Methods to see if a polynomial is irreducibleDoes the Rational Root Theorem ever guarantee that a polynomial is irreducible?Polynomial is irreducible over $mathbbQ$Prove that the polynomial $x^5+2x+1$ is irreducible over $mathbbZ$.Beginning of RomanceProve that $f(x)$ is irreducible iff its reciprocal polynomial $f^*(x)$ is irreducible.How to choose correct strategy for irreducibility testing in $mathbbZ[X]$?Help to understand this polynomial trickCriteria of irreducibility for polynomials with integer coefficientsIrreducibility and reducibilityWhy is this polynomial irreducible over $mathbbZ[i]$?How do I prove that this polynomial is irreducible?How to show that $x^6-72$ is irreducible over the rationals?Irreducibility of a quartic polynomialProve a polynomial is irreducible over $mathbb Q$Irreducible implies minimal polynomial?Irreducible monic polynomial in $mathbbQ[x]$$x^6+2x^3-3x^2+1$, irreducible over $mathbbQ$Irreducibility of polynomial over $mathbb Q$ and $mathbb Q(i)$Determine and construct irreducible polynomial (of degree >3) over finite fields.
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Methods to see if a polynomial is irreducible
Does the Rational Root Theorem ever guarantee that a polynomial is irreducible?Polynomial is irreducible over $mathbbQ$Prove that the polynomial $x^5+2x+1$ is irreducible over $mathbbZ$.Beginning of RomanceProve that $f(x)$ is irreducible iff its reciprocal polynomial $f^*(x)$ is irreducible.How to choose correct strategy for irreducibility testing in $mathbbZ[X]$?Help to understand this polynomial trickCriteria of irreducibility for polynomials with integer coefficientsIrreducibility and reducibilityWhy is this polynomial irreducible over $mathbbZ[i]$?How do I prove that this polynomial is irreducible?How to show that $x^6-72$ is irreducible over the rationals?Irreducibility of a quartic polynomialProve a polynomial is irreducible over $mathbb Q$Irreducible implies minimal polynomial?Irreducible monic polynomial in $mathbbQ[x]$$x^6+2x^3-3x^2+1$, irreducible over $mathbbQ$Irreducibility of polynomial over $mathbb Q$ and $mathbb Q(i)$Determine and construct irreducible polynomial (of degree >3) over finite fields.
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Given a polynomial over a field, what are the methods to see it is irreducible? Only two comes to my mind now. First is Eisenstein criterion. Another is that if a polynomial is irreducible mod p then it is irreducible. Are there any others?
abstract-algebra polynomials
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add a comment |
$begingroup$
Given a polynomial over a field, what are the methods to see it is irreducible? Only two comes to my mind now. First is Eisenstein criterion. Another is that if a polynomial is irreducible mod p then it is irreducible. Are there any others?
abstract-algebra polynomials
$endgroup$
4
$begingroup$
Well, Eisenstein's criterion is basically a version of reducing modulo a prime.
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– Akhil Mathew
Aug 9 '10 at 18:10
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See also this meta answer for grouped list of questions related to this topic math.meta.stackexchange.com/questions/1868/….
$endgroup$
– Sil
Aug 11 '18 at 11:13
add a comment |
$begingroup$
Given a polynomial over a field, what are the methods to see it is irreducible? Only two comes to my mind now. First is Eisenstein criterion. Another is that if a polynomial is irreducible mod p then it is irreducible. Are there any others?
abstract-algebra polynomials
$endgroup$
Given a polynomial over a field, what are the methods to see it is irreducible? Only two comes to my mind now. First is Eisenstein criterion. Another is that if a polynomial is irreducible mod p then it is irreducible. Are there any others?
abstract-algebra polynomials
abstract-algebra polynomials
edited May 4 '11 at 8:10
Bruno Stonek
6,44323992
6,44323992
asked Aug 9 '10 at 17:22
user977
4
$begingroup$
Well, Eisenstein's criterion is basically a version of reducing modulo a prime.
$endgroup$
– Akhil Mathew
Aug 9 '10 at 18:10
$begingroup$
See also this meta answer for grouped list of questions related to this topic math.meta.stackexchange.com/questions/1868/….
$endgroup$
– Sil
Aug 11 '18 at 11:13
add a comment |
4
$begingroup$
Well, Eisenstein's criterion is basically a version of reducing modulo a prime.
$endgroup$
– Akhil Mathew
Aug 9 '10 at 18:10
$begingroup$
See also this meta answer for grouped list of questions related to this topic math.meta.stackexchange.com/questions/1868/….
$endgroup$
– Sil
Aug 11 '18 at 11:13
4
4
$begingroup$
Well, Eisenstein's criterion is basically a version of reducing modulo a prime.
$endgroup$
– Akhil Mathew
Aug 9 '10 at 18:10
$begingroup$
Well, Eisenstein's criterion is basically a version of reducing modulo a prime.
$endgroup$
– Akhil Mathew
Aug 9 '10 at 18:10
$begingroup$
See also this meta answer for grouped list of questions related to this topic math.meta.stackexchange.com/questions/1868/….
$endgroup$
– Sil
Aug 11 '18 at 11:13
$begingroup$
See also this meta answer for grouped list of questions related to this topic math.meta.stackexchange.com/questions/1868/….
$endgroup$
– Sil
Aug 11 '18 at 11:13
add a comment |
6 Answers
6
active
oldest
votes
$begingroup$
To better understand the Eisenstein and related irreducibility tests you should learn about Newton polygons. It's the master theorem behind all these related results. A good place to start is Filaseta's notes - see the links below. Note: you may need to write to Filaseta to get access to his interesting book [1] on this topic.
[1] http://www.math.sc.edu/~filaseta/gradcourses/Math788F/latexbook/
[2] http://www.math.sc.edu/~filaseta/gradcourses/Math788F/NewtonPolygonsTalk.pdf
[3] Newton Polygon Applet
http://www.math.sc.edu/~filaseta/newton/newton.html
[4] Abhyankar, Shreeram S.
Historical ramblings in algebraic geometry and related algebra.
Amer. Math. Monthly 83 (1976), no. 6, 409-448.
http://links.jstor.org/sici?sici=0002-9890(197606/07)83:6%3C409:HRIAG...
$endgroup$
$begingroup$
I'm not able to access the first link you posted
$endgroup$
– crasic
Nov 24 '10 at 11:18
$begingroup$
@crasic: Did you try contacting him to obtain access?
$endgroup$
– Bill Dubuque
Nov 24 '10 at 14:08
$begingroup$
I would like to read this book too.
$endgroup$
– quanta
May 4 '11 at 11:42
5
$begingroup$
@Holdsworth88: So, ... two years later ... did you succeed or fail?
