Irreducibility check for polynomials not satisfying Eisenstein Criterion.
$begingroup$
My Question is to check Irreducibility for polynomials not satisfying Eisenstein Criterion.
As an Illustration, to check whether $x^{p-1}+.....+x+1$ for p a prime is irreducible or not, we replaced $x$ by $x+1$ and by using Eisenstein's Criterion for the resulting polynomial we conclude that resulting polynomial is irreducible and so is the original polynomial.
In general, for a given irreducible polynomial $f(x)$ with coefficients in a known U.F.D, is there some element $a$ such that we can apply Eisenstein's criterion to $f(x+a)$?
I am sure there would be no general structure for this but I expect there to be at least some special cases.
Any Reference/suggestion would be appreciated.
Thank You.
abstract-algebra ring-theory irreducible-polynomials
edited Aug 3 '13 at 14:31
Tom Oldfield
9,55812058
$endgroup$
add a comment |
$begingroup$
My Question is to check Irreducibility for polynomials not satisfying Eisenstein Criterion.
As an Illustration, to check whether $x^{p-1}+.....+x+1$ for p a prime is irreducible or not, we replaced $x$ by $x+1$ and by using Eisenstein's Criterion for the resulting polynomial we conclude that resulting polynomial is irreducible and so is the original polynomial.
In general, for a given irreducible polynomial $f(x)$ with coefficients in a known U.F.D, is there some element $a$ such that we can apply Eisenstein's criterion to $f(x+a)$?
I am sure there would be no general structure for this but I expect there to be at least some special cases.
Any Reference/suggestion would be appreciated.
Thank You.
abstract-algebra ring-theory irreducible-polynomials
edited Aug 3 '13 at 14:31
Tom Oldfield
9,55812058
$endgroup$
4
$begingroup$
For a polynomial with integral coefficients that is irreducible, there is usually going to be no integer $a$ that makes $f(x+a)$ Eisenstein with respect to some prime. When the Eisenstein criterion works it is a rather special situation, at least for degree above 2. The higher the degree gets, the more unexpected it is that you could use an Eisenstein translate (i.e., the Eisenstein condition on $f(x+a)$ for some $a$) to prove irreducibility of an irreducible polynomial $f(x)$. As an example, $x^4 + 10x^2 + 1$ is irreducible but it has no Eisenstein translation with respect to any prime.
$endgroup$
– KCd
Aug 3 '13 at 14:44
$begingroup$
@KCd i agree with "For a polynomial with integral coefficients that is irreducible, there is usually going to be no integer a that makes f(x+a) Eisenstein with respect to some prime" it seem to be correct. a good observation :)
$endgroup$
– user87543
Aug 3 '13 at 14:48
4
$begingroup$
That the Eisenstein criterion works for $1 + x + dots + x^{p-1}$ (after replacing $x$ with $x+1$) for any prime number $p$ and that this is so widely used as an example in books perhaps given a mistaken impression of the purpose and range of applicabilty of this irreducibility test. It is great for letting you construct irreducible polynomials over the rationals of any degree you wish (look: $x^n - 2$), but it sucks as a test that could be applied to a random irreducible polynomial over the rationals with integral coefficeints.
$endgroup$
– KCd
Aug 3 '13 at 14:49
add a comment |
$begingroup$
My Question is to check Irreducibility for polynomials not satisfying Eisenstein Criterion.
As an Illustration, to check whether $x^{p-1}+.....+x+1$ for p a prime is irreducible or not, we replaced $x$ by $x+1$ and by using Eisenstein's Criterion for the resulting polynomial we conclude that resulting polynomial is irreducible and so is the original polynomial.
In general, for a given irreducible polynomial $f(x)$ with coefficients in a known U.F.D, is there some element $a$ such that we can apply Eisenstein's criterion to $f(x+a)$?
I am sure there would be no general structure for this but I expect there to be at least some special cases.
Any Reference/suggestion would be appreciated.
Thank You.
abstract-algebra ring-theory irreducible-polynomials
edited Aug 3 '13 at 14:31
Tom Oldfield
9,55812058
$endgroup$
My Question is to check Irreducibility for polynomials not satisfying Eisenstein Criterion.
