Let k be a finite field. Is it true that the number of irreducible polynomials in k[x] is also finite?












3












$begingroup$


I know this question has been asked before and I understand that it can be proved using the same sort of proof as the one used to show that there's infinite primes, but are there other ways of showing this? Perhaps a counter example?






field-theory finite-fields irreducible-polynomials







share|cite|improve this question









$endgroup$












  • $begingroup$
    Here: math.stackexchange.com/q/80389/2838
    $endgroup$
    – Joe Johnson 126
    Dec 10 '18 at 22:03










  • $begingroup$
    For linking purposes here is a prior thread with Euclid's proof and a variant using the strong divisibility sequence $,(x^n-1)/(x-1) $
    $endgroup$
    – Bill Dubuque
    Dec 10 '18 at 22:07


















3












$begingroup$


I know this question has been asked before and I understand that it can be proved using the same sort of proof as the one used to show that there's infinite primes, but are there other ways of showing this? Perhaps a counter example?






field-theory finite-fields irreducible-polynomials







share|cite|improve this question









$endgroup$












  • $begingroup$
    Here: math.stackexchange.com/q/80389/2838
    $endgroup$
    – Joe Johnson 126
    Dec 10 '18 at 22:03










  • $begingroup$
    For linking purposes here is a prior thread with Euclid's proof and a variant using the strong divisibility sequence $,(x^n-1)/(x-1) $
    $endgroup$
    – Bill Dubuque
    Dec 10 '18 at 22:07
















3












3








3





$begingroup$


I know this question has been asked before and I understand that it can be proved using the same sort of proof as the one used to show that there's infinite primes, but are there other ways of showing this? Perhaps a counter example?






field-theory finite-fields irreducible-polynomials







share|cite|improve this question









$endgroup$




I know this question has been asked before and I understand that it can be proved using the same sort of proof as the one used to show that there's infinite primes, but are there other ways of showing this? Perhaps a counter example?






field-theory finite-fields irreducible-polynomials




field-theory finite-fields irreducible-polynomials






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 10 '18 at 21:52









Jon D.Jon D.

164




164












  • $begingroup$
    Here: math.stackexchange.com/q/80389/2838
    $endgroup$
    – Joe Johnson 126
    Dec 10 '18 at 22:03










  • $begingroup$
    For linking purposes here is a prior thread with Euclid's proof and a variant using the strong divisibility sequence $,(x^n-1)/(x-1) $
    $endgroup$
    – Bill Dubuque
    Dec 10 '18 at 22:07




















  • $begingroup$
    Here: math.stackexchange.com/q/80389/2838
    $endgroup$
    – Joe Johnson 126
    Dec 10 '18 at 22:03










  • $begingroup$
    For linking purposes here is a prior thread with Euclid's proof and a variant using the strong divisibility sequence $,(x^n-1)/(x-1) $
    $endgroup$
    – Bill Dubuque
    Dec 10 '18 at 22:07


















$begingroup$
Here: math.stackexchange.com/q/80389/2838
$endgroup$
– Joe Johnson 126
Dec 10 '18 at 22:03




$begingroup$
Here: math.stackexchange.com/q/80389/2838
$endgroup$
– Joe Johnson 126
Dec 10 '18 at 22:03












$begingroup$
For linking purposes here is a prior thread with Euclid's proof and a variant using the strong divisibility sequence $,(x^n-1)/(x-1) $
$endgroup$
– Bill Dubuque
Dec 10 '18 at 22:07






$begingroup$
For linking purposes here is a prior thread with Euclid's proof and a variant using the strong divisibility sequence $,(x^n-1)/(x-1) $
$endgroup$
– Bill Dubuque
Dec 10 '18 at 22:07












5 Answers
5






active

oldest

votes


















1












$begingroup$

One nice approach is to actually count the irreducible polynomials of given degree. For example, see the answers to this question: How many irreducible polynomials of degree $n$ exist over $mathbb{F}_p$? (this question uses prime fields, but exactly the same arguments and formulas work if $p$ is a power of a prime)



In particular, there is at least one irreducible polynomial of any given degree. Since there are infinitely many degrees, there are infinitely many irreducible polynomials.






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    Suppose there are only finitely many irreducible polynomials. Consider the splitting field $K$ of their product, which is finite dimensional over $k$, hence finite.



    Suppose $K'$ is an algebraic extension field of $K$; if $bin K'$, then $b$ is algebraic over $K$, hence also over $k$, so its minimal polynomial over $k$ is irreducible. But in $K$ there is a root of every irreducible polynomial in $k[x]$. Hence $bin K$.



    Therefore $K$ is algebraically closed.



    Let $K={a_1=0,a_2=1,a_3,dots,a_n}$. The polynomial
    $$
    (x-a_1)(x-a_2)(x-a_3)dots(x-a_n)+1
    $$
    has no root in $K$.



