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?










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$endgroup$







  • 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















35












$begingroup$


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?










share|cite|improve this question











$endgroup$







  • 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













35












35








35


38



$begingroup$


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?










share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








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












  • 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










6 Answers
6






active

oldest

votes


















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$begingroup$

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...






share|cite|improve this answer









$endgroup$












  • $begingroup$
    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


















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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.






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$endgroup$








  • 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


















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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.






share|cite|improve this answer









$endgroup$












  • $begingroup$
    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


















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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






share|cite|improve this answer









$endgroup$




















    9












    $begingroup$

    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





    share|cite|improve this answer











    $endgroup$




















      3












      $begingroup$

      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.





      share|cite|improve this answer











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        6 Answers
        6






        active

        oldest

        votes








        6 Answers
        6






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        20












        $begingroup$

        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...






        share|cite|improve this answer









        $endgroup$












        • $begingroup$
          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















        20












        $begingroup$

        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...






        share|cite|improve this answer









        $endgroup$












        • $begingroup$
          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













        20












        20








        20





        $begingroup$

        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...






        share|cite|improve this answer









        $endgroup$



        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...







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        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
















        • $begingroup$
          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











        15












        $begingroup$

        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.






        share|cite|improve this answer









        $endgroup$








        • 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















        15












        $begingroup$

        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.






        share|cite|improve this answer









        $endgroup$








        • 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













        15












        15








        15





        $begingroup$

        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.






        share|cite|improve this answer









        $endgroup$



        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.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        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












        • 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











        14












        $begingroup$

        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.






        share|cite|improve this answer









        $endgroup$












        • $begingroup$
          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















        14












        $begingroup$

        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.






        share|cite|improve this answer









        $endgroup$












        • $begingroup$
          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













        14












        14








        14





        $begingroup$

        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.






        share|cite|improve this answer









        $endgroup$



        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.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        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
















        • $begingroup$
          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











        10












        $begingroup$

        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






        share|cite|improve this answer









        $endgroup$

















          10












          $begingroup$

          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






          share|cite|improve this answer









          $endgroup$















            10












            10








            10





            $begingroup$

            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






            share|cite|improve this answer









            $endgroup$



            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







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Aug 9 '10 at 22:52









            Bill DubuqueBill Dubuque

            213k29195654




            213k29195654





















                9












                $begingroup$

                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





                share|cite|improve this answer











                $endgroup$

















                  9












                  $begingroup$

                  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





                  share|cite|improve this answer











                  $endgroup$















                    9












                    9








                    9





                    $begingroup$

                    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





                    share|cite|improve this answer











                    $endgroup$



                    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






                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Mar 15 at 22:03









                    user26857

                    39.4k124183




                    39.4k124183










                    answered Aug 10 '10 at 2:01









                    n0vakovicn0vakovic

                    499417




                    499417





















                        3












                        $begingroup$

                        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.





                        share|cite|improve this answer











                        $endgroup$

















                          3












                          $begingroup$

                          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.





                          share|cite|improve this answer











                          $endgroup$















                            3












                            3








                            3





                            $begingroup$

                            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.





                            share|cite|improve this answer











                            $endgroup$



                            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.






                            share|cite|improve this answer














                            share|cite|improve this answer



                            share|cite|improve this answer








                            edited Mar 15 at 16:04









                            Bill Dubuque

                            213k29195654




                            213k29195654










                            answered May 1 '18 at 8:06









                            SilSil

                            5,25721643




                            5,25721643



























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