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3

$begingroup$

Okay this might be a stupid question, but couldn't find any clear explanation on the internet, so here goes.

If I factor a polynomial into products of irreducible polynomials, are the products of these irreducible polynomials then again an irreducible polynomial?

For clarification, here is the problem im working on:

Let $J$ be the ideal of $mathbb{Q}[x]$ generated by two polynomials: $f(x)=(x+1)(x+2)(x^2 +1)$ $textit{and}$ $g(x)=(x+1)^2 (x+2)^2.$ Is the ideal $J$ prime? Is it maximal? Is it principal?

Clearly the common factors of $f$ and $g$ are $(x+1)(x+2)$, but would it be right to say that $p(x)=(x+1)(x+2)$ is an irreducible polynomial over $mathbb{Q}$?

If this is the right approach, I think I can handle the rest of the problem myself.

Thanks in advance :)

abstract-algebra polynomials irreducible-polynomials

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edited yesterday

Martin Sleziak

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asked yesterday

aaalgaaalg

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3

$begingroup$

Okay this might be a stupid question, but couldn't find any clear explanation on the internet, so here goes.

If I factor a polynomial into products of irreducible polynomials, are the products of these irreducible polynomials then again an irreducible polynomial?

For clarification, here is the problem im working on:

Let $J$ be the ideal of $mathbb{Q}[x]$ generated by two polynomials: $f(x)=(x+1)(x+2)(x^2 +1)$ $textit{and}$ $g(x)=(x+1)^2 (x+2)^2.$ Is the ideal $J$ prime? Is it maximal? Is it principal?

Clearly the common factors of $f$ and $g$ are $(x+1)(x+2)$, but would it be right to say that $p(x)=(x+1)(x+2)$ is an irreducible polynomial over $mathbb{Q}$?

If this is the right approach, I think I can handle the rest of the problem myself.

Thanks in advance :)

abstract-algebra polynomials irreducible-polynomials

share|cite|improve this question

edited yesterday

Martin Sleziak

45k10122277

asked yesterday

aaalgaaalg

182

New contributor

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

$endgroup$

add a comment | 

$begingroup$

Okay this might be a stupid question, but couldn't find any clear explanation on the internet, so here goes.

If I factor a polynomial into products of irreducible polynomials, are the products of these irreducible polynomials then again an irreducible polynomial?

For clarification, here is the problem im working on:

Let $J$ be the ideal of $mathbb{Q}[x]$ generated by two polynomials: $f(x)=(x+1)(x+2)(x^2 +1)$ $textit{and}$ $g(x)=(x+1)^2 (x+2)^2.$ Is the ideal $J$ prime? Is it maximal? Is it principal?

Clearly the common factors of $f$ and $g$ are $(x+1)(x+2)$, but would it be right to say that $p(x)=(x+1)(x+2)$ is an irreducible polynomial over $mathbb{Q}$?

If this is the right approach, I think I can handle the rest of the problem myself.

Thanks in advance :)

abstract-algebra polynomials irreducible-polynomials

share|cite|improve this question

edited yesterday

Martin Sleziak

45k10122277

asked yesterday

aaalgaaalg

182

New contributor

aaalg 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 question

edited yesterday

Martin Sleziak

45k10122277

asked yesterday

aaalgaaalg

182

New contributor

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

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edited yesterday

Martin Sleziak

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asked yesterday

aaalgaaalg

182

New contributor

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

edited yesterday

Martin Sleziak

45k10122277

edited yesterday

Martin Sleziak

45k10122277

asked yesterday

aaalgaaalg

182

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aaalgaaalg

182

2 Answers
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1

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If I factor a polynomial into products of irreducible polynomials, are the products of these irreducible polynomials then again an irreducible polynomial?

No, quite the opposite in fact. In fact, that's basically what irreducible means: $f$ is irreducible if $f=gh$ implies one of $g,h$ are constant, i.e. there's no nontrivial factorization. So if you take irreducible $g,h$ (which requires them to be have degree at least one) and multiply them together, you get a polynomial $f$ whose factorization is $gh$ .

Think of it like prime numbers - irreducible polynomials are like primes. Multiply two primes, you get a composite number, never, ever a prime number.

Clearly the common factors of $f$ and $g$ are $(x+1)(x+2)$, but would it be right to say that $p(x)=(x+1)(x+2)$ is an irreducible polynomial over $mathbb{Q}$?

It would not be irreducible by the previous discussion as a result.

share|cite|improve this answer

edited yesterday

Martin Sleziak

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answered yesterday

Eevee TrainerEevee Trainer

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2

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Simply $p(x)$ is not irreducible, because you have exhibited a factorisation. The factorisation you have given is a factorisation of $p(x)$ into irreducible factors.

share|cite|improve this answer

answered yesterday

Mark BennetMark Bennet

81.9k984183

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Aha thank you! So i should think of it as $p(x)=x^2 +3x +2$ being reducible and then argue from there, right?
  $endgroup$
  – aaalg
  yesterday
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @aaalg That's the way to go.
  $endgroup$
  – Mark Bennet
  yesterday
  

  
  
  
  

add a comment | 

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1

$begingroup$

If I factor a polynomial into products of irreducible polynomials, are the products of these irreducible polynomials then again an irreducible polynomial?