$endgroup$
– Eric Towers
Jun 29 '14 at 5:23
1
$begingroup$
Failure. He never replied.
$endgroup$
– Holdsworth88
Jul 3 '14 at 19:19
|
show 5 more comments
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One method for polynomials over $mathbbZ$ is to use complex analysis to say something about the location of the roots. Often Rouche's theorem is useful; this is how Perron's criterion is proven, which says that a monic polynomial $x^n + a_n-1 x^n-1 + ... + a_0$ with integer coefficients is irreducible if $|a_n-1| > 1 + |a_n-2| + ... + |a_0|$ and $a_0 neq 0$. A basic observation is that knowing a polynomial is reducible places constraints on where its roots can be; for example, if a monic polynomial with prime constant coefficient $p$ is reducible, one of its irreducible factors has constant term $pm p$ and the rest have constant term $pm 1$. It follows that the polynomial has at least one root inside the unit circle and at least one root outside.
An important thing to keep in mind here is that there exist irreducible polynomials over $mathbbZ$ which are reducible modulo every prime. For example, $x^4 + 16$ is such a polynomial. So the modular technique is not enough in general.
$endgroup$
1
$begingroup$
Incidentally, if a polynomial is irreducible in the algebraic closure of every prime field, then it is irreducible. (This is only interesting when one has a polynomial of multiple variables :)). This can be proved via an application of model theory: note that the statement "a given polynomial over $mathbbZ$ is irreducible in some field" can be phrased via first-order logic. So if it is true in algebraically closed fields of arbitrarily high characteristic, it is true over $mathbbC$, even.
$endgroup$
– Akhil Mathew
Aug 9 '10 at 23:16
add a comment |
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Here's an elementary trick that I occasionally find useful: Let $y=x+c$ for some fixed integer $c$, and write $f(x)=g(y)$. Then $f$ is irreducible if and only if $g$ is irreducible. You may be able to able to reduce $g$ modulo a prime and/or apply Eisenstein to show that $g$ is irreducible.
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Also works for $y=ax+c$.
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– Bruno Stonek
May 4 '11 at 8:09
1
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...if $a$ is a unit.
$endgroup$
– Bruno Stonek
Sep 8 '11 at 23:09
add a comment |
$begingroup$
Below is another method for irreducibility testing - excerpted from one of my old sci.math posts.
In 1918 Stackel published the following simple observation:
THEOREM If $ p(x) $ is a composite integer coefficient polynomial
then $ p(n) $ is composite for all $|n| > B $, for some bound $B$,
in fact $ p(n) $ has at most $ 2d $ prime values, where $ d = rm deg(p)$.
The simple proof can be found online in Mott & Rose [3], p. 8.
I highly recommend this delightful and stimulating 27 page paper
which discusses prime-producing polynomials and related topics.
Contrapositively, $ p(x) $ is prime (irreducible) if it assumes a prime value
for large enough $ |x| $. Conversely Bouniakowski conjectured (1857)
that prime $ p(x) $ assume infinitely many prime values (except in trivial
cases where the values of $p$ have an obvious common divisor, e.g. $ 2 | x(x+1)+2$ ).
As an example, Polya-Szego popularized A. Cohn's irreduciblity test, which
states that $ p(x) in mathbb Z[x]$ is prime if $ p(b) $
yields a prime in radix $b$ representation (so necessarily $0 le p_i < b$).
For example $f(x) = x^4 + 6 x^2 + 1 pmod p$ factors for all primes $p$,
yet $f(x)$ is prime since $f(8) = 10601$ octal $= 4481$ is prime.
Note: Cohn's test fails if, in radix $b$, negative digits are allowed, e.g.
$f(x) = x^3 - 9 x^2 + x-9 = (x-9)(x^2 + 1)$ but $f(10) = 101$ is prime.
For further discussion see my prior post [1], along with Murty's online paper [2].
[1] Dubuque, sci.math 2002-11-12, On prime producing polynomials
http://groups.google.com/groups?selm=y8zvg4m9yhm.fsf%40nestle.ai.mit.edu
[2] Murty, Ram. Prime numbers and irreducible polynomials.
Amer. Math. Monthly, Vol. 109 (2002), no. 5, 452-458.
http://www.mast.queensu.ca/~murty/polya4.dvi
[3] Mott, Joe L.; Rose, Kermit.
Prime producing cubic polynomials
Ideal theoretic methods in commutative algebra, 281-317.
Lecture Notes in Pure and Appl. Math., 220, Dekker, New York, 2001.
http://web.math.fsu.edu/~aluffi/archive/paper134.ps
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add a comment |
$begingroup$
Polynomials by Prasolov covers among others:
- Eisenstein's criterion
- Duma's criterion
- Irreducibility of polynomials attaining small values
- Hilbert's criterion
- Irreducibility of trinomials and fournomials
- A few algorithms for factorization
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add a comment |
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Reversing the polynomial. If you have polynomial with degree $geq 2$ and non zero constant coefficients (otherwise it would be reducible so it wouldn't be interesting anyway), then you can reverse the coefficients and check irreducibility on that reciprocal polynomial. For example instead of checking $f(x)=2x^4+2x^3+2x^2+2x+1$, you can instead check $x^4+2x^3+2x^2+2x+2$ (and see it is irreducible by Eisenstein). This corresponds to $x^4f(1/x)$.
Another useful criterion is one provided by Ram's Murty in paper already referenced in the other answer, similar to Cohn's irreducibility criteria, it states:
Murty's irreducibility criterion: Let $f(x)=a_mx^m+a_m-1x^m-1+dots+a_1x+a_0$ be a polynomial of degree $m$ in $mathbbZ[x]$ and set $$H=max_0leq ileq m-1 |a_i/a_m|.$$
If $f(n)$ is prime for some integer $ngeq H+2$, then $f(x)$ is irreducible in $mathbbZ[x]$.
You can see that for example $f(x)=x^3-11x^2+19x-17$ is irreducible because of that, if you try $n=24$.
Osada's criterion. Let $f(x)=x^n+a_1x^n-1+dots+a_n-1xpm p$ be a polynomial with integer coefficients, where $p$ is a prime. If $p>1+|a_1|+dots+|a_n-1|$, then $f(x)$ is irreducible over $mathbbZ$.
Following one is also simple to use, although I found it barely applicable, nvertheless interesting:
Bauer's criterion. Let $a_1 geq a_2 geq dots geq a_n$ be positive integers and $n geq 2$. Then the polynomial $p(x)=x^n-a_1x^n-1-a_2x^n-2-dots-a_n$ is irreducible over $mathbbZ$.
