Are there irreducible polynomials of every degree over every field?












1














Let $F$ be a field and $n geq 2$. Must there exist an irreducible polynomial of degree $n$ in $F[X]$?



When $F=mathbb{Q}$ the answer is certainly "yes," as you can apply Eisenstein's criterion to $X^n-2$, for example. When $F = mathbb{Z}/pmathbb{Z}$ it is well known that the answer is "yes" (there have been numerous StackExchange posts about that), although it takes more work to get there (and especially to actually construct an example).



What about general $F$?










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  • 4




    What about $F=mathbb{R}$ or $F=mathbb{C}$?
    – Mindlack
    1 hour ago
















1














Let $F$ be a field and $n geq 2$. Must there exist an irreducible polynomial of degree $n$ in $F[X]$?



When $F=mathbb{Q}$ the answer is certainly "yes," as you can apply Eisenstein's criterion to $X^n-2$, for example. When $F = mathbb{Z}/pmathbb{Z}$ it is well known that the answer is "yes" (there have been numerous StackExchange posts about that), although it takes more work to get there (and especially to actually construct an example).



What about general $F$?










share|cite|improve this question




















  • 4




    What about $F=mathbb{R}$ or $F=mathbb{C}$?
    – Mindlack
    1 hour ago














1












1








1







Let $F$ be a field and $n geq 2$. Must there exist an irreducible polynomial of degree $n$ in $F[X]$?



When $F=mathbb{Q}$ the answer is certainly "yes," as you can apply Eisenstein's criterion to $X^n-2$, for example. When $F = mathbb{Z}/pmathbb{Z}$ it is well known that the answer is "yes" (there have been numerous StackExchange posts about that), although it takes more work to get there (and especially to actually construct an example).



What about general $F$?










share|cite|improve this question















Let $F$ be a field and $n geq 2$. Must there exist an irreducible polynomial of degree $n$ in $F[X]$?



When $F=mathbb{Q}$ the answer is certainly "yes," as you can apply Eisenstein's criterion to $X^n-2$, for example. When $F = mathbb{Z}/pmathbb{Z}$ it is well known that the answer is "yes" (there have been numerous StackExchange posts about that), although it takes more work to get there (and especially to actually construct an example).



What about general $F$?







abstract-algebra polynomials irreducible-polynomials






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edited 42 mins ago









Bernard

118k639112




118k639112










asked 1 hour ago









Dave GaeblerDave Gaebler

1,350915




1,350915








  • 4




    What about $F=mathbb{R}$ or $F=mathbb{C}$?
    – Mindlack
    1 hour ago














  • 4




    What about $F=mathbb{R}$ or $F=mathbb{C}$?
    – Mindlack
    1 hour ago








4




4




What about $F=mathbb{R}$ or $F=mathbb{C}$?
– Mindlack
1 hour ago




What about $F=mathbb{R}$ or $F=mathbb{C}$?
– Mindlack
1 hour ago










2 Answers
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If $F$ is algebraically closed then every polynomial with coefficients in $F$ splits into a product of linear factors over $F$. So there do not exist any irreducible polynomials of degree $n>1$.






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    0














    There are no irreducible cubic polynomials over the real numbers. Nor of any odd degree greater than 1. This follows from the intermediate value theorem.



    In fact, the fundamental theorem of algebra says that the only irreducible polynomials over the real numbers are those of degree 1 and those of degree 2 with negative discriminant.






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    • Or even degree greater than 2. en.m.wikipedia.org/wiki/Irreducible_polynomial
      – badjohn
      23 mins ago











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    2 Answers
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    2 Answers
    2






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    active

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    active

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    6














    If $F$ is algebraically closed then every polynomial with coefficients in $F$ splits into a product of linear factors over $F$. So there do not exist any irreducible polynomials of degree $n>1$.






    share|cite|improve this answer


























      6














      If $F$ is algebraically closed then every polynomial with coefficients in $F$ splits into a product of linear factors over $F$. So there do not exist any irreducible polynomials of degree $n>1$.






      share|cite|improve this answer
























        6












        6








        6






        If $F$ is algebraically closed then every polynomial with coefficients in $F$ splits into a product of linear factors over $F$. So there do not exist any irreducible polynomials of degree $n>1$.






        share|cite|improve this answer












        If $F$ is algebraically closed then every polynomial with coefficients in $F$ splits into a product of linear factors over $F$. So there do not exist any irreducible polynomials of degree $n>1$.







        share|cite|improve this answer












        share|cite|improve this answer



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        answered 1 hour ago









        ServaesServaes

        22.5k33793




        22.5k33793























            0














            There are no irreducible cubic polynomials over the real numbers. Nor of any odd degree greater than 1. This follows from the intermediate value theorem.



            In fact, the fundamental theorem of algebra says that the only irreducible polynomials over the real numbers are those of degree 1 and those of degree 2 with negative discriminant.






            share|cite|improve this answer























            • Or even degree greater than 2. en.m.wikipedia.org/wiki/Irreducible_polynomial
              – badjohn
              23 mins ago
















            0














            There are no irreducible cubic polynomials over the real numbers. Nor of any odd degree greater than 1. This follows from the intermediate value theorem.



            In fact, the fundamental theorem of algebra says that the only irreducible polynomials over the real numbers are those of degree 1 and those of degree 2 with negative discriminant.






            share|cite|improve this answer























            • Or even degree greater than 2. en.m.wikipedia.org/wiki/Irreducible_polynomial
              – badjohn
              23 mins ago














            0












            0








            0






            There are no irreducible cubic polynomials over the real numbers. Nor of any odd degree greater than 1. This follows from the intermediate value theorem.



            In fact, the fundamental theorem of algebra says that the only irreducible polynomials over the real numbers are those of degree 1 and those of degree 2 with negative discriminant.






            share|cite|improve this answer














            There are no irreducible cubic polynomials over the real numbers. Nor of any odd degree greater than 1. This follows from the intermediate value theorem.



            In fact, the fundamental theorem of algebra says that the only irreducible polynomials over the real numbers are those of degree 1 and those of degree 2 with negative discriminant.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited 14 mins ago

























            answered 36 mins ago









            lhflhf

            163k10168388




            163k10168388












            • Or even degree greater than 2. en.m.wikipedia.org/wiki/Irreducible_polynomial
              – badjohn
              23 mins ago


















            • Or even degree greater than 2. en.m.wikipedia.org/wiki/Irreducible_polynomial
              – badjohn
              23 mins ago
















            Or even degree greater than 2. en.m.wikipedia.org/wiki/Irreducible_polynomial
            – badjohn
            23 mins ago




            Or even degree greater than 2. en.m.wikipedia.org/wiki/Irreducible_polynomial
            – badjohn
            23 mins ago


















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