Factoring of polynomial over the rationals (quick question) The 2019 Stack Overflow Developer Survey Results Are InFinding an irreducible polynomial over the integers.What are the prime ideals of $mathbbR[x_1,x_2,x_3,…]$How to prove how many ireducible polynomials are in a polynomial ring over a finite field.$X^3+2$ is irreducible in $mathbbF_7[X]$Confused on notions of maximal ideal and some notationFind all the maximal ideals in the ring $mathbbR[x]$.Show that the polynomial $x^2+y^2-1$ is irreducible over $mathbbQ[x,y]$Extending a principal prime of $mathbbZ[X]$ to a maximal oneFactor a polynomial over a finite fieldMaximal ideals of $Bbb F_2[x]$
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Factoring of polynomial over the rationals (quick question)
The 2019 Stack Overflow Developer Survey Results Are InFinding an irreducible polynomial over the integers.What are the prime ideals of $mathbbR[x_1,x_2,x_3,…]$How to prove how many ireducible polynomials are in a polynomial ring over a finite field.$X^3+2$ is irreducible in $mathbbF_7[X]$Confused on notions of maximal ideal and some notationFind all the maximal ideals in the ring $mathbbR[x]$.Show that the polynomial $x^2+y^2-1$ is irreducible over $mathbbQ[x,y]$Extending a principal prime of $mathbbZ[X]$ to a maximal oneFactor a polynomial over a finite fieldMaximal ideals of $Bbb F_2[x]$
$begingroup$
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 $mathbbQ[x]$ generated by two polynomials: $f(x)=(x+1)(x+2)(x^2 +1)$ $textitand$ $g(x)=(x+1)^2 (x+2)^2.$ 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 $mathbbQ$?
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
edited Apr 5 at 12:23
Martin Sleziak
45k10122277
asked Apr 5 at 10:09
aaalgaaalg
232
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 $mathbbQ[x]$ generated by two polynomials: $f(x)=(x+1)(x+2)(x^2 +1)$ $textitand$ $g(x)=(x+1)^2 (x+2)^2.$ 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 $mathbbQ$?
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
edited Apr 5 at 12:23
Martin Sleziak
45k10122277
asked Apr 5 at 10:09
aaalgaaalg
232
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 $mathbbQ[x]$ generated by two polynomials: $f(x)=(x+1)(x+2)(x^2 +1)$ $textitand$ $g(x)=(x+1)^2 (x+2)^2.$ 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 $mathbbQ$?
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
edited Apr 5 at 12:23
Martin Sleziak
45k10122277
asked Apr 5 at 10:09
aaalgaaalg
232
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.
$endgroup$
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 $mathbbQ[x]$ generated by two polynomials: $f(x)=(x+1)(x+2)(x^2 +1)$ $textitand$ $g(x)=(x+1)^2 (x+2)^2.$ 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 $mathbbQ$?
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
abstract-algebra polynomials irreducible-polynomials
edited Apr 5 at 12:23
Martin Sleziak
45k10122277
asked Apr 5 at 10:09
aaalgaaalg
232
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 Apr 5 at 12:23
Martin Sleziak
45k10122277
asked Apr 5 at 10:09
aaalgaaalg
232
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 Apr 5 at 12:23
Martin Sleziak
45k10122277
edited Apr 5 at 12:23
Martin Sleziak
45k10122277
edited Apr 5 at 12:23
Martin Sleziak
45k10122277
45k10122277
asked Apr 5 at 10:09
aaalgaaalg
232
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.
asked Apr 5 at 10:09
aaalgaaalg
232
asked Apr 5 at 10:09
aaalgaaalg
232
232
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.
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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Check out our Code of Conduct.
add a comment |
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2 Answers
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$begingroup$
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 $mathbbQ$?