Advanced criteria related to Newton polygons. These criteria are little bit more advanced to use, but the paper below provides plenty of corollaries in terms of prime powers (such as Eisenstein criterion, but in this case with multiple primes). Schönemann–Eisenstein–Dumas-Type Irreducibility Conditions that Use Arbitrarily Many Prime Numbers. An example: try to prove irreducibility of this one $$4x^6+108x^5+108x^4+108x^3+108x^2+108x+27.$$ The article above provides way to do this (it is first example in last section of examples).
Some related questions:
- How to choose correct strategy for irreducibility testing in $mathbbZ[X]$?
- Polynomial is irreducible over $mathbbQ$
- Prove that $f(x)$ is irreducible iff its reciprocal polynomial $f^*(x)$ is irreducible.
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add a comment |
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6 Answers
6
active
oldest
votes
6 Answers
6
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
To better understand the Eisenstein and related irreducibility tests you should learn about Newton polygons. It's the master theorem behind all these related results. A good place to start is Filaseta's notes - see the links below. Note: you may need to write to Filaseta to get access to his interesting book [1] on this topic.
[1] http://www.math.sc.edu/~filaseta/gradcourses/Math788F/latexbook/
[2] http://www.math.sc.edu/~filaseta/gradcourses/Math788F/NewtonPolygonsTalk.pdf
[3] Newton Polygon Applet
http://www.math.sc.edu/~filaseta/newton/newton.html
[4] Abhyankar, Shreeram S.
Historical ramblings in algebraic geometry and related algebra.
Amer. Math. Monthly 83 (1976), no. 6, 409-448.
http://links.jstor.org/sici?sici=0002-9890(197606/07)83:6%3C409:HRIAG...
$endgroup$
$begingroup$
I'm not able to access the first link you posted
$endgroup$
– crasic
Nov 24 '10 at 11:18
$begingroup$
@crasic: Did you try contacting him to obtain access?
$endgroup$
– Bill Dubuque
Nov 24 '10 at 14:08
$begingroup$
I would like to read this book too.
$endgroup$
– quanta
May 4 '11 at 11:42
5
$begingroup$
@Holdsworth88: So, ... two years later ... did you succeed or fail?
$endgroup$
– Eric Towers
Jun 29 '14 at 5:23
1
$begingroup$
Failure. He never replied.
$endgroup$
– Holdsworth88
Jul 3 '14 at 19:19
|
show 5 more comments
$begingroup$
To better understand the Eisenstein and related irreducibility tests you should learn about Newton polygons. It's the master theorem behind all these related results. A good place to start is Filaseta's notes - see the links below. Note: you may need to write to Filaseta to get access to his interesting book [1] on this topic.
[1] http://www.math.sc.edu/~filaseta/gradcourses/Math788F/latexbook/
[2] http://www.math.sc.edu/~filaseta/gradcourses/Math788F/NewtonPolygonsTalk.pdf
[3] Newton Polygon Applet
http://www.math.sc.edu/~filaseta/newton/newton.html
[4] Abhyankar, Shreeram S.
Historical ramblings in algebraic geometry and related algebra.
Amer. Math. Monthly 83 (1976), no. 6, 409-448.
http://links.jstor.org/sici?sici=0002-9890(197606/07)83:6%3C409:HRIAG...
$endgroup$
$begingroup$
I'm not able to access the first link you posted
$endgroup$
– crasic
Nov 24 '10 at 11:18
$begingroup$
@crasic: Did you try contacting him to obtain access?
$endgroup$
– Bill Dubuque
Nov 24 '10 at 14:08
$begingroup$
I would like to read this book too.
$endgroup$
– quanta
May 4 '11 at 11:42
5
$begingroup$
@Holdsworth88: So, ... two years later ... did you succeed or fail?
$endgroup$
– Eric Towers
Jun 29 '14 at 5:23
1
$begingroup$
Failure. He never replied.
$endgroup$
– Holdsworth88
Jul 3 '14 at 19:19
|
show 5 more comments
$begingroup$
To better understand the Eisenstein and related irreducibility tests you should learn about Newton polygons. It's the master theorem behind all these related results. A good place to start is Filaseta's notes - see the links below. Note: you may need to write to Filaseta to get access to his interesting book [1] on this topic.
[1] http://www.math.sc.edu/~filaseta/gradcourses/Math788F/latexbook/
[2] http://www.math.sc.edu/~filaseta/gradcourses/Math788F/NewtonPolygonsTalk.pdf
[3] Newton Polygon Applet
http://www.math.sc.edu/~filaseta/newton/newton.html
[4] Abhyankar, Shreeram S.
Historical ramblings in algebraic geometry and related algebra.
Amer. Math. Monthly 83 (1976), no. 6, 409-448.
http://links.jstor.org/sici?sici=0002-9890(197606/07)83:6%3C409:HRIAG...
$endgroup$
To better understand the Eisenstein and related irreducibility tests you should learn about Newton polygons. It's the master theorem behind all these related results. A good place to start is Filaseta's notes - see the links below. Note: you may need to write to Filaseta to get access to his interesting book [1] on this topic.
[1] http://www.math.sc.edu/~filaseta/gradcourses/Math788F/latexbook/
[2] http://www.math.sc.edu/~filaseta/gradcourses/Math788F/NewtonPolygonsTalk.pdf
[3] Newton Polygon Applet
http://www.math.sc.edu/~filaseta/newton/newton.html
[4] Abhyankar, Shreeram S.
Historical ramblings in algebraic geometry and related algebra.
Amer. Math. Monthly 83 (1976), no. 6, 409-448.
http://links.jstor.org/sici?sici=0002-9890(197606/07)83:6%3C409:HRIAG...
answered Aug 9 '10 at 17:32
Bill DubuqueBill Dubuque
213k29195654
213k29195654
$begingroup$
I'm not able to access the first link you posted
$endgroup$
– crasic
Nov 24 '10 at 11:18
$begingroup$
@crasic: Did you try contacting him to obtain access?
$endgroup$
– Bill Dubuque
Nov 24 '10 at 14:08
$begingroup$
I would like to read this book too.
$endgroup$
– quanta
May 4 '11 at 11:42
5
$begingroup$
@Holdsworth88: So, ... two years later ... did you succeed or fail?
$endgroup$
– Eric Towers
Jun 29 '14 at 5:23
1
$begingroup$
Failure. He never replied.
$endgroup$
– Holdsworth88
Jul 3 '14 at 19:19
|
show 5 more comments
$begingroup$
I'm not able to access the first link you posted
$endgroup$
– crasic
Nov 24 '10 at 11:18
$begingroup$
@crasic: Did you try contacting him to obtain access?