As an Illustration, to check whether $x^{p-1}+.....+x+1$ for p a prime is irreducible or not, we replaced $x$ by $x+1$ and by using Eisenstein's Criterion for the resulting polynomial we conclude that resulting polynomial is irreducible and so is the original polynomial.
In general, for a given irreducible polynomial $f(x)$ with coefficients in a known U.F.D, is there some element $a$ such that we can apply Eisenstein's criterion to $f(x+a)$?
I am sure there would be no general structure for this but I expect there to be at least some special cases.
Any Reference/suggestion would be appreciated.
Thank You.
abstract-algebra ring-theory irreducible-polynomials
abstract-algebra ring-theory irreducible-polynomials
edited Aug 3 '13 at 14:31
Tom Oldfield
9,55812058
edited Aug 3 '13 at 14:31
Tom Oldfield
9,55812058
edited Aug 3 '13 at 14:31
Tom Oldfield
9,55812058
edited Aug 3 '13 at 14:31
Tom Oldfield
9,55812058
edited Aug 3 '13 at 14:31
Tom Oldfield
9,55812058
9,55812058
asked Aug 3 '13 at 14:23
user87543
4
$begingroup$
For a polynomial with integral coefficients that is irreducible, there is usually going to be no integer $a$ that makes $f(x+a)$ Eisenstein with respect to some prime. When the Eisenstein criterion works it is a rather special situation, at least for degree above 2. The higher the degree gets, the more unexpected it is that you could use an Eisenstein translate (i.e., the Eisenstein condition on $f(x+a)$ for some $a$) to prove irreducibility of an irreducible polynomial $f(x)$. As an example, $x^4 + 10x^2 + 1$ is irreducible but it has no Eisenstein translation with respect to any prime.
$endgroup$
– KCd
Aug 3 '13 at 14:44
$begingroup$
@KCd i agree with "For a polynomial with integral coefficients that is irreducible, there is usually going to be no integer a that makes f(x+a) Eisenstein with respect to some prime" it seem to be correct. a good observation :)
$endgroup$
– user87543
Aug 3 '13 at 14:48
4
$begingroup$
That the Eisenstein criterion works for $1 + x + dots + x^{p-1}$ (after replacing $x$ with $x+1$) for any prime number $p$ and that this is so widely used as an example in books perhaps given a mistaken impression of the purpose and range of applicabilty of this irreducibility test. It is great for letting you construct irreducible polynomials over the rationals of any degree you wish (look: $x^n - 2$), but it sucks as a test that could be applied to a random irreducible polynomial over the rationals with integral coefficeints.
$endgroup$
– KCd
Aug 3 '13 at 14:49
add a comment |
4
$begingroup$
For a polynomial with integral coefficients that is irreducible, there is usually going to be no integer $a$ that makes $f(x+a)$ Eisenstein with respect to some prime. When the Eisenstein criterion works it is a rather special situation, at least for degree above 2. The higher the degree gets, the more unexpected it is that you could use an Eisenstein translate (i.e., the Eisenstein condition on $f(x+a)$ for some $a$) to prove irreducibility of an irreducible polynomial $f(x)$. As an example, $x^4 + 10x^2 + 1$ is irreducible but it has no Eisenstein translation with respect to any prime.
$endgroup$
– KCd
Aug 3 '13 at 14:44
$begingroup$
@KCd i agree with "For a polynomial with integral coefficients that is irreducible, there is usually going to be no integer a that makes f(x+a) Eisenstein with respect to some prime" it seem to be correct. a good observation :)
$endgroup$
– user87543
Aug 3 '13 at 14:48
4
$begingroup$
That the Eisenstein criterion works for $1 + x + dots + x^{p-1}$ (after replacing $x$ with $x+1$) for any prime number $p$ and that this is so widely used as an example in books perhaps given a mistaken impression of the purpose and range of applicabilty of this irreducibility test. It is great for letting you construct irreducible polynomials over the rationals of any degree you wish (look: $x^n - 2$), but it sucks as a test that could be applied to a random irreducible polynomial over the rationals with integral coefficeints.
