha)$ for $alpha$ root of irreducible polynomial with odd degreeIf one root is in then all roots are in

0

$begingroup$

Prove or Disprove: For every irreducible cubic polynomial $f(x)in Bbb Q[x]$, there exist a subfield $F$ of $Bbb C$ such that $F nsubseteq Bbb R$ and $F simeq Bbb Q[x]/langle f(x) rangle $

Here, $deg f$ is odd implies $f$ has a real root $alpha$ and by irreducibility, $alpha$ must be irrational. Thus $$F(alpha) simeq Bbb Q[x]/langle f(x) rangle $$ and $F(alpha) subset Bbb R subset Bbb C$

If $beta, gamma $ are other roots of $f$ (in some extention) ,then $$F(beta) simeq Bbb Q[x]/langle f(x) rangle simeq F(gamma) $$ If $beta , gamma in Bbb C$ then $F(gamma) nsubseteq Bbb R$. So every $f$ has the desired property, so the statement is true

But the answer given is "the statement is false". So my question is :

Is there any such $f$ with all three roots are irrational ? How to disprove the statement ?

abstract-algebra field-theory

share|cite|improve this question

asked Mar 21 at 12:44

Chinnapparaj RChinnapparaj R

5,8682928

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  See this piece of classical mathematics
  $endgroup$
  – user647486
  Mar 21 at 12:49
  

  
  
  
  

add a comment | 

0

$begingroup$

Prove or Disprove: For every irreducible cubic polynomial $f(x)in Bbb Q[x]$, there exist a subfield $F$ of $Bbb C$ such that $F nsubseteq Bbb R$ and $F simeq Bbb Q[x]/langle f(x) rangle $

Here, $deg f$ is odd implies $f$ has a real root $alpha$ and by irreducibility, $alpha$ must be irrational. Thus $$F(alpha) simeq Bbb Q[x]/langle f(x) rangle $$ and $F(alpha) subset Bbb R subset Bbb C$

If $beta, gamma $ are other roots of $f$ (in some extention) ,then $$F(beta) simeq Bbb Q[x]/langle f(x) rangle simeq F(gamma) $$ If $beta , gamma in Bbb C$ then $F(gamma) nsubseteq Bbb R$. So every $f$ has the desired property, so the statement is true

But the answer given is "the statement is false". So my question is :

Is there any such $f$ with all three roots are irrational ? How to disprove the statement ?

abstract-algebra field-theory

share|cite|improve this question

asked Mar 21 at 12:44

Chinnapparaj RChinnapparaj R

5,8682928

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  See this piece of classical mathematics
  $endgroup$
  – user647486
  Mar 21 at 12:49
  

  
  
  
  

add a comment | 

$begingroup$

Prove or Disprove: For every irreducible cubic polynomial $f(x)in Bbb Q[x]$, there exist a subfield $F$ of $Bbb C$ such that $F nsubseteq Bbb R$ and $F simeq Bbb Q[x]/langle f(x) rangle $

Here, $deg f$ is odd implies $f$ has a real root $alpha$ and by irreducibility, $alpha$ must be irrational. Thus $$F(alpha) simeq Bbb Q[x]/langle f(x) rangle $$ and $F(alpha) subset Bbb R subset Bbb C$

If $beta, gamma $ are other roots of $f$ (in some extention) ,then $$F(beta) simeq Bbb Q[x]/langle f(x) rangle simeq F(gamma) $$ If $beta , gamma in Bbb C$ then $F(gamma) nsubseteq Bbb R$. So every $f$ has the desired property, so the statement is true

But the answer given is "the statement is false". So my question is :

Is there any such $f$ with all three roots are irrational ? How to disprove the statement ?

abstract-algebra field-theory

share|cite|improve this question

asked Mar 21 at 12:44

Chinnapparaj RChinnapparaj R

5,8682928

$endgroup$

share|cite|improve this question

asked Mar 21 at 12:44

Chinnapparaj RChinnapparaj R

5,8682928

share|cite|improve this question

asked Mar 21 at 12:44

Chinnapparaj RChinnapparaj R

5,8682928

asked Mar 21 at 12:44

Chinnapparaj RChinnapparaj R

5,8682928

asked Mar 21 at 12:44

Chinnapparaj RChinnapparaj R

5,8682928

3 Answers
3

active

oldest

votes

3

$begingroup$

The answer to your title question is yes. For example,
$$
x^3-3x+1=0
$$
has three real solutions and none of them is rational by the rational root test.

share|cite|improve this answer

answered Mar 21 at 12:54

Dietrich BurdeDietrich Burde

81.6k648106

$endgroup$

add a comment | 

2

$begingroup$

I believe the submitter has misinterpreted the question. The expression '$F nsubseteq Bbb R$' has nothing to do with irrationality per se; it expresses the notion that $F$ is a field contained in $Bbb C$ but NOT in $Bbb R$ - i.e that $F$ necessarily contains complex-valued elements whose imaginary parts are non-zero. So the question is, whether EVERY cubic has at least one non-real complex root - which plainly is not true, as other answers show.

share|cite|improve this answer

answered Mar 21 at 13:02

PMarPMar

211

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You might add a few words about "up to isomorphism".
  $endgroup$
  – hardmath
  Mar 21 at 14:33
  

  
  
  
  

add a comment | 

1

$begingroup$

Let's take a nice cubic polynomial with 3 real roots (like $x(x-2)(x+2) = x^3 - 4x$), then add a small constant (like $2$) to it. What we get is $x^3 - 4x + 2$, which is a polynomial with three real roots (calculating the value of the polynomial at $-3, 0, 1, 2$ shows that it still has three real roots) and by Eisenstein it is irreducible.