$endgroup$
– Bill Dubuque
Nov 24 '10 at 14:08
$begingroup$
I would like to read this book too.
$endgroup$
– quanta
May 4 '11 at 11:42
5
$begingroup$
@Holdsworth88: So, ... two years later ... did you succeed or fail?
$endgroup$
– Eric Towers
Jun 29 '14 at 5:23
1
$begingroup$
Failure. He never replied.
$endgroup$
– Holdsworth88
Jul 3 '14 at 19:19
$begingroup$
I'm not able to access the first link you posted
$endgroup$
– crasic
Nov 24 '10 at 11:18
$begingroup$
I'm not able to access the first link you posted
$endgroup$
– crasic
Nov 24 '10 at 11:18
$begingroup$
@crasic: Did you try contacting him to obtain access?
$endgroup$
– Bill Dubuque
Nov 24 '10 at 14:08
$begingroup$
@crasic: Did you try contacting him to obtain access?
$endgroup$
– Bill Dubuque
Nov 24 '10 at 14:08
$begingroup$
I would like to read this book too.
$endgroup$
– quanta
May 4 '11 at 11:42
$begingroup$
I would like to read this book too.
$endgroup$
– quanta
May 4 '11 at 11:42
5
5
$begingroup$
@Holdsworth88: So, ... two years later ... did you succeed or fail?
$endgroup$
– Eric Towers
Jun 29 '14 at 5:23
$begingroup$
@Holdsworth88: So, ... two years later ... did you succeed or fail?
$endgroup$
– Eric Towers
Jun 29 '14 at 5:23
1
1
$begingroup$
Failure. He never replied.
$endgroup$
– Holdsworth88
Jul 3 '14 at 19:19
$begingroup$
Failure. He never replied.
$endgroup$
– Holdsworth88
Jul 3 '14 at 19:19
|
show 5 more comments
$begingroup$
One method for polynomials over $mathbbZ$ is to use complex analysis to say something about the location of the roots. Often Rouche's theorem is useful; this is how Perron's criterion is proven, which says that a monic polynomial $x^n + a_n-1 x^n-1 + ... + a_0$ with integer coefficients is irreducible if $|a_n-1| > 1 + |a_n-2| + ... + |a_0|$ and $a_0 neq 0$. A basic observation is that knowing a polynomial is reducible places constraints on where its roots can be; for example, if a monic polynomial with prime constant coefficient $p$ is reducible, one of its irreducible factors has constant term $pm p$ and the rest have constant term $pm 1$. It follows that the polynomial has at least one root inside the unit circle and at least one root outside.
An important thing to keep in mind here is that there exist irreducible polynomials over $mathbbZ$ which are reducible modulo every prime. For example, $x^4 + 16$ is such a polynomial. So the modular technique is not enough in general.
$endgroup$
1
$begingroup$
Incidentally, if a polynomial is irreducible in the algebraic closure of every prime field, then it is irreducible. (This is only interesting when one has a polynomial of multiple variables :)). This can be proved via an application of model theory: note that the statement "a given polynomial over $mathbbZ$ is irreducible in some field" can be phrased via first-order logic. So if it is true in algebraically closed fields of arbitrarily high characteristic, it is true over $mathbbC$, even.
$endgroup$
– Akhil Mathew
Aug 9 '10 at 23:16
add a comment |
$begingroup$
One method for polynomials over $mathbbZ$ is to use complex analysis to say something about the location of the roots. Often Rouche's theorem is useful; this is how Perron's criterion is proven, which says that a monic polynomial $x^n + a_n-1 x^n-1 + ... + a_0$ with integer coefficients is irreducible if $|a_n-1| > 1 + |a_n-2| + ... + |a_0|$ and $a_0 neq 0$. A basic observation is that knowing a polynomial is reducible places constraints on where its roots can be; for example, if a monic polynomial with prime constant coefficient $p$ is reducible, one of its irreducible factors has constant term $pm p$ and the rest have constant term $pm 1$. It follows that the polynomial has at least one root inside the unit circle and at least one root outside.
An important thing to keep in mind here is that there exist irreducible polynomials over $mathbbZ$ which are reducible modulo every prime. For example, $x^4 + 16$ is such a polynomial. So the modular technique is not enough in general.
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1
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Incidentally, if a polynomial is irreducible in the algebraic closure of every prime field, then it is irreducible. (This is only interesting when one has a polynomial of multiple variables :)). This can be proved via an application of model theory: note that the statement "a given polynomial over $mathbbZ$ is irreducible in some field" can be phrased via first-order logic. So if it is true in algebraically closed fields of arbitrarily high characteristic, it is true over $mathbbC$, even.
$endgroup$
– Akhil Mathew
Aug 9 '10 at 23:16
add a comment |
$begingroup$
One method for polynomials over $mathbbZ$ is to use complex analysis to say something about the location of the roots. Often Rouche's theorem is useful; this is how Perron's criterion is proven, which says that a monic polynomial $x^n + a_n-1 x^n-1 + ... + a_0$ with integer coefficients is irreducible if $|a_n-1| > 1 + |a_n-2| + ... + |a_0|$ and $a_0 neq 0$. A basic observation is that knowing a polynomial is reducible places constraints on where its roots can be; for example, if a monic polynomial with prime constant coefficient $p$ is reducible, one of its irreducible factors has constant term $pm p$ and the rest have constant term $pm 1$. It follows that the polynomial has at least one root inside the unit circle and at least one root outside.
An important thing to keep in mind here is that there exist irreducible polynomials over $mathbbZ$ which are reducible modulo every prime. For example, $x^4 + 16$ is such a polynomial. So the modular technique is not enough in general.
$endgroup$
One method for polynomials over $mathbbZ$ is to use complex analysis to say something about the location of the roots. Often Rouche's theorem is useful; this is how Perron's criterion is proven, which says that a monic polynomial $x^n + a_n-1 x^n-1 + ... + a_0$ with integer coefficients is irreducible if $|a_n-1| > 1 + |a_n-2| + ... + |a_0|$ and $a_0 neq 0$. A basic observation is that knowing a polynomial is reducible places constraints on where its roots can be; for example, if a monic polynomial with prime constant coefficient $p$ is reducible, one of its irreducible factors has constant term $pm p$ and the rest have constant term $pm 1$. It follows that the polynomial has at least one root inside the unit circle and at least one root outside.
