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.
abstract-algebra algebraic-geometry reference-request commutative-algebra
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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.
abstract-algebra algebraic-geometry reference-request commutative-algebra
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.
â user26857
Aug 24 at 12:48
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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.
abstract-algebra algebraic-geometry reference-request commutative-algebra
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.
abstract-algebra algebraic-geometry reference-request commutative-algebra
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.
â user26857
Aug 24 at 12:48
add a comment |Â
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.
â 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.
â 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.
â user26857
Aug 24 at 12:48
add a comment |Â
1 Answer
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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.
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
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.
add a comment |Â
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.
add a comment |Â
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.
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.
answered Aug 24 at 18:17
Ben Blum-Smith
9,68822981
9,68822981
add a comment |Â
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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.
â user26857
Aug 24 at 12:48