It would not be irreducible by the previous discussion as a result.
edited Apr 5 at 12:25
Martin Sleziak
45k10122277
answered Apr 5 at 10:14
Eevee TrainerEevee Trainer
10.4k31742
$endgroup$
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.
answered Apr 5 at 10:13
Mark BennetMark Bennet
82k984183
$endgroup$
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
Apr 5 at 10:58
$begingroup$
@aaalg That's the way to go.
$endgroup$
– Mark Bennet
Apr 5 at 11:09
add a comment |
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$begingroup$
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 $mathbbQ$?
It would not be irreducible by the previous discussion as a result.
edited Apr 5 at 12:25
Martin Sleziak
45k10122277
answered Apr 5 at 10:14
Eevee TrainerEevee Trainer
10.4k31742
$endgroup$
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 $mathbbQ$?
It would not be irreducible by the previous discussion as a result.
edited Apr 5 at 12:25
Martin Sleziak
45k10122277
answered Apr 5 at 10:14
Eevee TrainerEevee Trainer
10.4k31742
$endgroup$
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 $mathbbQ$?
It would not be irreducible by the previous discussion as a result.
edited Apr 5 at 12:25
Martin Sleziak
45k10122277
answered Apr 5 at 10:14
Eevee TrainerEevee Trainer
10.4k31742
$endgroup$
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 $mathbbQ$?
It would not be irreducible by the previous discussion as a result.
edited Apr 5 at 12:25
Martin Sleziak
45k10122277
answered Apr 5 at 10:14
Eevee TrainerEevee Trainer
10.4k31742
edited Apr 5 at 12:25
Martin Sleziak
45k10122277
edited Apr 5 at 12:25
Martin Sleziak
45k10122277
edited Apr 5 at 12:25
Martin Sleziak
45k10122277
45k10122277
answered Apr 5 at 10:14
Eevee TrainerEevee Trainer
10.4k31742
answered Apr 5 at 10:14
Eevee TrainerEevee Trainer
10.4k31742
answered Apr 5 at 10:14
Eevee TrainerEevee Trainer
10.4k31742
10.4k31742
add a comment |
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.
answered Apr 5 at 10:13
Mark BennetMark Bennet
82k984183
$endgroup$
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
Apr 5 at 10:58
$begingroup$
@aaalg That's the way to go.
$endgroup$
– Mark Bennet
Apr 5 at 11:09
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.
answered Apr 5 at 10:13
Mark BennetMark Bennet
82k984183
$endgroup$
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
Apr 5 at 10:58
$begingroup$
@aaalg That's the way to go.
$endgroup$
– Mark Bennet
Apr 5 at 11:09
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.
answered Apr 5 at 10:13
Mark BennetMark Bennet
82k984183
$endgroup$
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.
answered Apr 5 at 10:13
Mark BennetMark Bennet
82k984183
answered Apr 5 at 10:13
Mark BennetMark Bennet
82k984183
answered Apr 5 at 10:13
Mark BennetMark Bennet
82k984183
answered Apr 5 at 10:13
Mark BennetMark Bennet
82k984183
82k984183
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
Apr 5 at 10:58
$begingroup$
@aaalg That's the way to go.
$endgroup$
– Mark Bennet
Apr 5 at 11:09
add a comment |
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
Apr 5 at 10:58
$begingroup$
@aaalg That's the way to go.
$endgroup$
– Mark Bennet
Apr 5 at 11:09
1
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
Apr 5 at 10:58
$begingroup$
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
Apr 5 at 10:58
$begingroup$
@aaalg That's the way to go.
$endgroup$
– Mark Bennet
Apr 5 at 11:09
$begingroup$
@aaalg That's the way to go.
$endgroup$
– Mark Bennet
Apr 5 at 11:09
add a comment |
aaalg is a new contributor. Be nice, and check out our Code of Conduct.
aaalg is a new contributor. Be nice, and check out our Code of Conduct.
aaalg is a new contributor. Be nice, and check out our Code of Conduct.
aaalg is a new contributor. Be nice, and check out our Code of Conduct.
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