An important thing to keep in mind here is that there exist irreducible polynomials over $mathbbZ$ which are reducible modulo every prime. For example, $x^4 + 16$ is such a polynomial. So the modular technique is not enough in general.
answered Aug 9 '10 at 18:23
Qiaochu YuanQiaochu Yuan
281k32593938
281k32593938
1
$begingroup$
Incidentally, if a polynomial is irreducible in the algebraic closure of every prime field, then it is irreducible. (This is only interesting when one has a polynomial of multiple variables :)). This can be proved via an application of model theory: note that the statement "a given polynomial over $mathbbZ$ is irreducible in some field" can be phrased via first-order logic. So if it is true in algebraically closed fields of arbitrarily high characteristic, it is true over $mathbbC$, even.
$endgroup$
– Akhil Mathew
Aug 9 '10 at 23:16
add a comment |
1
$begingroup$
Incidentally, if a polynomial is irreducible in the algebraic closure of every prime field, then it is irreducible. (This is only interesting when one has a polynomial of multiple variables :)). This can be proved via an application of model theory: note that the statement "a given polynomial over $mathbbZ$ is irreducible in some field" can be phrased via first-order logic. So if it is true in algebraically closed fields of arbitrarily high characteristic, it is true over $mathbbC$, even.
$endgroup$
– Akhil Mathew
Aug 9 '10 at 23:16
1
1
$begingroup$
Incidentally, if a polynomial is irreducible in the algebraic closure of every prime field, then it is irreducible. (This is only interesting when one has a polynomial of multiple variables :)). This can be proved via an application of model theory: note that the statement "a given polynomial over $mathbbZ$ is irreducible in some field" can be phrased via first-order logic. So if it is true in algebraically closed fields of arbitrarily high characteristic, it is true over $mathbbC$, even.
$endgroup$
– Akhil Mathew
Aug 9 '10 at 23:16
$begingroup$
Incidentally, if a polynomial is irreducible in the algebraic closure of every prime field, then it is irreducible. (This is only interesting when one has a polynomial of multiple variables :)). This can be proved via an application of model theory: note that the statement "a given polynomial over $mathbbZ$ is irreducible in some field" can be phrased via first-order logic. So if it is true in algebraically closed fields of arbitrarily high characteristic, it is true over $mathbbC$, even.
$endgroup$
– Akhil Mathew
Aug 9 '10 at 23:16
add a comment |
$begingroup$
Here's an elementary trick that I occasionally find useful: Let $y=x+c$ for some fixed integer $c$, and write $f(x)=g(y)$. Then $f$ is irreducible if and only if $g$ is irreducible. You may be able to able to reduce $g$ modulo a prime and/or apply Eisenstein to show that $g$ is irreducible.
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Also works for $y=ax+c$.
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– Bruno Stonek
May 4 '11 at 8:09
1
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...if $a$ is a unit.
$endgroup$
– Bruno Stonek
Sep 8 '11 at 23:09
add a comment |
$begingroup$
Here's an elementary trick that I occasionally find useful: Let $y=x+c$ for some fixed integer $c$, and write $f(x)=g(y)$. Then $f$ is irreducible if and only if $g$ is irreducible. You may be able to able to reduce $g$ modulo a prime and/or apply Eisenstein to show that $g$ is irreducible.
$endgroup$
$begingroup$
Also works for $y=ax+c$.
$endgroup$
– Bruno Stonek
May 4 '11 at 8:09
1
$begingroup$
...if $a$ is a unit.
$endgroup$
– Bruno Stonek
Sep 8 '11 at 23:09
add a comment |
$begingroup$
Here's an elementary trick that I occasionally find useful: Let $y=x+c$ for some fixed integer $c$, and write $f(x)=g(y)$. Then $f$ is irreducible if and only if $g$ is irreducible. You may be able to able to reduce $g$ modulo a prime and/or apply Eisenstein to show that $g$ is irreducible.
$endgroup$
Here's an elementary trick that I occasionally find useful: Let $y=x+c$ for some fixed integer $c$, and write $f(x)=g(y)$. Then $f$ is irreducible if and only if $g$ is irreducible. You may be able to able to reduce $g$ modulo a prime and/or apply Eisenstein to show that $g$ is irreducible.
answered Apr 6 '11 at 5:26
user9187
$begingroup$
Also works for $y=ax+c$.
$endgroup$
– Bruno Stonek
May 4 '11 at 8:09
1
$begingroup$
...if $a$ is a unit.
$endgroup$
– Bruno Stonek
Sep 8 '11 at 23:09
add a comment |
$begingroup$
Also works for $y=ax+c$.
$endgroup$
– Bruno Stonek
May 4 '11 at 8:09
1
$begingroup$
...if $a$ is a unit.
$endgroup$
– Bruno Stonek
Sep 8 '11 at 23:09
$begingroup$
Also works for $y=ax+c$.
$endgroup$
– Bruno Stonek
May 4 '11 at 8:09
$begingroup$
Also works for $y=ax+c$.
$endgroup$
– Bruno Stonek
May 4 '11 at 8:09
1
1
$begingroup$
...if $a$ is a unit.
$endgroup$
– Bruno Stonek
Sep 8 '11 at 23:09
$begingroup$
...if $a$ is a unit.
$endgroup$
– Bruno Stonek
Sep 8 '11 at 23:09
add a comment |
$begingroup$
Below is another method for irreducibility testing - excerpted from one of my old sci.math posts.
In 1918 Stackel published the following simple observation:
THEOREM If $ p(x) $ is a composite integer coefficient polynomial
then $ p(n) $ is composite for all $|n| > B $, for some bound $B$,
in fact $ p(n) $ has at most $ 2d $ prime values, where $ d = rm deg(p)$.
The simple proof can be found online in Mott & Rose [3], p. 8.
I highly recommend this delightful and stimulating 27 page paper
which discusses prime-producing polynomials and related topics.
Contrapositively, $ p(x) $ is prime (irreducible) if it assumes a prime value
for large enough $ |x| $. Conversely Bouniakowski conjectured (1857)
that prime $ p(x) $ assume infinitely many prime values (except in trivial
cases where the values of $p$ have an obvious common divisor, e.g. $ 2 | x(x+1)+2$ ).
As an example, Polya-Szego popularized A. Cohn's irreduciblity test, which
states that $ p(x) in mathbb Z[x]$ is prime if $ p(b) $
yields a prime in radix $b$ representation (so necessarily $0 le p_i < b$).
For example $f(x) = x^4 + 6 x^2 + 1 pmod p$ factors for all primes $p$,
yet $f(x)$ is prime since $f(8) = 10601$ octal $= 4481$ is prime.
Note: Cohn's test fails if, in radix $b$, negative digits are allowed, e.g.
$f(x) = x^3 - 9 x^2 + x-9 = (x-9)(x^2 + 1)$ but $f(10) = 101$ is prime.