    Contradiction.






    share|cite|improve this answer









    $endgroup$





















      0












      $begingroup$

      This is a bit of an overkill, and I'm not sure if you can make it non-circular, but here's one way to convince yourself that this is true.



      For every finite field extension $Ksupseteq F$, there is an irreducible $pin F[x]$ such that $K$ is isomorphic (over $F$) to $F[x]/p$.






      share|cite|improve this answer









      $endgroup$





















        0












        $begingroup$

        No, there exist irreducible polynomial in every degree, due to the following formula:




        Let $p$ be a prime number, $q=p^n$. In $ mathbf F_q[X]$, we have the factorisation:
        $$ X^{q^n}-X=prod_{dmid n}prod_{:deg P=d,\text{ irreducible}}mkern-12mu P$$







        share|cite|improve this answer









        $endgroup$





















          0












          $begingroup$

          So after more reading, I believe I can show it is false by a counter-example.



          Consider f(x) = xn + 3 ∈ ℤ5[x] with n ∈ ℕ.
          Then Eisenstein's criterion (with p = 3) says that f(x) is irreducible in ℚ[x] and therefore, f(x) is irreducible in ℤ5[x] since ℤ5[x] ⊂ ℚ[x]. This gives infinitely many irreducible polynomials in ℤ5[x] because there's infinite possibilities for the value of the exponent n.



          Therefore, the initial statement is false.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            That approach is already in this older dupe.
            $endgroup$
            – Bill Dubuque
            Dec 11 '18 at 0:30











          Your Answer





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          5 Answers
          5






          active

          oldest

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          5 Answers
          5






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          1












          $begingroup$

          One nice approach is to actually count the irreducible polynomials of given degree. For example, see the answers to this question: How many irreducible polynomials of degree $n$ exist over $mathbb{F}_p$? (this question uses prime fields, but exactly the same arguments and formulas work if $p$ is a power of a prime)



          In particular, there is at least one irreducible polynomial of any given degree. Since there are infinitely many degrees, there are infinitely many irreducible polynomials.






          share|cite|improve this answer









          $endgroup$


















            1












            $begingroup$

            One nice approach is to actually count the irreducible polynomials of given degree. For example, see the answers to this question: How many irreducible polynomials of degree $n$ exist over $mathbb{F}_p$? (this question uses prime fields, but exactly the same arguments and formulas work if $p$ is a power of a prime)



            In particular, there is at least one irreducible polynomial of any given degree. Since there are infinitely many degrees, there are infinitely many irreducible polynomials.






            share|cite|improve this answer









            $endgroup$
















              1












              1








              1





              $begingroup$

              One nice approach is to actually count the irreducible polynomials of given degree. For example, see the answers to this question: How many irreducible polynomials of degree $n$ exist over $mathbb{F}_p$? (this question uses prime fields, but exactly the same arguments and formulas work if $p$ is a power of a prime)



              In particular, there is at least one irreducible polynomial of any given degree. Since there are infinitely many degrees, there are infinitely many irreducible polynomials.






              share|cite|improve this answer









              $endgroup$



              One nice approach is to actually count the irreducible polynomials of given degree. For example, see the answers to this question: How many irreducible polynomials of degree $n$ exist over $mathbb{F}_p$? (this question uses prime fields, but exactly the same arguments and formulas work if $p$ is a power of a prime)



              In particular, there is at least one irreducible polynomial of any given degree. Since there are infinitely many degrees, there are infinitely many irreducible polynomials.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Dec 10 '18 at 22:05









              SladeSlade

              25.1k12665




              25.1k12665























                  1












                  $begingroup$

                  Suppose there are only finitely many irreducible polynomials. Consider the splitting field $K$ of their product, which is finite dimensional over $k$, hence finite.



                  Suppose $K'$ is an algebraic extension field of $K$; if $bin K'$, then $b$ is algebraic over $K$, hence also over $k$, so its minimal polynomial over $k$ is irreducible. But in $K$ there is a root of every irreducible polynomial in $k[x]$. Hence $bin K$.



                  Therefore $K$ is algebraically closed.



                  Let $K={a_1=0,a_2=1,a_3,dots,a_n}$. The polynomial
                  $$
                  (x-a_1)(x-a_2)(x-a_3)dots(x-a_n)+1
                  $$
                  has no root in $K$.



                  Contradiction.






                  share|cite|improve this answer









                  $endgroup$


















                    1












                    $begingroup$

                    Suppose there are only finitely many irreducible polynomials. Consider the splitting field $K$ of their product, which is finite dimensional over $k$, hence finite.