One might need some tweaking to make this work (my original attempt used $x^3 - x$ instead, which doesn't work quite as nicely), but it's a simple way to make examples relatively consistently.

share|cite|improve this answer

answered Mar 21 at 12:54

ArthurArthur

122k7122210

$endgroup$

add a comment | 

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3

$begingroup$

The answer to your title question is yes. For example,
$$
x^3-3x+1=0
$$
has three real solutions and none of them is rational by the rational root test.

share|cite|improve this answer

answered Mar 21 at 12:54

Dietrich BurdeDietrich Burde

81.6k648106

$endgroup$

add a comment | 

3

$begingroup$

The answer to your title question is yes. For example,
$$
x^3-3x+1=0
$$
has three real solutions and none of them is rational by the rational root test.

share|cite|improve this answer

answered Mar 21 at 12:54

Dietrich BurdeDietrich Burde

81.6k648106

$endgroup$

add a comment | 

$begingroup$

The answer to your title question is yes. For example,
$$
x^3-3x+1=0
$$
has three real solutions and none of them is rational by the rational root test.

share|cite|improve this answer

answered Mar 21 at 12:54

Dietrich BurdeDietrich Burde

81.6k648106

$endgroup$

share|cite|improve this answer

answered Mar 21 at 12:54

Dietrich BurdeDietrich Burde

81.6k648106

answered Mar 21 at 12:54

Dietrich BurdeDietrich Burde

81.6k648106

answered Mar 21 at 12:54

Dietrich BurdeDietrich Burde

81.6k648106

2

$begingroup$

I believe the submitter has misinterpreted the question. The expression '$F nsubseteq Bbb R$' has nothing to do with irrationality per se; it expresses the notion that $F$ is a field contained in $Bbb C$ but NOT in $Bbb R$ - i.e that $F$ necessarily contains complex-valued elements whose imaginary parts are non-zero. So the question is, whether EVERY cubic has at least one non-real complex root - which plainly is not true, as other answers show.

share|cite|improve this answer

answered Mar 21 at 13:02

PMarPMar

211

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You might add a few words about "up to isomorphism".
  $endgroup$
  – hardmath
  Mar 21 at 14:33
  

  
  
  
  

add a comment | 

2

$begingroup$

I believe the submitter has misinterpreted the question. The expression '$F nsubseteq Bbb R$' has nothing to do with irrationality per se; it expresses the notion that $F$ is a field contained in $Bbb C$ but NOT in $Bbb R$ - i.e that $F$ necessarily contains complex-valued elements whose imaginary parts are non-zero. So the question is, whether EVERY cubic has at least one non-real complex root - which plainly is not true, as other answers show.

share|cite|improve this answer

answered Mar 21 at 13:02

PMarPMar

211

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You might add a few words about "up to isomorphism".
  $endgroup$
  – hardmath
  Mar 21 at 14:33
  

  
  
  
  

add a comment | 

$begingroup$

I believe the submitter has misinterpreted the question. The expression '$F nsubseteq Bbb R$' has nothing to do with irrationality per se; it expresses the notion that $F$ is a field contained in $Bbb C$ but NOT in $Bbb R$ - i.e that $F$ necessarily contains complex-valued elements whose imaginary parts are non-zero. So the question is, whether EVERY cubic has at least one non-real complex root - which plainly is not true, as other answers show.

share|cite|improve this answer

answered Mar 21 at 13:02

PMarPMar

211

$endgroup$

share|cite|improve this answer

answered Mar 21 at 13:02

PMarPMar

211

answered Mar 21 at 13:02

PMarPMar

211

answered Mar 21 at 13:02

PMarPMar

211

1

$begingroup$

Let's take a nice cubic polynomial with 3 real roots (like $x(x-2)(x+2) = x^3 - 4x$), then add a small constant (like $2$) to it. What we get is $x^3 - 4x + 2$, which is a polynomial with three real roots (calculating the value of the polynomial at $-3, 0, 1, 2$ shows that it still has three real roots) and by Eisenstein it is irreducible.

One might need some tweaking to make this work (my original attempt used $x^3 - x$ instead, which doesn't work quite as nicely), but it's a simple way to make examples relatively consistently.

share|cite|improve this answer

answered Mar 21 at 12:54

ArthurArthur

122k7122210

$endgroup$

add a comment | 

1

$begingroup$

Let's take a nice cubic polynomial with 3 real roots (like $x(x-2)(x+2) = x^3 - 4x$), then add a small constant (like $2$) to it. What we get is $x^3 - 4x + 2$, which is a polynomial with three real roots (calculating the value of the polynomial at $-3, 0, 1, 2$ shows that it still has three real roots) and by Eisenstein it is irreducible.

One might need some tweaking to make this work (my original attempt used $x^3 - x$ instead, which doesn't work quite as nicely), but it's a simple way to make examples relatively consistently.

share|cite|improve this answer

answered Mar 21 at 12:54

ArthurArthur

122k7122210

$endgroup$

add a comment | 

$begingroup$

Let's take a nice cubic polynomial with 3 real roots (like $x(x-2)(x+2) = x^3 - 4x$), then add a small constant (like $2$) to it. What we get is $x^3 - 4x + 2$, which is a polynomial with three real roots (calculating the value of the polynomial at $-3, 0, 1, 2$ shows that it still has three real roots) and by Eisenstein it is irreducible.

One might need some tweaking to make this work (my original attempt used $x^3 - x$ instead, which doesn't work quite as nicely), but it's a simple way to make examples relatively consistently.

share|cite|improve this answer

answered Mar 21 at 12:54

ArthurArthur

122k7122210

$endgroup$

share|cite|improve this answer

answered Mar 21 at 12:54

ArthurArthur

122k7122210

answered Mar 21 at 12:54

ArthurArthur

122k7122210

answered Mar 21 at 12:54

ArthurArthur

122k7122210