For further discussion see my prior post [1], along with Murty's online paper [2].
[1] Dubuque, sci.math 2002-11-12, On prime producing polynomials
http://groups.google.com/groups?selm=y8zvg4m9yhm.fsf%40nestle.ai.mit.edu
[2] Murty, Ram. Prime numbers and irreducible polynomials.
Amer. Math. Monthly, Vol. 109 (2002), no. 5, 452-458.
http://www.mast.queensu.ca/~murty/polya4.dvi
[3] Mott, Joe L.; Rose, Kermit.
Prime producing cubic polynomials
Ideal theoretic methods in commutative algebra, 281-317.
Lecture Notes in Pure and Appl. Math., 220, Dekker, New York, 2001.
http://web.math.fsu.edu/~aluffi/archive/paper134.ps
$endgroup$
add a comment |
$begingroup$
Below is another method for irreducibility testing - excerpted from one of my old sci.math posts.
In 1918 Stackel published the following simple observation:
THEOREM If $ p(x) $ is a composite integer coefficient polynomial
then $ p(n) $ is composite for all $|n| > B $, for some bound $B$,
in fact $ p(n) $ has at most $ 2d $ prime values, where $ d = rm deg(p)$.
The simple proof can be found online in Mott & Rose [3], p. 8.
I highly recommend this delightful and stimulating 27 page paper
which discusses prime-producing polynomials and related topics.
Contrapositively, $ p(x) $ is prime (irreducible) if it assumes a prime value
for large enough $ |x| $. Conversely Bouniakowski conjectured (1857)
that prime $ p(x) $ assume infinitely many prime values (except in trivial
cases where the values of $p$ have an obvious common divisor, e.g. $ 2 | x(x+1)+2$ ).
As an example, Polya-Szego popularized A. Cohn's irreduciblity test, which
states that $ p(x) in mathbb Z[x]$ is prime if $ p(b) $
yields a prime in radix $b$ representation (so necessarily $0 le p_i < b$).
For example $f(x) = x^4 + 6 x^2 + 1 pmod p$ factors for all primes $p$,
yet $f(x)$ is prime since $f(8) = 10601$ octal $= 4481$ is prime.
Note: Cohn's test fails if, in radix $b$, negative digits are allowed, e.g.
$f(x) = x^3 - 9 x^2 + x-9 = (x-9)(x^2 + 1)$ but $f(10) = 101$ is prime.
For further discussion see my prior post [1], along with Murty's online paper [2].
[1] Dubuque, sci.math 2002-11-12, On prime producing polynomials
http://groups.google.com/groups?selm=y8zvg4m9yhm.fsf%40nestle.ai.mit.edu
[2] Murty, Ram. Prime numbers and irreducible polynomials.
Amer. Math. Monthly, Vol. 109 (2002), no. 5, 452-458.
http://www.mast.queensu.ca/~murty/polya4.dvi
[3] Mott, Joe L.; Rose, Kermit.
Prime producing cubic polynomials
Ideal theoretic methods in commutative algebra, 281-317.
Lecture Notes in Pure and Appl. Math., 220, Dekker, New York, 2001.
http://web.math.fsu.edu/~aluffi/archive/paper134.ps
$endgroup$
add a comment |
$begingroup$
Below is another method for irreducibility testing - excerpted from one of my old sci.math posts.
In 1918 Stackel published the following simple observation:
THEOREM If $ p(x) $ is a composite integer coefficient polynomial
then $ p(n) $ is composite for all $|n| > B $, for some bound $B$,
in fact $ p(n) $ has at most $ 2d $ prime values, where $ d = rm deg(p)$.
The simple proof can be found online in Mott & Rose [3], p. 8.
I highly recommend this delightful and stimulating 27 page paper
which discusses prime-producing polynomials and related topics.
Contrapositively, $ p(x) $ is prime (irreducible) if it assumes a prime value
for large enough $ |x| $. Conversely Bouniakowski conjectured (1857)
that prime $ p(x) $ assume infinitely many prime values (except in trivial
cases where the values of $p$ have an obvious common divisor, e.g. $ 2 | x(x+1)+2$ ).
As an example, Polya-Szego popularized A. Cohn's irreduciblity test, which
states that $ p(x) in mathbb Z[x]$ is prime if $ p(b) $
yields a prime in radix $b$ representation (so necessarily $0 le p_i < b$).
For example $f(x) = x^4 + 6 x^2 + 1 pmod p$ factors for all primes $p$,
yet $f(x)$ is prime since $f(8) = 10601$ octal $= 4481$ is prime.
Note: Cohn's test fails if, in radix $b$, negative digits are allowed, e.g.
$f(x) = x^3 - 9 x^2 + x-9 = (x-9)(x^2 + 1)$ but $f(10) = 101$ is prime.
For further discussion see my prior post [1], along with Murty's online paper [2].
[1] Dubuque, sci.math 2002-11-12, On prime producing polynomials
http://groups.google.com/groups?selm=y8zvg4m9yhm.fsf%40nestle.ai.mit.edu
[2] Murty, Ram. Prime numbers and irreducible polynomials.
Amer. Math. Monthly, Vol. 109 (2002), no. 5, 452-458.
http://www.mast.queensu.ca/~murty/polya4.dvi
[3] Mott, Joe L.; Rose, Kermit.
Prime producing cubic polynomials
Ideal theoretic methods in commutative algebra, 281-317.
Lecture Notes in Pure and Appl. Math., 220, Dekker, New York, 2001.
http://web.math.fsu.edu/~aluffi/archive/paper134.ps
$endgroup$
Below is another method for irreducibility testing - excerpted from one of my old sci.math posts.
In 1918 Stackel published the following simple observation:
THEOREM If $ p(x) $ is a composite integer coefficient polynomial
then $ p(n) $ is composite for all $|n| > B $, for some bound $B$,
in fact $ p(n) $ has at most $ 2d $ prime values, where $ d = rm deg(p)$.
The simple proof can be found online in Mott & Rose [3], p. 8.
I highly recommend this delightful and stimulating 27 page paper
which discusses prime-producing polynomials and related topics.
Contrapositively, $ p(x) $ is prime (irreducible) if it assumes a prime value
for large enough $ |x| $. Conversely Bouniakowski conjectured (1857)
that prime $ p(x) $ assume infinitely many prime values (except in trivial
cases where the values of $p$ have an obvious common divisor, e.g. $ 2 | x(x+1)+2$ ).
As an example, Polya-Szego popularized A. Cohn's irreduciblity test, which
states that $ p(x) in mathbb Z[x]$ is prime if $ p(b) $
yields a prime in radix $b$ representation (so necessarily $0 le p_i < b$).
For example $f(x) = x^4 + 6 x^2 + 1 pmod p$ factors for all primes $p$,
yet $f(x)$ is prime since $f(8) = 10601$ octal $= 4481$ is prime.
Note: Cohn's test fails if, in radix $b$, negative digits are allowed, e.g.
$f(x) = x^3 - 9 x^2 + x-9 = (x-9)(x^2 + 1)$ but $f(10) = 101$ is prime.
For further discussion see my prior post [1], along with Murty's online paper [2].
[1] Dubuque, sci.math 2002-11-12, On prime producing polynomials
http://groups.google.com/groups?selm=y8zvg4m9yhm.fsf%40nestle.ai.mit.edu
[2] Murty, Ram. Prime numbers and irreducible polynomials.
Amer. Math. Monthly, Vol. 109 (2002), no. 5, 452-458.
http://www.mast.queensu.ca/~murty/polya4.dvi
[3] Mott, Joe L.; Rose, Kermit.
Prime producing cubic polynomials
Ideal theoretic methods in commutative algebra, 281-317.
Lecture Notes in Pure and Appl. Math., 220, Dekker, New York, 2001.
http://web.math.fsu.edu/~aluffi/archive/paper134.ps
answered Aug 9 '10 at 22:52
Bill DubuqueBill Dubuque
213k29195654
213k29195654
add a comment |
add a comment |
$begingroup$
Polynomials by Prasolov covers among others:
- Eisenstein's criterion
- Duma's criterion
- Irreducibility of polynomials attaining small values
- Hilbert's criterion
- Irreducibility of trinomials and fournomials
- A few algorithms for factorization
$endgroup$
add a comment |
$begingroup$
Polynomials by Prasolov covers among others:
- Eisenstein's criterion
- Duma's criterion
- Irreducibility of polynomials attaining small values
- Hilbert's criterion
- Irreducibility of trinomials and fournomials
- A few algorithms for factorization
$endgroup$
add a comment |
$begingroup$
Polynomials by Prasolov covers among others:
- Eisenstein's criterion
- Duma's criterion
- Irreducibility of polynomials attaining small values
- Hilbert's criterion
- Irreducibility of trinomials and fournomials
- A few algorithms for factorization
$endgroup$
Polynomials by Prasolov covers among others:
- Eisenstein's criterion
- Duma's criterion
- Irreducibility of polynomials attaining small values
- Hilbert's criterion
- Irreducibility of trinomials and fournomials
- A few algorithms for factorization
edited Mar 15 at 22:03
user26857
39.4k124183
39.4k124183
answered Aug 10 '10 at 2:01
n0vakovicn0vakovic
499417
499417
add a comment |
add a comment |
$begingroup$
Reversing the polynomial. If you have polynomial with degree $geq 2$ and non zero constant coefficients (otherwise it would be reducible so it wouldn't be interesting anyway), then you can reverse the coefficients and check irreducibility on that reciprocal polynomial. For example instead of checking $f(x)=2x^4+2x^3+2x^2+2x+1$, you can instead check $x^4+2x^3+2x^2+2x+2$ (and see it is irreducible by Eisenstein). This corresponds to $x^4f(1/x)$.
Another useful criterion is one provided by Ram's Murty in paper already referenced in the other answer, similar to Cohn's irreducibility criteria, it states:
Murty's irreducibility criterion: Let $f(x)=a_mx^m+a_m-1x^m-1+dots+a_1x+a_0$ be a polynomial of degree $m$ in $mathbbZ[x]$ and set $$H=max_0leq ileq m-1 |a_i/a_m|.$$
If $f(n)$ is prime for some integer $ngeq H+2$, then $f(x)$ is irreducible in $mathbbZ[x]$.
You can see that for example $f(x)=x^3-11x^2+19x-17$ is irreducible because of that, if you try $n=24$.
Osada's criterion. Let $f(x)=x^n+a_1x^n-1+dots+a_n-1xpm p$ be a polynomial with integer coefficients, where $p$ is a prime. If $p>1+|a_1|+dots+|a_n-1|$, then $f(x)$ is irreducible over $mathbbZ$.
Following one is also simple to use, although I found it barely applicable, nvertheless interesting:
Bauer's criterion. Let $a_1 geq a_2 geq dots geq a_n$ be positive integers and $n geq 2$. Then the polynomial $p(x)=x^n-a_1x^n-1-a_2x^n-2-dots-a_n$ is irreducible over $mathbbZ$.
Advanced criteria related to Newton polygons. These criteria are little bit more advanced to use, but the paper below provides plenty of corollaries in terms of prime powers (such as Eisenstein criterion, but in this case with multiple primes). Schönemann–Eisenstein–Dumas-Type Irreducibility Conditions that Use Arbitrarily Many Prime Numbers. An example: try to prove irreducibility of this one $$4x^6+108x^5+108x^4+108x^3+108x^2+108x+27.$$ The article above provides way to do this (it is first example in last section of examples).
Some related questions:
- How to choose correct strategy for irreducibility testing in $mathbbZ[X]$?
- Polynomial is irreducible over $mathbbQ$
- Prove that $f(x)$ is irreducible iff its reciprocal polynomial $f^*(x)$ is irreducible.
$endgroup$
add a comment |
$begingroup$
Reversing the polynomial. If you have polynomial with degree $geq 2$ and non zero constant coefficients (otherwise it would be reducible so it wouldn't be interesting anyway), then you can reverse the coefficients and check irreducibility on that reciprocal polynomial. For example instead of checking $f(x)=2x^4+2x^3+2x^2+2x+1$, you can instead check $x^4+2x^3+2x^2+2x+2$ (and see it is irreducible by Eisenstein). This corresponds to $x^4f(1/x)$.
Another useful criterion is one provided by Ram's Murty in paper already referenced in the other answer, similar to Cohn's irreducibility criteria, it states:
Murty's irreducibility criterion: Let $f(x)=a_mx^m+a_m-1x^m-1+dots+a_1x+a_0$ be a polynomial of degree $m$ in $mathbbZ[x]$ and set $$H=max_0leq ileq m-1 |a_i/a_m|.$$
If $f(n)$ is prime for some integer $ngeq H+2$, then $f(x)$ is irreducible in $mathbbZ[x]$.
You can see that for example $f(x)=x^3-11x^2+19x-17$ is irreducible because of that, if you try $n=24$.
Osada's criterion. Let $f(x)=x^n+a_1x^n-1+dots+a_n-1xpm p$ be a polynomial with integer coefficients, where $p$ is a prime. If $p>1+|a_1|+dots+|a_n-1|$, then $f(x)$ is irreducible over $mathbbZ$.
