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Polynomials and Irreducible polynomials

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1

$begingroup$

So I'm trying to prove this Proposition:

Every polynomial of degree greater than $0$ is divisible by an irreducible polynomial.

Proof:

Consider a polynomial $P in mathbb R [x]$ of degree $geq 1$

Let $S$ denote the set of non-trivial factors of $P$.

Suppose that $S$ does not contain any irreducible polynomials.

By the well ordering principle let $P_0$ denote the polynomial of least degree in $S$

By our assumption $P_0$ is not irreducible $therefore P_0=P_1P_2$ for some

$P_1,P_2 in mathbb R[x]$

This contradicts the definition of $P_0$ being the polynomial of least degree to be a factor of $P$

Thus we are forced to conclude the polynomial of least degree is irreducible and the claim follows.

$blacksquare$

My main issue is can I use the well ordering principle in this fashion?

thanks

polynomials irreducible-polynomials

share|cite|improve this question

edited Mar 16 at 11:21

Sil

5,31221643

asked Dec 26 '18 at 11:16

PolynomialCPolynomialC

876

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $S$ must be the set of monic (non-trivial factors of $P$). With this, you can say that $P_1,P_2$ are monic (and have degree $geq1$)
  $endgroup$
  – Martín Vacas Vignolo
  Dec 26 '18 at 11:26
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Why must the factors be monic?
  $endgroup$
  – PolynomialC
  Dec 26 '18 at 11:28
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  No. $S$ is a subset of set of all factors. The key is that in your proof you can not say that $deg P_i geq 1$
  $endgroup$
  – Martín Vacas Vignolo
  Dec 26 '18 at 11:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Is this to make sure the set is of finite size and thus bounded below? Cheers for the input.
  $endgroup$
  – PolynomialC
  Dec 26 '18 at 11:33
  

  
  
  
  

add a comment | 

1

$begingroup$

So I'm trying to prove this Proposition:

Every polynomial of degree greater than $0$ is divisible by an irreducible polynomial.

Proof:

Consider a polynomial $P in mathbb R [x]$ of degree $geq 1$

Let $S$ denote the set of non-trivial factors of $P$.

Suppose that $S$ does not contain any irreducible polynomials.

By the well ordering principle let $P_0$ denote the polynomial of least degree in $S$

By our assumption $P_0$ is not irreducible $therefore P_0=P_1P_2$ for some

$P_1,P_2 in mathbb R[x]$

This contradicts the definition of $P_0$ being the polynomial of least degree to be a factor of $P$

Thus we are forced to conclude the polynomial of least degree is irreducible and the claim follows.

$blacksquare$

My main issue is can I use the well ordering principle in this fashion?

thanks

polynomials irreducible-polynomials

share|cite|improve this question

edited Mar 16 at 11:21

Sil

5,31221643

asked Dec 26 '18 at 11:16

PolynomialCPolynomialC

876

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $S$ must be the set of monic (non-trivial factors of $P$). With this, you can say that $P_1,P_2$ are monic (and have degree $geq1$)
  $endgroup$
  – Martín Vacas Vignolo
  Dec 26 '18 at 11:26
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Why must the factors be monic?
  $endgroup$
  – PolynomialC
  Dec 26 '18 at 11:28
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  No. $S$ is a subset of set of all factors. The key is that in your proof you can not say that $deg P_i geq 1$
  $endgroup$
  – Martín Vacas Vignolo
  Dec 26 '18 at 11:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Is this to make sure the set is of finite size and thus bounded below? Cheers for the input.
  $endgroup$
  – PolynomialC
  Dec 26 '18 at 11:33
  

  
  
  
  

add a comment | 

$begingroup$

So I'm trying to prove this Proposition:

Every polynomial of degree greater than $0$ is divisible by an irreducible polynomial.

Proof:

Consider a polynomial $P in mathbb R [x]$ of degree $geq 1$

Let $S$ denote the set of non-trivial factors of $P$.

Suppose that $S$ does not contain any irreducible polynomials.

