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 $Bbb F_2[x]$
How to notate time signature switching consistently every measure
I see my dog run
What do hard-Brexiteers want with respect to the Irish border?
How can I autofill dates in Excel excluding Sunday?
Worn-tile Scrabble
How to type this arrow in math mode?
Is there a symbol for a right arrow with a square in the middle?
How are circuits which use complex ICs normally simulated?
"as much details as you can remember"
How to save as into a customized destination on macOS?
How to manage monthly salary
Origin of "cooter" meaning "vagina"
What is the most effective way of iterating a std::vector and why?
One word riddle: Vowel in the middle
Are children permitted to help build the Beis Hamikdash?
Building a conditional check constraint
Landlord wants to switch my lease to a "Land contract" to "get back at the city"
If a Druid sees an animal’s corpse, can they wild shape into that animal?
How to support a colleague who finds meetings extremely tiring?
What is the closest word meaning "respect for time / mindful"
Button changing it's text & action. Good or terrible?
Deal with toxic manager when you can't quit
Is "plugging out" electronic devices an American expression?
Have you ever entered Singapore using a different passport or name?
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
New contributor
$endgroup$
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
New contributor
$endgroup$
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
New contributor
$endgroup$
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
New contributor
New contributor
edited Apr 5 at 12:23
Martin Sleziak
45k10122277
45k10122277
New contributor
asked Apr 5 at 10:09
aaalgaaalg
232
232
New contributor
New contributor
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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.
$endgroup$
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.
$endgroup$
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 |
Your Answer
StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");
StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);
else
createEditor();
);
function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);
);
aaalg is a new contributor. Be nice, and check out our Code of Conduct.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3175756%2ffactoring-of-polynomial-over-the-rationals-quick-question%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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.
$endgroup$
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.
$endgroup$
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.
$endgroup$
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
45k10122277
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.
$endgroup$
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.
$endgroup$
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.
$endgroup$
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
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.
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3175756%2ffactoring-of-polynomial-over-the-rationals-quick-question%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown