$K[X^2,X^3]subset K[X]$ is a Noetherian domain and all its prime ideals are maximal
$begingroup$
Consider $K$ field and consider the ring $R=K[X^2,X^3]subset K[X]$. It is clear that $R$ is not a Dedekind domain, since with the element $X$ one immediately see that it is not integrally closed. But $R$ is a Noetherian domain and every non trivial prime ideal is maximal. I have a proof for the last two part, but it is a bit heavy and maybe someone can help me finding a better proof!
Thanks in advance!
abstract-algebra algebraic-geometry commutative-algebra noetherian krull-dimension
edited Jan 23 at 17:21
user26857
39.3k124183
asked Jan 22 at 14:43
Lei FeimaLei Feima
867
$endgroup$
|
show 6 more comments
$begingroup$
Consider $K$ field and consider the ring $R=K[X^2,X^3]subset K[X]$. It is clear that $R$ is not a Dedekind domain, since with the element $X$ one immediately see that it is not integrally closed. But $R$ is a Noetherian domain and every non trivial prime ideal is maximal. I have a proof for the last two part, but it is a bit heavy and maybe someone can help me finding a better proof!
Thanks in advance!
abstract-algebra algebraic-geometry commutative-algebra noetherian krull-dimension
edited Jan 23 at 17:21
user26857
39.3k124183
asked Jan 22 at 14:43
Lei FeimaLei Feima
867
$endgroup$
2
$begingroup$
Have you tried considering $R = K[X^2, X^3] cong frac{K[Y, Z]}{(Y^3 - Z^2)}$?
$endgroup$
– André 3000
Jan 22 at 16:25
1
$begingroup$
@André3000 so the polynomial is irreducible and the ideal generated by it is prime. Moreover, $K[Y,Z]$ should be Noetherian by Hilbert's Basissatz, so the quotient is Noetherian. Right? The dimension $1$ (i.e., all its prime ideal are maximal) is not so evident for me...
$endgroup$
– Lei Feima
Jan 22 at 16:35
2
$begingroup$
@LeiFeima: $langle Y^{3}-Z^{2} rangle$ is a prime ideal of $K[Y, Z]$, and any proper chain of prime ideals of $K[Y, Z]/langle Y^{3}-Z^{2} rangle$ lifts to a chain of prime ideals of $K[Y, Z]$ which begins $0 subset langle Y^{3}-Z^{2} rangle subset cdots$. Since $K[Y, Z]$ has Krull dimension $2$, it follows that $K[Y, Z]/langle Y^{3}-Z^{2} rangle$ has Krull dimension at most $1$. One can see that it has Krull dimension exactly $1$ by (e.g.) considering the projection of the maximal ideal $langle Y, Zrangle$ of $K[Y, Z]$.
$endgroup$
– Alex Wertheim
Jan 22 at 18:37
3
$begingroup$
I also have an alternative approach for showing that $K[X^{2}, X^{3}]$ has dimension $1$ which only uses basic facts on integrality, if you're interested.
$endgroup$
– Alex Wertheim
Jan 22 at 18:39
1
$begingroup$
@LeiFeima It depends on how elementary you'd like your proof to be. For instance, $R$ is the coordinate ring of a curve, so its dimension is $1$. Or since the ideal $(Y^3 - Z^2)$ is principal, it has codimension $1$ by Krull's Hauptidealsatz. Or as Alex Wertheim suggests, $R$ is an integral (quadratic) extension of $K[Y]$. Integral extensions preserve dimension by the Cohen-Seidenberg theorems, so $R$ has the same dimension as $K[Y]$. I suspect you may be able to reproduce this result for integral quadratic extensions by hand, if you don't want to appeal to these theorems.
$endgroup$
– André 3000
Jan 22 at 22:26
|
show 6 more comments
$begingroup$
Consider $K$ field and consider the ring $R=K[X^2,X^3]subset K[X]$. It is clear that $R$ is not a Dedekind domain, since with the element $X$ one immediately see that it is not integrally closed. But $R$ is a Noetherian domain and every non trivial prime ideal is maximal. I have a proof for the last two part, but it is a bit heavy and maybe someone can help me finding a better proof!