$endgroup$
– KCd
Aug 3 '13 at 14:49
4
4
$begingroup$
For a polynomial with integral coefficients that is irreducible, there is usually going to be no integer $a$ that makes $f(x+a)$ Eisenstein with respect to some prime. When the Eisenstein criterion works it is a rather special situation, at least for degree above 2. The higher the degree gets, the more unexpected it is that you could use an Eisenstein translate (i.e., the Eisenstein condition on $f(x+a)$ for some $a$) to prove irreducibility of an irreducible polynomial $f(x)$. As an example, $x^4 + 10x^2 + 1$ is irreducible but it has no Eisenstein translation with respect to any prime.
$endgroup$
– KCd
Aug 3 '13 at 14:44
$begingroup$
For a polynomial with integral coefficients that is irreducible, there is usually going to be no integer $a$ that makes $f(x+a)$ Eisenstein with respect to some prime. When the Eisenstein criterion works it is a rather special situation, at least for degree above 2. The higher the degree gets, the more unexpected it is that you could use an Eisenstein translate (i.e., the Eisenstein condition on $f(x+a)$ for some $a$) to prove irreducibility of an irreducible polynomial $f(x)$. As an example, $x^4 + 10x^2 + 1$ is irreducible but it has no Eisenstein translation with respect to any prime.
$endgroup$
– KCd
Aug 3 '13 at 14:44
$begingroup$
@KCd i agree with "For a polynomial with integral coefficients that is irreducible, there is usually going to be no integer a that makes f(x+a) Eisenstein with respect to some prime" it seem to be correct. a good observation :)
$endgroup$
– user87543
Aug 3 '13 at 14:48
$begingroup$
@KCd i agree with "For a polynomial with integral coefficients that is irreducible, there is usually going to be no integer a that makes f(x+a) Eisenstein with respect to some prime" it seem to be correct. a good observation :)
$endgroup$
– user87543
Aug 3 '13 at 14:48
4
4
$begingroup$
That the Eisenstein criterion works for $1 + x + dots + x^{p-1}$ (after replacing $x$ with $x+1$) for any prime number $p$ and that this is so widely used as an example in books perhaps given a mistaken impression of the purpose and range of applicabilty of this irreducibility test. It is great for letting you construct irreducible polynomials over the rationals of any degree you wish (look: $x^n - 2$), but it sucks as a test that could be applied to a random irreducible polynomial over the rationals with integral coefficeints.
$endgroup$
– KCd
Aug 3 '13 at 14:49
$begingroup$
That the Eisenstein criterion works for $1 + x + dots + x^{p-1}$ (after replacing $x$ with $x+1$) for any prime number $p$ and that this is so widely used as an example in books perhaps given a mistaken impression of the purpose and range of applicabilty of this irreducibility test. It is great for letting you construct irreducible polynomials over the rationals of any degree you wish (look: $x^n - 2$), but it sucks as a test that could be applied to a random irreducible polynomial over the rationals with integral coefficeints.
$endgroup$
– KCd
Aug 3 '13 at 14:49
add a comment |
1 Answer
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$begingroup$
Do Eisenstein and linear translations suffice to determine irreducibility of any integer polynomial?
The answer is no.
Given $f(x)=a_nx^n+a_{n-1}x^{n-1}+ldots +a_0$ there is little choice in the linear translation to be applied.
In the special case $a_n=1$, $a_{n-1}=0$, the coefficient of $x^{n-1}$ in $f(x+a)$ is $na$, so we must take $aequiv 0pmod p$, i.e. we stay with the given $f$, unless $p|n$.
On th eother hand, if $p|n$ then the coefficient of $x^{n-2}$ becomes $a_{n-2}+{pchoose 2}a$ and this is $equiv a_{n-2}pmod p$ unless $p=2$. Thus if we exhibit any irreducible polynomial with $n$ odd, $a_n=1$, $a_{n-1}=0$ and such that Eisenstein cannot show its irreducibility, then neither Eisenstein plus linear translations can show irreducibility. One such polynomial is $$f(x)=x^3+x+1inmathbb Z[x].$$
answered Aug 3 '13 at 14:46
Hagen von EitzenHagen von Eitzen
283k23273508
$endgroup$
$begingroup$
I agree with the Question/Answer "Do Eisenstein and linear translations suffice to determine irreducibility of any integer polynomial? The answer is no." I am looking for special cases only when Eisenstein or Eisenstein plus linear translations suffice for the check :)
$endgroup$
– user87543
Aug 3 '13 at 15:00
add a comment |
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$begingroup$
Do Eisenstein and linear translations suffice to determine irreducibility of any integer polynomial?