                    Suppose $K'$ is an algebraic extension field of $K$; if $bin K'$, then $b$ is algebraic over $K$, hence also over $k$, so its minimal polynomial over $k$ is irreducible. But in $K$ there is a root of every irreducible polynomial in $k[x]$. Hence $bin K$.



                    Therefore $K$ is algebraically closed.



                    Let $K={a_1=0,a_2=1,a_3,dots,a_n}$. The polynomial
                    $$
                    (x-a_1)(x-a_2)(x-a_3)dots(x-a_n)+1
                    $$
                    has no root in $K$.



                    Contradiction.






                    share|cite|improve this answer









                    $endgroup$
















                      1












                      1








                      1





                      $begingroup$

                      Suppose there are only finitely many irreducible polynomials. Consider the splitting field $K$ of their product, which is finite dimensional over $k$, hence finite.



                      Suppose $K'$ is an algebraic extension field of $K$; if $bin K'$, then $b$ is algebraic over $K$, hence also over $k$, so its minimal polynomial over $k$ is irreducible. But in $K$ there is a root of every irreducible polynomial in $k[x]$. Hence $bin K$.



                      Therefore $K$ is algebraically closed.



                      Let $K={a_1=0,a_2=1,a_3,dots,a_n}$. The polynomial
                      $$
                      (x-a_1)(x-a_2)(x-a_3)dots(x-a_n)+1
                      $$
                      has no root in $K$.



                      Contradiction.






                      share|cite|improve this answer









                      $endgroup$



                      Suppose there are only finitely many irreducible polynomials. Consider the splitting field $K$ of their product, which is finite dimensional over $k$, hence finite.



                      Suppose $K'$ is an algebraic extension field of $K$; if $bin K'$, then $b$ is algebraic over $K$, hence also over $k$, so its minimal polynomial over $k$ is irreducible. But in $K$ there is a root of every irreducible polynomial in $k[x]$. Hence $bin K$.



                      Therefore $K$ is algebraically closed.



                      Let $K={a_1=0,a_2=1,a_3,dots,a_n}$. The polynomial
                      $$
                      (x-a_1)(x-a_2)(x-a_3)dots(x-a_n)+1
                      $$
                      has no root in $K$.



                      Contradiction.







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered Dec 10 '18 at 22:38









                      egregegreg

                      181k1485203




                      181k1485203























                          0












                          $begingroup$

                          This is a bit of an overkill, and I'm not sure if you can make it non-circular, but here's one way to convince yourself that this is true.



                          For every finite field extension $Ksupseteq F$, there is an irreducible $pin F[x]$ such that $K$ is isomorphic (over $F$) to $F[x]/p$.






                          share|cite|improve this answer









                          $endgroup$


















                            0












                            $begingroup$

                            This is a bit of an overkill, and I'm not sure if you can make it non-circular, but here's one way to convince yourself that this is true.



                            For every finite field extension $Ksupseteq F$, there is an irreducible $pin F[x]$ such that $K$ is isomorphic (over $F$) to $F[x]/p$.






                            share|cite|improve this answer









                            $endgroup$
















                              0












                              0








                              0





                              $begingroup$

                              This is a bit of an overkill, and I'm not sure if you can make it non-circular, but here's one way to convince yourself that this is true.



                              For every finite field extension $Ksupseteq F$, there is an irreducible $pin F[x]$ such that $K$ is isomorphic (over $F$) to $F[x]/p$.






                              share|cite|improve this answer









                              $endgroup$



                              This is a bit of an overkill, and I'm not sure if you can make it non-circular, but here's one way to convince yourself that this is true.



                              For every finite field extension $Ksupseteq F$, there is an irreducible $pin F[x]$ such that $K$ is isomorphic (over $F$) to $F[x]/p$.







                              share|cite|improve this answer












                              share|cite|improve this answer



                              share|cite|improve this answer










                              answered Dec 10 '18 at 22:21









                              tomasztomasz

                              23.6k23479




                              23.6k23479























                                  0












                                  $begingroup$

                                  No, there exist irreducible polynomial in every degree, due to the following formula:




                                  Let $p$ be a prime number, $q=p^n$. In $ mathbf F_q[X]$, we have the factorisation:
                                  $$ X^{q^n}-X=prod_{dmid n}prod_{:deg P=d,\text{ irreducible}}mkern-12mu P$$







                                  share|cite|improve this answer









                                  $endgroup$


















                                    0












                                    $begingroup$

                                    No, there exist irreducible polynomial in every degree, due to the following formula:




                                    Let $p$ be a prime number, $q=p^n$. In $ mathbf F_q[X]$, we have the factorisation:
                                    $$ X^{q^n}-X=prod_{dmid n}prod_{:deg P=d,\text{ irreducible}}mkern-12mu P$$







