Finitely generated $k$-algebra of Krull-dimension one has infinitely many primes

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I am looking for a reference or a quick proof of the statement that




Every finitely generated $k$-algebra of Krull-dimension one has infinitely many prime ideals.




Here $k$ is a field.







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    A finitely generated k-algebra is a finite extension of a polynomial ring over k and this has infinitely many irreducible non-associate polynomials.
    – user26857
    Aug 24 at 12:48















up vote
3
down vote

favorite












I am looking for a reference or a quick proof of the statement that




Every finitely generated $k$-algebra of Krull-dimension one has infinitely many prime ideals.




Here $k$ is a field.







share|cite|improve this question
















  • 2




    A finitely generated k-algebra is a finite extension of a polynomial ring over k and this has infinitely many irreducible non-associate polynomials.
    – user26857
    Aug 24 at 12:48













up vote
3
down vote

favorite









up vote
3
down vote

favorite











I am looking for a reference or a quick proof of the statement that




Every finitely generated $k$-algebra of Krull-dimension one has infinitely many prime ideals.




Here $k$ is a field.







share|cite|improve this question












I am looking for a reference or a quick proof of the statement that




Every finitely generated $k$-algebra of Krull-dimension one has infinitely many prime ideals.




Here $k$ is a field.









share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Aug 24 at 10:05









windsheaf

503311




503311







  • 2




    A finitely generated k-algebra is a finite extension of a polynomial ring over k and this has infinitely many irreducible non-associate polynomials.
    – user26857
    Aug 24 at 12:48













  • 2




    A finitely generated k-algebra is a finite extension of a polynomial ring over k and this has infinitely many irreducible non-associate polynomials.
    – user26857
    Aug 24 at 12:48








2




2




A finitely generated k-algebra is a finite extension of a polynomial ring over k and this has infinitely many irreducible non-associate polynomials.
– user26857
Aug 24 at 12:48





A finitely generated k-algebra is a finite extension of a polynomial ring over k and this has infinitely many irreducible non-associate polynomials.
– user26857
Aug 24 at 12:48











1 Answer
1






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oldest

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4
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accepted










Expanding user26857's comment into an answer:



Let $R$ be a finitely generated $k$-algebra of dimension at least one, $k$ an arbitrary field. (The result does not require the dimension 1 hypothesis but does require that $R$ not be dimension zero.)



By Noether normalization, $R$ has a subring $S$ such that (a) $S$ is a polynomial ring (i.e. generated over $k$ by a finite number of algebraically independent elements), and (b) $R$ is finite over $S$. Because the dimension of $R$ is at least one, $S$ has at least one indeterminate.



(b) implies $R$ is integral over $S$, and the lying-over theorem then implies $R$ has at least one prime ideal for each prime ideal of $S$.



Thus the problem is reduced to proving the polynomial ring $S$ (with at least one indeterminate) has infinitely many prime ideals. Since any irreducible polynomial generates a prime ideal, and no two distinct monic polynomials can generate the same ideal, it will be sufficient to exhibit infinitely many irreducible monic polynomials.



At this point, Euclid's proof of the infinitude of the primes can be reappropriated to complete the proof. Given any finite list of monic irreducible polynomials $f_1,dots,f_r$, the polynomial $f_1dots f_r + 1$ is also monic, and it cannot have an irreducible factor already in the list $f_1,dots,f_r$. Thus, any of its monic irreducible factors will extend the list. Thus there are infinitely many distinct monic irreducible polynomials in any polynomial ring, and therefore infinitely many prime ideals.






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    1 Answer
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    active

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    1 Answer
    1






    active

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    active

    oldest

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    active

    oldest

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    up vote
    4
    down vote



    accepted










    Expanding user26857's comment into an answer:



    Let $R$ be a finitely generated $k$-algebra of dimension at least one, $k$ an arbitrary field. (The result does not require the dimension 1 hypothesis but does require that $R$ not be dimension zero.)



    By Noether normalization, $R$ has a subring $S$ such that (a) $S$ is a polynomial ring (i.e. generated over $k$ by a finite number of algebraically independent elements), and (b) $R$ is finite over $S$. Because the dimension of $R$ is at least one, $S$ has at least one indeterminate.



    (b) implies $R$ is integral over $S$, and the lying-over theorem then implies $R$ has at least one prime ideal for each prime ideal of $S$.



    Thus the problem is reduced to proving the polynomial ring $S$ (with at least one indeterminate) has infinitely many prime ideals. Since any irreducible polynomial generates a prime ideal, and no two distinct monic polynomials can generate the same ideal, it will be sufficient to exhibit infinitely many irreducible monic polynomials.