Following one is also simple to use, although I found it barely applicable, nvertheless interesting:
Bauer's criterion. Let $a_1 geq a_2 geq dots geq a_n$ be positive integers and $n geq 2$. Then the polynomial $p(x)=x^n-a_1x^n-1-a_2x^n-2-dots-a_n$ is irreducible over $mathbbZ$.
Advanced criteria related to Newton polygons. These criteria are little bit more advanced to use, but the paper below provides plenty of corollaries in terms of prime powers (such as Eisenstein criterion, but in this case with multiple primes). Schönemann–Eisenstein–Dumas-Type Irreducibility Conditions that Use Arbitrarily Many Prime Numbers. An example: try to prove irreducibility of this one $$4x^6+108x^5+108x^4+108x^3+108x^2+108x+27.$$ The article above provides way to do this (it is first example in last section of examples).
Some related questions:
- How to choose correct strategy for irreducibility testing in $mathbbZ[X]$?
- Polynomial is irreducible over $mathbbQ$
- Prove that $f(x)$ is irreducible iff its reciprocal polynomial $f^*(x)$ is irreducible.
$endgroup$
add a comment |
$begingroup$
Reversing the polynomial. If you have polynomial with degree $geq 2$ and non zero constant coefficients (otherwise it would be reducible so it wouldn't be interesting anyway), then you can reverse the coefficients and check irreducibility on that reciprocal polynomial. For example instead of checking $f(x)=2x^4+2x^3+2x^2+2x+1$, you can instead check $x^4+2x^3+2x^2+2x+2$ (and see it is irreducible by Eisenstein). This corresponds to $x^4f(1/x)$.
Another useful criterion is one provided by Ram's Murty in paper already referenced in the other answer, similar to Cohn's irreducibility criteria, it states:
Murty's irreducibility criterion: Let $f(x)=a_mx^m+a_m-1x^m-1+dots+a_1x+a_0$ be a polynomial of degree $m$ in $mathbbZ[x]$ and set $$H=max_0leq ileq m-1 |a_i/a_m|.$$
If $f(n)$ is prime for some integer $ngeq H+2$, then $f(x)$ is irreducible in $mathbbZ[x]$.
You can see that for example $f(x)=x^3-11x^2+19x-17$ is irreducible because of that, if you try $n=24$.
Osada's criterion. Let $f(x)=x^n+a_1x^n-1+dots+a_n-1xpm p$ be a polynomial with integer coefficients, where $p$ is a prime. If $p>1+|a_1|+dots+|a_n-1|$, then $f(x)$ is irreducible over $mathbbZ$.
Following one is also simple to use, although I found it barely applicable, nvertheless interesting:
Bauer's criterion. Let $a_1 geq a_2 geq dots geq a_n$ be positive integers and $n geq 2$. Then the polynomial $p(x)=x^n-a_1x^n-1-a_2x^n-2-dots-a_n$ is irreducible over $mathbbZ$.
Advanced criteria related to Newton polygons. These criteria are little bit more advanced to use, but the paper below provides plenty of corollaries in terms of prime powers (such as Eisenstein criterion, but in this case with multiple primes). Schönemann–Eisenstein–Dumas-Type Irreducibility Conditions that Use Arbitrarily Many Prime Numbers. An example: try to prove irreducibility of this one $$4x^6+108x^5+108x^4+108x^3+108x^2+108x+27.$$ The article above provides way to do this (it is first example in last section of examples).
Some related questions:
- How to choose correct strategy for irreducibility testing in $mathbbZ[X]$?
- Polynomial is irreducible over $mathbbQ$
- Prove that $f(x)$ is irreducible iff its reciprocal polynomial $f^*(x)$ is irreducible.
$endgroup$
Reversing the polynomial. If you have polynomial with degree $geq 2$ and non zero constant coefficients (otherwise it would be reducible so it wouldn't be interesting anyway), then you can reverse the coefficients and check irreducibility on that reciprocal polynomial. For example instead of checking $f(x)=2x^4+2x^3+2x^2+2x+1$, you can instead check $x^4+2x^3+2x^2+2x+2$ (and see it is irreducible by Eisenstein). This corresponds to $x^4f(1/x)$.
Another useful criterion is one provided by Ram's Murty in paper already referenced in the other answer, similar to Cohn's irreducibility criteria, it states:
Murty's irreducibility criterion: Let $f(x)=a_mx^m+a_m-1x^m-1+dots+a_1x+a_0$ be a polynomial of degree $m$ in $mathbbZ[x]$ and set $$H=max_0leq ileq m-1 |a_i/a_m|.$$
If $f(n)$ is prime for some integer $ngeq H+2$, then $f(x)$ is irreducible in $mathbbZ[x]$.
You can see that for example $f(x)=x^3-11x^2+19x-17$ is irreducible because of that, if you try $n=24$.
Osada's criterion. Let $f(x)=x^n+a_1x^n-1+dots+a_n-1xpm p$ be a polynomial with integer coefficients, where $p$ is a prime. If $p>1+|a_1|+dots+|a_n-1|$, then $f(x)$ is irreducible over $mathbbZ$.
Following one is also simple to use, although I found it barely applicable, nvertheless interesting:
Bauer's criterion. Let $a_1 geq a_2 geq dots geq a_n$ be positive integers and $n geq 2$. Then the polynomial $p(x)=x^n-a_1x^n-1-a_2x^n-2-dots-a_n$ is irreducible over $mathbbZ$.
Advanced criteria related to Newton polygons. These criteria are little bit more advanced to use, but the paper below provides plenty of corollaries in terms of prime powers (such as Eisenstein criterion, but in this case with multiple primes). Schönemann–Eisenstein–Dumas-Type Irreducibility Conditions that Use Arbitrarily Many Prime Numbers. An example: try to prove irreducibility of this one $$4x^6+108x^5+108x^4+108x^3+108x^2+108x+27.$$ The article above provides way to do this (it is first example in last section of examples).
Some related questions:
- How to choose correct strategy for irreducibility testing in $mathbbZ[X]$?
- Polynomial is irreducible over $mathbbQ$
- Prove that $f(x)$ is irreducible iff its reciprocal polynomial $f^*(x)$ is irreducible.
edited Mar 15 at 16:04
Bill Dubuque
213k29195654
213k29195654
answered May 1 '18 at 8:06
SilSil
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Well, Eisenstein's criterion is basically a version of reducing modulo a prime.
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– Akhil Mathew
Aug 9 '10 at 18:10
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See also this meta answer for grouped list of questions related to this topic math.meta.stackexchange.com/questions/1868/….
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– Sil
Aug 11 '18 at 11:13