Irreducible polynomials of degree greater than 4 over finite fields












2














I want to build a field with p^n elements. I know that this can be done by finding a irreducible (on Z_p) polynomial f of degree n and the result would be the Z_p/f.
My question is finding this irreducible polynomial. I know that if it has degree <= 3, then it's irreducible iff it has no roots. But what if I want to construct a field with 81 = 3^4 elements? How can I find an irreducible polynomial of degree 4?






number-theory polynomials finite-fields irreducible-polynomials







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    I want to build a field with p^n elements. I know that this can be done by finding a irreducible (on Z_p) polynomial f of degree n and the result would be the Z_p/f.
    My question is finding this irreducible polynomial. I know that if it has degree <= 3, then it's irreducible iff it has no roots. But what if I want to construct a field with 81 = 3^4 elements? How can I find an irreducible polynomial of degree 4?






    number-theory polynomials finite-fields irreducible-polynomials







    share|cite|improve this question

























      2












      2








      2







      I want to build a field with p^n elements. I know that this can be done by finding a irreducible (on Z_p) polynomial f of degree n and the result would be the Z_p/f.
      My question is finding this irreducible polynomial. I know that if it has degree <= 3, then it's irreducible iff it has no roots. But what if I want to construct a field with 81 = 3^4 elements? How can I find an irreducible polynomial of degree 4?






      number-theory polynomials finite-fields irreducible-polynomials







      share|cite|improve this question













      I want to build a field with p^n elements. I know that this can be done by finding a irreducible (on Z_p) polynomial f of degree n and the result would be the Z_p/f.
      My question is finding this irreducible polynomial. I know that if it has degree <= 3, then it's irreducible iff it has no roots. But what if I want to construct a field with 81 = 3^4 elements? How can I find an irreducible polynomial of degree 4?






      number-theory polynomials finite-fields irreducible-polynomials




      number-theory polynomials finite-fields irreducible-polynomials






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      asked 1 hour ago









      J. Dionisio

      7811




      7811






















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          Find the irreducible quadratics. Multiply them together. Those fourth degree polynomials won't do. Now try some others at random (or systematically, following a list in some natural order). When you find one with no roots you're done.



          This is mildly tedious, but you'll get good at the arithmetic, which may come in handy in other computations in the future.



          You can also ask Wolfram alpha to factor polynomials modulo $3$.






          share|cite|improve this answer































            2














            The elements of $GF(p^n)$ are the zeros of the polynomial $x^{p^n}-x$. This polynomials decomposes into irreducible polynomials of degree $d$ over $GF(p)$ where $d$ divides $n$. It can be shown that this decomposition contains at least one polynomial of degree $n$ which ensures the existence of a finite field with $p^n$ elements. In a CAS you usually have access to such irreducible polynomials.






            share|cite|improve this answer





























              2














              Since $3$ is a primitive root modulo $5$, the fifth roots of unity are in ${bf F}_{81}$, but not in a proper subfield. This means that the cyclotomic polynomial $Phi_5(X)=X^4+X^3+X^2+X+1$ is irreducible modulo $3$.






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                3 Answers
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                active

                oldest

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                3 Answers
                3






                active

                oldest

                votes









                active

                oldest

                votes






                active

                oldest

                votes









                2














                Find the irreducible quadratics. Multiply them together. Those fourth degree polynomials won't do. Now try some others at random (or systematically, following a list in some natural order). When you find one with no roots you're done.



                This is mildly tedious, but you'll get good at the arithmetic, which may come in handy in other computations in the future.



                You can also ask Wolfram alpha to factor polynomials modulo $3$.






                share|cite|improve this answer




























                  2














                  Find the irreducible quadratics. Multiply them together. Those fourth degree polynomials won't do. Now try some others at random (or systematically, following a list in some natural order). When you find one with no roots you're done.



                  This is mildly tedious, but you'll get good at the arithmetic, which may come in handy in other computations in the future.



                  You can also ask Wolfram alpha to factor polynomials modulo $3$.






                  share|cite|improve this answer


























                    2












                    2








                    2






                    Find the irreducible quadratics. Multiply them together. Those fourth degree polynomials won't do. Now try some others at random (or systematically, following a list in some natural order). When you find one with no roots you're done.