Thanks in advance!
abstract-algebra algebraic-geometry commutative-algebra noetherian krull-dimension
edited Jan 23 at 17:21
user26857
39.3k124183
asked Jan 22 at 14:43
Lei FeimaLei Feima
867
$endgroup$
Consider $K$ field and consider the ring $R=K[X^2,X^3]subset K[X]$. It is clear that $R$ is not a Dedekind domain, since with the element $X$ one immediately see that it is not integrally closed. But $R$ is a Noetherian domain and every non trivial prime ideal is maximal. I have a proof for the last two part, but it is a bit heavy and maybe someone can help me finding a better proof!
Thanks in advance!
abstract-algebra algebraic-geometry commutative-algebra noetherian krull-dimension
abstract-algebra algebraic-geometry commutative-algebra noetherian krull-dimension
edited Jan 23 at 17:21
user26857
39.3k124183
asked Jan 22 at 14:43
Lei FeimaLei Feima
867
edited Jan 23 at 17:21
user26857
39.3k124183
asked Jan 22 at 14:43
Lei FeimaLei Feima
867
edited Jan 23 at 17:21
user26857
39.3k124183
edited Jan 23 at 17:21
user26857
39.3k124183
edited Jan 23 at 17:21
user26857
39.3k124183
39.3k124183
asked Jan 22 at 14:43
Lei FeimaLei Feima
867
asked Jan 22 at 14:43
Lei FeimaLei Feima
867
asked Jan 22 at 14:43
Lei FeimaLei Feima
867
867
2
$begingroup$
Have you tried considering $R = K[X^2, X^3] cong frac{K[Y, Z]}{(Y^3 - Z^2)}$?
$endgroup$
– André 3000
Jan 22 at 16:25
1
$begingroup$
@André3000 so the polynomial is irreducible and the ideal generated by it is prime. Moreover, $K[Y,Z]$ should be Noetherian by Hilbert's Basissatz, so the quotient is Noetherian. Right? The dimension $1$ (i.e., all its prime ideal are maximal) is not so evident for me...
$endgroup$
– Lei Feima
Jan 22 at 16:35
2
$begingroup$
@LeiFeima: $langle Y^{3}-Z^{2} rangle$ is a prime ideal of $K[Y, Z]$, and any proper chain of prime ideals of $K[Y, Z]/langle Y^{3}-Z^{2} rangle$ lifts to a chain of prime ideals of $K[Y, Z]$ which begins $0 subset langle Y^{3}-Z^{2} rangle subset cdots$. Since $K[Y, Z]$ has Krull dimension $2$, it follows that $K[Y, Z]/langle Y^{3}-Z^{2} rangle$ has Krull dimension at most $1$. One can see that it has Krull dimension exactly $1$ by (e.g.) considering the projection of the maximal ideal $langle Y, Zrangle$ of $K[Y, Z]$.
$endgroup$
– Alex Wertheim
Jan 22 at 18:37
3
$begingroup$
I also have an alternative approach for showing that $K[X^{2}, X^{3}]$ has dimension $1$ which only uses basic facts on integrality, if you're interested.
$endgroup$
– Alex Wertheim
Jan 22 at 18:39
1
$begingroup$
@LeiFeima It depends on how elementary you'd like your proof to be. For instance, $R$ is the coordinate ring of a curve, so its dimension is $1$. Or since the ideal $(Y^3 - Z^2)$ is principal, it has codimension $1$ by Krull's Hauptidealsatz. Or as Alex Wertheim suggests, $R$ is an integral (quadratic) extension of $K[Y]$. Integral extensions preserve dimension by the Cohen-Seidenberg theorems, so $R$ has the same dimension as $K[Y]$. I suspect you may be able to reproduce this result for integral quadratic extensions by hand, if you don't want to appeal to these theorems.
$endgroup$
– André 3000
Jan 22 at 22:26
|
show 6 more comments
2
$begingroup$
Have you tried considering $R = K[X^2, X^3] cong frac{K[Y, Z]}{(Y^3 - Z^2)}$?
$endgroup$
– André 3000
Jan 22 at 16:25
1
$begingroup$
@André3000 so the polynomial is irreducible and the ideal generated by it is prime. Moreover, $K[Y,Z]$ should be Noetherian by Hilbert's Basissatz, so the quotient is Noetherian. Right? The dimension $1$ (i.e., all its prime ideal are maximal) is not so evident for me...
$endgroup$
– Lei Feima
Jan 22 at 16:35
2
$begingroup$
@LeiFeima: $langle Y^{3}-Z^{2} rangle$ is a prime ideal of $K[Y, Z]$, and any proper chain of prime ideals of $K[Y, Z]/langle Y^{3}-Z^{2} rangle$ lifts to a chain of prime ideals of $K[Y, Z]$ which begins $0 subset langle Y^{3}-Z^{2} rangle subset cdots$. Since $K[Y, Z]$ has Krull dimension $2$, it follows that $K[Y, Z]/langle Y^{3}-Z^{2} rangle$ has Krull dimension at most $1$. One can see that it has Krull dimension exactly $1$ by (e.g.) considering the projection of the maximal ideal $langle Y, Zrangle$ of $K[Y, Z]$.
$endgroup$
– Alex Wertheim
Jan 22 at 18:37
3
$begingroup$
I also have an alternative approach for showing that $K[X^{2}, X^{3}]$ has dimension $1$ which only uses basic facts on integrality, if you're interested.
$endgroup$
– Alex Wertheim
Jan 22 at 18:39
1
$begingroup$
@LeiFeima It depends on how elementary you'd like your proof to be. For instance, $R$ is the coordinate ring of a curve, so its dimension is $1$. Or since the ideal $(Y^3 - Z^2)$ is principal, it has codimension $1$ by Krull's Hauptidealsatz. Or as Alex Wertheim suggests, $R$ is an integral (quadratic) extension of $K[Y]$. Integral extensions preserve dimension by the Cohen-Seidenberg theorems, so $R$ has the same dimension as $K[Y]$. I suspect you may be able to reproduce this result for integral quadratic extensions by hand, if you don't want to appeal to these theorems.
$endgroup$
– André 3000
Jan 22 at 22:26
2
2
$begingroup$
Have you tried considering $R = K[X^2, X^3] cong frac{K[Y, Z]}{(Y^3 - Z^2)}$?
$endgroup$
– André 3000
Jan 22 at 16:25
$begingroup$
Have you tried considering $R = K[X^2, X^3] cong frac{K[Y, Z]}{(Y^3 - Z^2)}$?
$endgroup$
– André 3000
Jan 22 at 16:25
1
1
$begingroup$
@André3000 so the polynomial is irreducible and the ideal generated by it is prime. Moreover, $K[Y,Z]$ should be Noetherian by Hilbert's Basissatz, so the quotient is Noetherian. Right? The dimension $1$ (i.e., all its prime ideal are maximal) is not so evident for me...
$endgroup$
– Lei Feima
Jan 22 at 16:35
$begingroup$
@André3000 so the polynomial is irreducible and the ideal generated by it is prime. Moreover, $K[Y,Z]$ should be Noetherian by Hilbert's Basissatz, so the quotient is Noetherian. Right? The dimension $1$ (i.e., all its prime ideal are maximal) is not so evident for me...
$endgroup$
– Lei Feima
Jan 22 at 16:35
2
2
$begingroup$
@LeiFeima: $langle Y^{3}-Z^{2} rangle$ is a prime ideal of $K[Y, Z]$, and any proper chain of prime ideals of $K[Y, Z]/langle Y^{3}-Z^{2} rangle$ lifts to a chain of prime ideals of $K[Y, Z]$ which begins $0 subset langle Y^{3}-Z^{2} rangle subset cdots$. Since $K[Y, Z]$ has Krull dimension $2$, it follows that $K[Y, Z]/langle Y^{3}-Z^{2} rangle$ has Krull dimension at most $1$. One can see that it has Krull dimension exactly $1$ by (e.g.) considering the projection of the maximal ideal $langle Y, Zrangle$ of $K[Y, Z]$.
$endgroup$
– Alex Wertheim
Jan 22 at 18:37
$begingroup$
@LeiFeima: $langle Y^{3}-Z^{2} rangle$ is a prime ideal of $K[Y, Z]$, and any proper chain of prime ideals of $K[Y, Z]/langle Y^{3}-Z^{2} rangle$ lifts to a chain of prime ideals of $K[Y, Z]$ which begins $0 subset langle Y^{3}-Z^{2} rangle subset cdots$. Since $K[Y, Z]$ has Krull dimension $2$, it follows that $K[Y, Z]/langle Y^{3}-Z^{2} rangle$ has Krull dimension at most $1$. One can see that it has Krull dimension exactly $1$ by (e.g.) considering the projection of the maximal ideal $langle Y, Zrangle$ of $K[Y, Z]$.
$endgroup$
– Alex Wertheim
Jan 22 at 18:37
3
3
$begingroup$
I also have an alternative approach for showing that $K[X^{2}, X^{3}]$ has dimension $1$ which only uses basic facts on integrality, if you're interested.
$endgroup$
– Alex Wertheim
Jan 22 at 18:39
$begingroup$
I also have an alternative approach for showing that $K[X^{2}, X^{3}]$ has dimension $1$ which only uses basic facts on integrality, if you're interested.
$endgroup$
– Alex Wertheim
Jan 22 at 18:39
1
1
$begingroup$
@LeiFeima It depends on how elementary you'd like your proof to be. For instance, $R$ is the coordinate ring of a curve, so its dimension is $1$. Or since the ideal $(Y^3 - Z^2)$ is principal, it has codimension $1$ by Krull's Hauptidealsatz. Or as Alex Wertheim suggests, $R$ is an integral (quadratic) extension of $K[Y]$. Integral extensions preserve dimension by the Cohen-Seidenberg theorems, so $R$ has the same dimension as $K[Y]$. I suspect you may be able to reproduce this result for integral quadratic extensions by hand, if you don't want to appeal to these theorems.
$endgroup$
– André 3000
Jan 22 at 22:26
$begingroup$
@LeiFeima It depends on how elementary you'd like your proof to be. For instance, $R$ is the coordinate ring of a curve, so its dimension is $1$. Or since the ideal $(Y^3 - Z^2)$ is principal, it has codimension $1$ by Krull's Hauptidealsatz. Or as Alex Wertheim suggests, $R$ is an integral (quadratic) extension of $K[Y]$. Integral extensions preserve dimension by the Cohen-Seidenberg theorems, so $R$ has the same dimension as $K[Y]$. I suspect you may be able to reproduce this result for integral quadratic extensions by hand, if you don't want to appeal to these theorems.
$endgroup$
– André 3000
Jan 22 at 22:26
|
show 6 more comments
2 Answers
2
active
oldest
votes
$begingroup$
Suppose $mathfrak{p}$ is a nonzero prime ideal of $A := K[X^{2}, X^{3}]$; we want to show that $mathfrak{p}$ is maximal. Note that $K[X^{2}]$ is a subring of $A$, and $K[X^{2}]$ is a principal ideal domain, since it is isomorphic to $K[X]$ via the morphism of $K$-algebras $K[X] to K[X^{2}], X mapsto X^{2}$. Note that $mathfrak{m} := mathfrak{p} cap K[X^{2}]$ is a prime ideal of $K[X^{2}]$, since it is the preimage of $mathfrak{p}$ under the inclusion morphism $K[X^{2}] hookrightarrow A$. If $mathfrak{m}$ is nonzero, then $mathfrak{m}$ is maximal, since $K[X^{2}]$ is a PID. Moreover, since the inclusion $K[X^{2}] hookrightarrow A$ is integral, so too is the induced (injective) morphism $K[X^{2}]/mathfrak{m} hookrightarrow A/mathfrak{p}$. Since $K[X^{2}]/mathfrak{m}$ is a field and $A$ is a domain, $A$ must be a field as well (this is, e.g., Proposition 5.7 in Atiyah Macdonald).
It therefore suffices to show that $mathfrak{p} cap K[X^{2}]$ is nonzero for any nonzero prime ideal $mathfrak{p}$ of $A$. This amounts to showing that any nonzero $mathfrak{p}$ contains a polynomial whose monomial terms all have even degree. Take $f(X) in mathfrak{p}$ nonzero, and write $f(X) = g(X) + h(X)$, where $g$ has monomial terms of even degree only, and $h$ has monomial terms of odd degree only. Then $f(-X) = g(X) - h(X)$, and $f(-X) in A$, so $f(-X)f(X) = g(X)^{2} - h(X)^{2} in mathfrak{p}$, which clearly has monomial terms of even degree only.
answered Jan 23 at 2:58
Alex WertheimAlex Wertheim
16.1k22848
$endgroup$
$begingroup$
Thanks a lot! This is exactly the proof that I had in mind when I posed the question, even if I showed that $pcap K[X^2]$ is nonempty in a slightly different way...
