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When does a polynomial with integer coefficients have only irrational roots?

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1

$begingroup$

Given a polynomial $P$ with integer coefficients, is there any simple criterion (other than explicitly calculating the roots) with which I can check whether its roots are irrational?

polynomials roots irrational-numbers

share|cite|improve this question

edited Mar 13 at 17:35

J. W. Tanner

3,4601320

asked Mar 13 at 17:21

tsttst

741412

$endgroup$

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  Look up rational root theorem.
  $endgroup$
  – Parcly Taxel
  Mar 13 at 17:22
  

  
  
  
  

add a comment | 

1

$begingroup$

Given a polynomial $P$ with integer coefficients, is there any simple criterion (other than explicitly calculating the roots) with which I can check whether its roots are irrational?

polynomials roots irrational-numbers

share|cite|improve this question

edited Mar 13 at 17:35

J. W. Tanner

3,4601320

asked Mar 13 at 17:21

tsttst

741412

$endgroup$

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  Look up rational root theorem.
  $endgroup$
  – Parcly Taxel
  Mar 13 at 17:22
  

  
  
  
  

add a comment | 

$begingroup$

Given a polynomial $P$ with integer coefficients, is there any simple criterion (other than explicitly calculating the roots) with which I can check whether its roots are irrational?

polynomials roots irrational-numbers

share|cite|improve this question

edited Mar 13 at 17:35

J. W. Tanner

3,4601320

asked Mar 13 at 17:21

tsttst

741412

$endgroup$

share|cite|improve this question

edited Mar 13 at 17:35

J. W. Tanner

3,4601320

asked Mar 13 at 17:21

tsttst

741412

share|cite|improve this question

edited Mar 13 at 17:35

J. W. Tanner

3,4601320

asked Mar 13 at 17:21

tsttst

741412

edited Mar 13 at 17:35

J. W. Tanner

3,4601320

edited Mar 13 at 17:35

J. W. Tanner

3,4601320

asked Mar 13 at 17:21

tsttst

741412

asked Mar 13 at 17:21

tsttst

741412

1 Answer
1

active

oldest

votes

0

$begingroup$

Before improvements you can obtain a bound in an elementary way

$$f(x) = sum_n=0^N a_n x^n in mathbbZ[x]$$
Then $$g(x)= a_N^N-1 f(x/a_N)= x^N+sum_n=0^N-1 b_nx^n in mathbbZ[x] $$

• $M = sup_n |b_n|$. If $|x| > N M$ then $g(x) ne 0$.




• If $g(t) = 0, t=p/q in mathbbQ$ then $t^N = -sum_n=0^N-1 b_n t^n $ $$implies mathbbZ[t] subset sum_n=0^N-1 c_n t^n, c_n in mathbbZ subset q^-N mathbbZ$$ This is a contradiction if $t not in mathbbZ$. Thus $t in mathbbZ$ and $t in -NM,ldots,NM$.

share|cite|improve this answer

edited Mar 13 at 19:32

answered Mar 13 at 18:20

reunsreuns

21.5k21352

$endgroup$

add a comment | 

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0

$begingroup$

Before improvements you can obtain a bound in an elementary way

$$f(x) = sum_n=0^N a_n x^n in mathbbZ[x]$$
Then $$g(x)= a_N^N-1 f(x/a_N)= x^N+sum_n=0^N-1 b_nx^n in mathbbZ[x] $$

• $M = sup_n |b_n|$. If $|x| > N M$ then $g(x) ne 0$.




• If $g(t) = 0, t=p/q in mathbbQ$ then $t^N = -sum_n=0^N-1 b_n t^n $ $$implies mathbbZ[t] subset sum_n=0^N-1 c_n t^n, c_n in mathbbZ subset q^-N mathbbZ$$ This is a contradiction if $t not in mathbbZ$. Thus $t in mathbbZ$ and $t in -NM,ldots,NM$.

share|cite|improve this answer

edited Mar 13 at 19:32

answered Mar 13 at 18:20

reunsreuns

21.5k21352

$endgroup$

add a comment | 

0

$begingroup$

Before improvements you can obtain a bound in an elementary way

$$f(x) = sum_n=0^N a_n x^n in mathbbZ[x]$$
Then $$g(x)= a_N^N-1 f(x/a_N)= x^N+sum_n=0^N-1 b_nx^n in mathbbZ[x] $$

• $M = sup_n |b_n|$. If $|x| > N M$ then $g(x) ne 0$.




• If $g(t) = 0, t=p/q in mathbbQ$ then $t^N = -sum_n=0^N-1 b_n t^n $ $$implies mathbbZ[t] subset sum_n=0^N-1 c_n t^n, c_n in mathbbZ subset q^-N mathbbZ$$ This is a contradiction if $t not in mathbbZ$. Thus $t in mathbbZ$ and $t in -NM,ldots,NM$.

share|cite|improve this answer

edited Mar 13 at 19:32

answered Mar 13 at 18:20

reunsreuns

21.5k21352

$endgroup$

add a comment | 

$begingroup$

Before improvements you can obtain a bound in an elementary way

$$f(x) = sum_n=0^N a_n x^n in mathbbZ[x]$$
Then $$g(x)= a_N^N-1 f(x/a_N)= x^N+sum_n=0^N-1 b_nx^n in mathbbZ[x] $$

• $M = sup_n |b_n|$. If $|x| > N M$ then $g(x) ne 0$.




• If $g(t) = 0, t=p/q in mathbbQ$ then $t^N = -sum_n=0^N-1 b_n t^n $ $$implies mathbbZ[t] subset sum_n=0^N-1 c_n t^n, c_n in mathbbZ subset q^-N mathbbZ$$ This is a contradiction if $t not in mathbbZ$. Thus $t in mathbbZ$ and $t in -NM,ldots,NM$.

share|cite|improve this answer

edited Mar 13 at 19:32

answered Mar 13 at 18:20

reunsreuns

21.5k21352

$endgroup$

share|cite|improve this answer

edited Mar 13 at 19:32

answered Mar 13 at 18:20

reunsreuns

21.5k21352

answered Mar 13 at 18:20

reunsreuns

21.5k21352

answered Mar 13 at 18:20

reunsreuns

21.5k21352