                    This is mildly tedious, but you'll get good at the arithmetic, which may come in handy in other computations in the future.



                    You can also ask Wolfram alpha to factor polynomials modulo $3$.






                    share|cite|improve this answer














                    Find the irreducible quadratics. Multiply them together. Those fourth degree polynomials won't do. Now try some others at random (or systematically, following a list in some natural order). When you find one with no roots you're done.



                    This is mildly tedious, but you'll get good at the arithmetic, which may come in handy in other computations in the future.



                    You can also ask Wolfram alpha to factor polynomials modulo $3$.







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited 18 mins ago

























                    answered 1 hour ago









                    Ethan Bolker

                    41.4k547108




                    41.4k547108























                        2














                        The elements of $GF(p^n)$ are the zeros of the polynomial $x^{p^n}-x$. This polynomials decomposes into irreducible polynomials of degree $d$ over $GF(p)$ where $d$ divides $n$. It can be shown that this decomposition contains at least one polynomial of degree $n$ which ensures the existence of a finite field with $p^n$ elements. In a CAS you usually have access to such irreducible polynomials.






                        share|cite|improve this answer


























                          2














                          The elements of $GF(p^n)$ are the zeros of the polynomial $x^{p^n}-x$. This polynomials decomposes into irreducible polynomials of degree $d$ over $GF(p)$ where $d$ divides $n$. It can be shown that this decomposition contains at least one polynomial of degree $n$ which ensures the existence of a finite field with $p^n$ elements. In a CAS you usually have access to such irreducible polynomials.






                          share|cite|improve this answer
























                            2












                            2








                            2






                            The elements of $GF(p^n)$ are the zeros of the polynomial $x^{p^n}-x$. This polynomials decomposes into irreducible polynomials of degree $d$ over $GF(p)$ where $d$ divides $n$. It can be shown that this decomposition contains at least one polynomial of degree $n$ which ensures the existence of a finite field with $p^n$ elements. In a CAS you usually have access to such irreducible polynomials.






                            share|cite|improve this answer












                            The elements of $GF(p^n)$ are the zeros of the polynomial $x^{p^n}-x$. This polynomials decomposes into irreducible polynomials of degree $d$ over $GF(p)$ where $d$ divides $n$. It can be shown that this decomposition contains at least one polynomial of degree $n$ which ensures the existence of a finite field with $p^n$ elements. In a CAS you usually have access to such irreducible polynomials.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered 1 hour ago









                            Wuestenfux

                            3,5711411




                            3,5711411























                                2














                                Since $3$ is a primitive root modulo $5$, the fifth roots of unity are in ${bf F}_{81}$, but not in a proper subfield. This means that the cyclotomic polynomial $Phi_5(X)=X^4+X^3+X^2+X+1$ is irreducible modulo $3$.






                                share|cite|improve this answer


























                                  2














                                  Since $3$ is a primitive root modulo $5$, the fifth roots of unity are in ${bf F}_{81}$, but not in a proper subfield. This means that the cyclotomic polynomial $Phi_5(X)=X^4+X^3+X^2+X+1$ is irreducible modulo $3$.






                                  share|cite|improve this answer
























                                    2












                                    2








                                    2






                                    Since $3$ is a primitive root modulo $5$, the fifth roots of unity are in ${bf F}_{81}$, but not in a proper subfield. This means that the cyclotomic polynomial $Phi_5(X)=X^4+X^3+X^2+X+1$ is irreducible modulo $3$.






                                    share|cite|improve this answer












                                    Since $3$ is a primitive root modulo $5$, the fifth roots of unity are in ${bf F}_{81}$, but not in a proper subfield. This means that the cyclotomic polynomial $Phi_5(X)=X^4+X^3+X^2+X+1$ is irreducible modulo $3$.







                                    share|cite|improve this answer












                                    share|cite|improve this answer



                                    share|cite|improve this answer










                                    answered 1 hour ago









                                    Rene Schoof

                                    25613




                                    25613






























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