No, quite the opposite in fact. In fact, that's basically what irreducible means: $f$ is irreducible if $f=gh$ implies one of $g,h$ are constant, i.e. there's no nontrivial factorization. So if you take irreducible $g,h$ (which requires them to be have degree at least one) and multiply them together, you get a polynomial $f$ whose factorization is $gh$ .

Think of it like prime numbers - irreducible polynomials are like primes. Multiply two primes, you get a composite number, never, ever a prime number.

Clearly the common factors of $f$ and $g$ are $(x+1)(x+2)$, but would it be right to say that $p(x)=(x+1)(x+2)$ is an irreducible polynomial over $mathbb{Q}$?

It would not be irreducible by the previous discussion as a result.

share|cite|improve this answer

edited yesterday

Martin Sleziak

45k10122277

answered yesterday

Eevee TrainerEevee Trainer

10k31740

$endgroup$

add a comment | 

1

$begingroup$

If I factor a polynomial into products of irreducible polynomials, are the products of these irreducible polynomials then again an irreducible polynomial?

No, quite the opposite in fact. In fact, that's basically what irreducible means: $f$ is irreducible if $f=gh$ implies one of $g,h$ are constant, i.e. there's no nontrivial factorization. So if you take irreducible $g,h$ (which requires them to be have degree at least one) and multiply them together, you get a polynomial $f$ whose factorization is $gh$ .

Think of it like prime numbers - irreducible polynomials are like primes. Multiply two primes, you get a composite number, never, ever a prime number.

Clearly the common factors of $f$ and $g$ are $(x+1)(x+2)$, but would it be right to say that $p(x)=(x+1)(x+2)$ is an irreducible polynomial over $mathbb{Q}$?

It would not be irreducible by the previous discussion as a result.

share|cite|improve this answer

edited yesterday

Martin Sleziak

45k10122277

answered yesterday

Eevee TrainerEevee Trainer

10k31740

$endgroup$

add a comment | 

$begingroup$

If I factor a polynomial into products of irreducible polynomials, are the products of these irreducible polynomials then again an irreducible polynomial?

No, quite the opposite in fact. In fact, that's basically what irreducible means: $f$ is irreducible if $f=gh$ implies one of $g,h$ are constant, i.e. there's no nontrivial factorization. So if you take irreducible $g,h$ (which requires them to be have degree at least one) and multiply them together, you get a polynomial $f$ whose factorization is $gh$ .

Think of it like prime numbers - irreducible polynomials are like primes. Multiply two primes, you get a composite number, never, ever a prime number.

Clearly the common factors of $f$ and $g$ are $(x+1)(x+2)$, but would it be right to say that $p(x)=(x+1)(x+2)$ is an irreducible polynomial over $mathbb{Q}$?

It would not be irreducible by the previous discussion as a result.

share|cite|improve this answer

edited yesterday

Martin Sleziak

45k10122277

answered yesterday

Eevee TrainerEevee Trainer

10k31740

$endgroup$

share|cite|improve this answer

edited yesterday

Martin Sleziak

45k10122277

answered yesterday

Eevee TrainerEevee Trainer

10k31740

edited yesterday

Martin Sleziak

45k10122277

edited yesterday

Martin Sleziak

45k10122277

answered yesterday

Eevee TrainerEevee Trainer

10k31740

answered yesterday

Eevee TrainerEevee Trainer

10k31740

2

$begingroup$

Simply $p(x)$ is not irreducible, because you have exhibited a factorisation. The factorisation you have given is a factorisation of $p(x)$ into irreducible factors.

share|cite|improve this answer

answered yesterday

Mark BennetMark Bennet

81.9k984183

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Aha thank you! So i should think of it as $p(x)=x^2 +3x +2$ being reducible and then argue from there, right?
  $endgroup$
  – aaalg
  yesterday
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @aaalg That's the way to go.
  $endgroup$
  – Mark Bennet
  yesterday
  

  
  
  
  

add a comment | 

2

$begingroup$

Simply $p(x)$ is not irreducible, because you have exhibited a factorisation. The factorisation you have given is a factorisation of $p(x)$ into irreducible factors.

share|cite|improve this answer

answered yesterday

Mark BennetMark Bennet

81.9k984183

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Aha thank you! So i should think of it as $p(x)=x^2 +3x +2$ being reducible and then argue from there, right?
  $endgroup$
  – aaalg
  yesterday
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @aaalg That's the way to go.
  $endgroup$
  – Mark Bennet
  yesterday
  

  
  
  
  

add a comment | 

$begingroup$

Simply $p(x)$ is not irreducible, because you have exhibited a factorisation. The factorisation you have given is a factorisation of $p(x)$ into irreducible factors.

share|cite|improve this answer

answered yesterday

Mark BennetMark Bennet

81.9k984183

$endgroup$

share|cite|improve this answer

answered yesterday

Mark BennetMark Bennet

81.9k984183

answered yesterday

Mark BennetMark Bennet

81.9k984183

answered yesterday

Mark BennetMark Bennet

81.9k984183