                                    share|cite|improve this answer









                                    $endgroup$
















                                      0












                                      0








                                      0





                                      $begingroup$

                                      No, there exist irreducible polynomial in every degree, due to the following formula:




                                      Let $p$ be a prime number, $q=p^n$. In $ mathbf F_q[X]$, we have the factorisation:
                                      $$ X^{q^n}-X=prod_{dmid n}prod_{:deg P=d,\text{ irreducible}}mkern-12mu P$$







                                      share|cite|improve this answer









                                      $endgroup$



                                      No, there exist irreducible polynomial in every degree, due to the following formula:




                                      Let $p$ be a prime number, $q=p^n$. In $ mathbf F_q[X]$, we have the factorisation:
                                      $$ X^{q^n}-X=prod_{dmid n}prod_{:deg P=d,\text{ irreducible}}mkern-12mu P$$








                                      share|cite|improve this answer












                                      share|cite|improve this answer



                                      share|cite|improve this answer










                                      answered Dec 10 '18 at 22:27









                                      BernardBernard

                                      120k740113




                                      120k740113























                                          0












                                          $begingroup$

                                          So after more reading, I believe I can show it is false by a counter-example.



                                          Consider f(x) = xn + 3 ∈ ℤ5[x] with n ∈ ℕ.
                                          Then Eisenstein's criterion (with p = 3) says that f(x) is irreducible in ℚ[x] and therefore, f(x) is irreducible in ℤ5[x] since ℤ5[x] ⊂ ℚ[x]. This gives infinitely many irreducible polynomials in ℤ5[x] because there's infinite possibilities for the value of the exponent n.



                                          Therefore, the initial statement is false.






                                          share|cite|improve this answer









                                          $endgroup$













                                          • $begingroup$
                                            That approach is already in this older dupe.
                                            $endgroup$
                                            – Bill Dubuque
                                            Dec 11 '18 at 0:30
















                                          0












                                          $begingroup$

                                          So after more reading, I believe I can show it is false by a counter-example.



                                          Consider f(x) = xn + 3 ∈ ℤ5[x] with n ∈ ℕ.
                                          Then Eisenstein's criterion (with p = 3) says that f(x) is irreducible in ℚ[x] and therefore, f(x) is irreducible in ℤ5[x] since ℤ5[x] ⊂ ℚ[x]. This gives infinitely many irreducible polynomials in ℤ5[x] because there's infinite possibilities for the value of the exponent n.



                                          Therefore, the initial statement is false.






                                          share|cite|improve this answer









                                          $endgroup$













                                          • $begingroup$
                                            That approach is already in this older dupe.
                                            $endgroup$
                                            – Bill Dubuque
                                            Dec 11 '18 at 0:30














                                          0












                                          0








                                          0





                                          $begingroup$

                                          So after more reading, I believe I can show it is false by a counter-example.



                                          Consider f(x) = xn + 3 ∈ ℤ5[x] with n ∈ ℕ.
                                          Then Eisenstein's criterion (with p = 3) says that f(x) is irreducible in ℚ[x] and therefore, f(x) is irreducible in ℤ5[x] since ℤ5[x] ⊂ ℚ[x]. This gives infinitely many irreducible polynomials in ℤ5[x] because there's infinite possibilities for the value of the exponent n.



                                          Therefore, the initial statement is false.






                                          share|cite|improve this answer









                                          $endgroup$



                                          So after more reading, I believe I can show it is false by a counter-example.



                                          Consider f(x) = xn + 3 ∈ ℤ5[x] with n ∈ ℕ.
                                          Then Eisenstein's criterion (with p = 3) says that f(x) is irreducible in ℚ[x] and therefore, f(x) is irreducible in ℤ5[x] since ℤ5[x] ⊂ ℚ[x]. This gives infinitely many irreducible polynomials in ℤ5[x] because there's infinite possibilities for the value of the exponent n.



                                          Therefore, the initial statement is false.







                                          share|cite|improve this answer












                                          share|cite|improve this answer



                                          share|cite|improve this answer










                                          answered Dec 10 '18 at 23:51









                                          Jon D.Jon D.

                                          164




                                          164












                                          • $begingroup$
                                            That approach is already in this older dupe.
                                            $endgroup$
                                            – Bill Dubuque
                                            Dec 11 '18 at 0:30


















                                          • $begingroup$
                                            That approach is already in this older dupe.
                                            $endgroup$
                                            – Bill Dubuque
                                            Dec 11 '18 at 0:30
















                                          $begingroup$
                                          That approach is already in this older dupe.
                                          $endgroup$
                                          – Bill Dubuque
                                          Dec 11 '18 at 0:30




                                          $begingroup$
                                          That approach is already in this older dupe.
                                          $endgroup$
                                          – Bill Dubuque
                                          Dec 11 '18 at 0:30


















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