The answer is no.
Given $f(x)=a_nx^n+a_{n-1}x^{n-1}+ldots +a_0$ there is little choice in the linear translation to be applied.
In the special case $a_n=1$, $a_{n-1}=0$, the coefficient of $x^{n-1}$ in $f(x+a)$ is $na$, so we must take $aequiv 0pmod p$, i.e. we stay with the given $f$, unless $p|n$.
On th eother hand, if $p|n$ then the coefficient of $x^{n-2}$ becomes $a_{n-2}+{pchoose 2}a$ and this is $equiv a_{n-2}pmod p$ unless $p=2$. Thus if we exhibit any irreducible polynomial with $n$ odd, $a_n=1$, $a_{n-1}=0$ and such that Eisenstein cannot show its irreducibility, then neither Eisenstein plus linear translations can show irreducibility. One such polynomial is $$f(x)=x^3+x+1inmathbb Z[x].$$
answered Aug 3 '13 at 14:46
Hagen von EitzenHagen von Eitzen
283k23273508
$endgroup$
$begingroup$
I agree with the Question/Answer "Do Eisenstein and linear translations suffice to determine irreducibility of any integer polynomial? The answer is no." I am looking for special cases only when Eisenstein or Eisenstein plus linear translations suffice for the check :)
$endgroup$
– user87543
Aug 3 '13 at 15:00
add a comment |
$begingroup$
Do Eisenstein and linear translations suffice to determine irreducibility of any integer polynomial?
The answer is no.
Given $f(x)=a_nx^n+a_{n-1}x^{n-1}+ldots +a_0$ there is little choice in the linear translation to be applied.
In the special case $a_n=1$, $a_{n-1}=0$, the coefficient of $x^{n-1}$ in $f(x+a)$ is $na$, so we must take $aequiv 0pmod p$, i.e. we stay with the given $f$, unless $p|n$.
On th eother hand, if $p|n$ then the coefficient of $x^{n-2}$ becomes $a_{n-2}+{pchoose 2}a$ and this is $equiv a_{n-2}pmod p$ unless $p=2$. Thus if we exhibit any irreducible polynomial with $n$ odd, $a_n=1$, $a_{n-1}=0$ and such that Eisenstein cannot show its irreducibility, then neither Eisenstein plus linear translations can show irreducibility. One such polynomial is $$f(x)=x^3+x+1inmathbb Z[x].$$
answered Aug 3 '13 at 14:46
Hagen von EitzenHagen von Eitzen
283k23273508
$endgroup$
$begingroup$
I agree with the Question/Answer "Do Eisenstein and linear translations suffice to determine irreducibility of any integer polynomial? The answer is no." I am looking for special cases only when Eisenstein or Eisenstein plus linear translations suffice for the check :)
$endgroup$
– user87543
Aug 3 '13 at 15:00
add a comment |
$begingroup$
Do Eisenstein and linear translations suffice to determine irreducibility of any integer polynomial?
The answer is no.
Given $f(x)=a_nx^n+a_{n-1}x^{n-1}+ldots +a_0$ there is little choice in the linear translation to be applied.
In the special case $a_n=1$, $a_{n-1}=0$, the coefficient of $x^{n-1}$ in $f(x+a)$ is $na$, so we must take $aequiv 0pmod p$, i.e. we stay with the given $f$, unless $p|n$.
On th eother hand, if $p|n$ then the coefficient of $x^{n-2}$ becomes $a_{n-2}+{pchoose 2}a$ and this is $equiv a_{n-2}pmod p$ unless $p=2$. Thus if we exhibit any irreducible polynomial with $n$ odd, $a_n=1$, $a_{n-1}=0$ and such that Eisenstein cannot show its irreducibility, then neither Eisenstein plus linear translations can show irreducibility. One such polynomial is $$f(x)=x^3+x+1inmathbb Z[x].$$
answered Aug 3 '13 at 14:46
Hagen von EitzenHagen von Eitzen
283k23273508
$endgroup$
Do Eisenstein and linear translations suffice to determine irreducibility of any integer polynomial?