By the well ordering principle let $P_0$ denote the polynomial of least degree in $S$

By our assumption $P_0$ is not irreducible $therefore P_0=P_1P_2$ for some

$P_1,P_2 in mathbb R[x]$

This contradicts the definition of $P_0$ being the polynomial of least degree to be a factor of $P$

Thus we are forced to conclude the polynomial of least degree is irreducible and the claim follows.

$blacksquare$

My main issue is can I use the well ordering principle in this fashion?

thanks

polynomials irreducible-polynomials

share|cite|improve this question

edited Mar 16 at 11:21

Sil

5,31221643

asked Dec 26 '18 at 11:16

PolynomialCPolynomialC

876

$endgroup$

share|cite|improve this question

edited Mar 16 at 11:21

Sil

5,31221643

asked Dec 26 '18 at 11:16

PolynomialCPolynomialC

876

share|cite|improve this question

edited Mar 16 at 11:21

Sil

5,31221643

asked Dec 26 '18 at 11:16

PolynomialCPolynomialC

876

edited Mar 16 at 11:21

Sil

5,31221643

edited Mar 16 at 11:21

Sil

5,31221643

asked Dec 26 '18 at 11:16

PolynomialCPolynomialC

876

asked Dec 26 '18 at 11:16

PolynomialCPolynomialC

876

1 Answer
1

active

oldest

votes

0

$begingroup$

If you want to be precise, you need to apply the well-ordering principle to a non-empty set of $mathbb N$.

For that, define $D = n in mathbb N : textthere is a factor of $P$ of degree $n$ $.
Then $D$ is not empty because $P$ is a factor of $P$.
Let $m = min D$. Then there is a polynomial $Q$ such that $Q$ is a factor of $P$ of degree $m$. The minimality of $m$ implies that $Q$ is irreducible.

share|cite|improve this answer

edited Dec 26 '18 at 11:46

answered Dec 26 '18 at 11:28

lhflhf

166k11172402

$endgroup$

add a comment | 

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0

$begingroup$

If you want to be precise, you need to apply the well-ordering principle to a non-empty set of $mathbb N$.

For that, define $D = n in mathbb N : textthere is a factor of $P$ of degree $n$ $.
Then $D$ is not empty because $P$ is a factor of $P$.
Let $m = min D$. Then there is a polynomial $Q$ such that $Q$ is a factor of $P$ of degree $m$. The minimality of $m$ implies that $Q$ is irreducible.

share|cite|improve this answer

edited Dec 26 '18 at 11:46

answered Dec 26 '18 at 11:28

lhflhf

166k11172402

$endgroup$

add a comment | 

0

$begingroup$

If you want to be precise, you need to apply the well-ordering principle to a non-empty set of $mathbb N$.

For that, define $D = n in mathbb N : textthere is a factor of $P$ of degree $n$ $.
Then $D$ is not empty because $P$ is a factor of $P$.
Let $m = min D$. Then there is a polynomial $Q$ such that $Q$ is a factor of $P$ of degree $m$. The minimality of $m$ implies that $Q$ is irreducible.

share|cite|improve this answer

edited Dec 26 '18 at 11:46

answered Dec 26 '18 at 11:28

lhflhf

166k11172402

$endgroup$

add a comment | 

$begingroup$

If you want to be precise, you need to apply the well-ordering principle to a non-empty set of $mathbb N$.

For that, define $D = n in mathbb N : textthere is a factor of $P$ of degree $n$ $.
Then $D$ is not empty because $P$ is a factor of $P$.
Let $m = min D$. Then there is a polynomial $Q$ such that $Q$ is a factor of $P$ of degree $m$. The minimality of $m$ implies that $Q$ is irreducible.

share|cite|improve this answer

edited Dec 26 '18 at 11:46

answered Dec 26 '18 at 11:28

lhflhf

166k11172402

$endgroup$

share|cite|improve this answer

edited Dec 26 '18 at 11:46

answered Dec 26 '18 at 11:28

lhflhf

166k11172402

answered Dec 26 '18 at 11:28

lhflhf

166k11172402

answered Dec 26 '18 at 11:28

lhflhf

166k11172402