    At this point, Euclid's proof of the infinitude of the primes can be reappropriated to complete the proof. Given any finite list of monic irreducible polynomials $f_1,dots,f_r$, the polynomial $f_1dots f_r + 1$ is also monic, and it cannot have an irreducible factor already in the list $f_1,dots,f_r$. Thus, any of its monic irreducible factors will extend the list. Thus there are infinitely many distinct monic irreducible polynomials in any polynomial ring, and therefore infinitely many prime ideals.






    share|cite|improve this answer
























      up vote
      4
      down vote



      accepted










      Expanding user26857's comment into an answer:



      Let $R$ be a finitely generated $k$-algebra of dimension at least one, $k$ an arbitrary field. (The result does not require the dimension 1 hypothesis but does require that $R$ not be dimension zero.)



      By Noether normalization, $R$ has a subring $S$ such that (a) $S$ is a polynomial ring (i.e. generated over $k$ by a finite number of algebraically independent elements), and (b) $R$ is finite over $S$. Because the dimension of $R$ is at least one, $S$ has at least one indeterminate.



      (b) implies $R$ is integral over $S$, and the lying-over theorem then implies $R$ has at least one prime ideal for each prime ideal of $S$.



      Thus the problem is reduced to proving the polynomial ring $S$ (with at least one indeterminate) has infinitely many prime ideals. Since any irreducible polynomial generates a prime ideal, and no two distinct monic polynomials can generate the same ideal, it will be sufficient to exhibit infinitely many irreducible monic polynomials.



      At this point, Euclid's proof of the infinitude of the primes can be reappropriated to complete the proof. Given any finite list of monic irreducible polynomials $f_1,dots,f_r$, the polynomial $f_1dots f_r + 1$ is also monic, and it cannot have an irreducible factor already in the list $f_1,dots,f_r$. Thus, any of its monic irreducible factors will extend the list. Thus there are infinitely many distinct monic irreducible polynomials in any polynomial ring, and therefore infinitely many prime ideals.






      share|cite|improve this answer






















        up vote
        4
        down vote



        accepted







        up vote
        4
        down vote



        accepted






        Expanding user26857's comment into an answer:



        Let $R$ be a finitely generated $k$-algebra of dimension at least one, $k$ an arbitrary field. (The result does not require the dimension 1 hypothesis but does require that $R$ not be dimension zero.)



        By Noether normalization, $R$ has a subring $S$ such that (a) $S$ is a polynomial ring (i.e. generated over $k$ by a finite number of algebraically independent elements), and (b) $R$ is finite over $S$. Because the dimension of $R$ is at least one, $S$ has at least one indeterminate.



        (b) implies $R$ is integral over $S$, and the lying-over theorem then implies $R$ has at least one prime ideal for each prime ideal of $S$.



        Thus the problem is reduced to proving the polynomial ring $S$ (with at least one indeterminate) has infinitely many prime ideals. Since any irreducible polynomial generates a prime ideal, and no two distinct monic polynomials can generate the same ideal, it will be sufficient to exhibit infinitely many irreducible monic polynomials.



        At this point, Euclid's proof of the infinitude of the primes can be reappropriated to complete the proof. Given any finite list of monic irreducible polynomials $f_1,dots,f_r$, the polynomial $f_1dots f_r + 1$ is also monic, and it cannot have an irreducible factor already in the list $f_1,dots,f_r$. Thus, any of its monic irreducible factors will extend the list. Thus there are infinitely many distinct monic irreducible polynomials in any polynomial ring, and therefore infinitely many prime ideals.






        share|cite|improve this answer












        Expanding user26857's comment into an answer:



        Let $R$ be a finitely generated $k$-algebra of dimension at least one, $k$ an arbitrary field. (The result does not require the dimension 1 hypothesis but does require that $R$ not be dimension zero.)



        By Noether normalization, $R$ has a subring $S$ such that (a) $S$ is a polynomial ring (i.e. generated over $k$ by a finite number of algebraically independent elements), and (b) $R$ is finite over $S$. Because the dimension of $R$ is at least one, $S$ has at least one indeterminate.



        (b) implies $R$ is integral over $S$, and the lying-over theorem then implies $R$ has at least one prime ideal for each prime ideal of $S$.



        Thus the problem is reduced to proving the polynomial ring $S$ (with at least one indeterminate) has infinitely many prime ideals. Since any irreducible polynomial generates a prime ideal, and no two distinct monic polynomials can generate the same ideal, it will be sufficient to exhibit infinitely many irreducible monic polynomials.



        At this point, Euclid's proof of the infinitude of the primes can be reappropriated to complete the proof. Given any finite list of monic irreducible polynomials $f_1,dots,f_r$, the polynomial $f_1dots f_r + 1$ is also monic, and it cannot have an irreducible factor already in the list $f_1,dots,f_r$. Thus, any of its monic irreducible factors will extend the list. Thus there are infinitely many distinct monic irreducible polynomials in any polynomial ring, and therefore infinitely many prime ideals.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Aug 24 at 18:17









        Ben Blum-Smith

        9,68822981




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