The answer is no.
Given $f(x)=a_nx^n+a_{n-1}x^{n-1}+ldots +a_0$ there is little choice in the linear translation to be applied.
In the special case $a_n=1$, $a_{n-1}=0$, the coefficient of $x^{n-1}$ in $f(x+a)$ is $na$, so we must take $aequiv 0pmod p$, i.e. we stay with the given $f$, unless $p|n$.
On th eother hand, if $p|n$ then the coefficient of $x^{n-2}$ becomes $a_{n-2}+{pchoose 2}a$ and this is $equiv a_{n-2}pmod p$ unless $p=2$. Thus if we exhibit any irreducible polynomial with $n$ odd, $a_n=1$, $a_{n-1}=0$ and such that Eisenstein cannot show its irreducibility, then neither Eisenstein plus linear translations can show irreducibility. One such polynomial is $$f(x)=x^3+x+1inmathbb Z[x].$$
answered Aug 3 '13 at 14:46
Hagen von EitzenHagen von Eitzen
283k23273508
answered Aug 3 '13 at 14:46
Hagen von EitzenHagen von Eitzen
283k23273508
answered Aug 3 '13 at 14:46
Hagen von EitzenHagen von Eitzen
283k23273508
answered Aug 3 '13 at 14:46
Hagen von EitzenHagen von Eitzen
283k23273508
283k23273508
$begingroup$
I agree with the Question/Answer "Do Eisenstein and linear translations suffice to determine irreducibility of any integer polynomial? The answer is no." I am looking for special cases only when Eisenstein or Eisenstein plus linear translations suffice for the check :)
$endgroup$
– user87543
Aug 3 '13 at 15:00
add a comment |
$begingroup$
I agree with the Question/Answer "Do Eisenstein and linear translations suffice to determine irreducibility of any integer polynomial? The answer is no." I am looking for special cases only when Eisenstein or Eisenstein plus linear translations suffice for the check :)
$endgroup$
– user87543
Aug 3 '13 at 15:00
$begingroup$
I agree with the Question/Answer "Do Eisenstein and linear translations suffice to determine irreducibility of any integer polynomial? The answer is no." I am looking for special cases only when Eisenstein or Eisenstein plus linear translations suffice for the check :)
$endgroup$
– user87543
Aug 3 '13 at 15:00
$begingroup$
I agree with the Question/Answer "Do Eisenstein and linear translations suffice to determine irreducibility of any integer polynomial? The answer is no." I am looking for special cases only when Eisenstein or Eisenstein plus linear translations suffice for the check :)
$endgroup$
– user87543
Aug 3 '13 at 15:00
add a comment |
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$begingroup$
For a polynomial with integral coefficients that is irreducible, there is usually going to be no integer $a$ that makes $f(x+a)$ Eisenstein with respect to some prime. When the Eisenstein criterion works it is a rather special situation, at least for degree above 2. The higher the degree gets, the more unexpected it is that you could use an Eisenstein translate (i.e., the Eisenstein condition on $f(x+a)$ for some $a$) to prove irreducibility of an irreducible polynomial $f(x)$. As an example, $x^4 + 10x^2 + 1$ is irreducible but it has no Eisenstein translation with respect to any prime.
$endgroup$
– KCd
Aug 3 '13 at 14:44
$begingroup$
@KCd i agree with "For a polynomial with integral coefficients that is irreducible, there is usually going to be no integer a that makes f(x+a) Eisenstein with respect to some prime" it seem to be correct. a good observation :)
$endgroup$
– user87543
Aug 3 '13 at 14:48
4
$begingroup$
That the Eisenstein criterion works for $1 + x + dots + x^{p-1}$ (after replacing $x$ with $x+1$) for any prime number $p$ and that this is so widely used as an example in books perhaps given a mistaken impression of the purpose and range of applicabilty of this irreducibility test. It is great for letting you construct irreducible polynomials over the rationals of any degree you wish (look: $x^n - 2$), but it sucks as a test that could be applied to a random irreducible polynomial over the rationals with integral coefficeints.
$endgroup$
– KCd
Aug 3 '13 at 14:49