$endgroup$
– Lei Feima
Jan 23 at 9:00
1
$begingroup$
@LeiFeima: great minds think alike. :) I suppose it's good to have it recorded here, if nothing else.
$endgroup$
– Alex Wertheim
Jan 23 at 9:04
$begingroup$
Just for the sake of completeness and clarity: I considered the $p$ prime ideal of $R$ and $p':=pcap K[X^3]$. Since $R$ is integral over $K[X^3]$, I consider a nonzero $cin p$ such that, for $b_i in K[X^3]$ with $b_0 neq 0$, $c^n+c^{n-1}b_{n-1}+...+b_0 =0$. Then $c(c^{n-1}+b_{n-1}c^{n-2}+...+b_1)=-b_0 in p'$.
$endgroup$
– Lei Feima
Jan 23 at 14:05
add a comment |
$begingroup$
Merely because $K[X]$ is an integral extension of $K[X^2,X^3]$.
More precisely, the going up theorem ensures any prime of $K[X^2,X^3]$ is a contraction of $K[X]$, and you must know that an ideal $mathfrak{m}subseteq K[X]$ is maximal if and only if $mathfrak{m}cap K[X^2,X^3]subseteq K[X^2,X^3]$ is maximal.
answered Jan 23 at 3:52
Cube BearCube Bear
597211
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
Suppose $mathfrak{p}$ is a nonzero prime ideal of $A := K[X^{2}, X^{3}]$; we want to show that $mathfrak{p}$ is maximal. Note that $K[X^{2}]$ is a subring of $A$, and $K[X^{2}]$ is a principal ideal domain, since it is isomorphic to $K[X]$ via the morphism of $K$-algebras $K[X] to K[X^{2}], X mapsto X^{2}$. Note that $mathfrak{m} := mathfrak{p} cap K[X^{2}]$ is a prime ideal of $K[X^{2}]$, since it is the preimage of $mathfrak{p}$ under the inclusion morphism $K[X^{2}] hookrightarrow A$. If $mathfrak{m}$ is nonzero, then $mathfrak{m}$ is maximal, since $K[X^{2}]$ is a PID. Moreover, since the inclusion $K[X^{2}] hookrightarrow A$ is integral, so too is the induced (injective) morphism $K[X^{2}]/mathfrak{m} hookrightarrow A/mathfrak{p}$. Since $K[X^{2}]/mathfrak{m}$ is a field and $A$ is a domain, $A$ must be a field as well (this is, e.g., Proposition 5.7 in Atiyah Macdonald).
It therefore suffices to show that $mathfrak{p} cap K[X^{2}]$ is nonzero for any nonzero prime ideal $mathfrak{p}$ of $A$. This amounts to showing that any nonzero $mathfrak{p}$ contains a polynomial whose monomial terms all have even degree. Take $f(X) in mathfrak{p}$ nonzero, and write $f(X) = g(X) + h(X)$, where $g$ has monomial terms of even degree only, and $h$ has monomial terms of odd degree only. Then $f(-X) = g(X) - h(X)$, and $f(-X) in A$, so $f(-X)f(X) = g(X)^{2} - h(X)^{2} in mathfrak{p}$, which clearly has monomial terms of even degree only.
answered Jan 23 at 2:58
Alex WertheimAlex Wertheim
16.1k22848
$endgroup$
$begingroup$
Thanks a lot! This is exactly the proof that I had in mind when I posed the question, even if I showed that $pcap K[X^2]$ is nonempty in a slightly different way...
$endgroup$
– Lei Feima
Jan 23 at 9:00
1
$begingroup$
@LeiFeima: great minds think alike. :) I suppose it's good to have it recorded here, if nothing else.
$endgroup$
– Alex Wertheim
Jan 23 at 9:04
$begingroup$
Just for the sake of completeness and clarity: I considered the $p$ prime ideal of $R$ and $p':=pcap K[X^3]$. Since $R$ is integral over $K[X^3]$, I consider a nonzero $cin p$ such that, for $b_i in K[X^3]$ with $b_0 neq 0$, $c^n+c^{n-1}b_{n-1}+...+b_0 =0$. Then $c(c^{n-1}+b_{n-1}c^{n-2}+...+b_1)=-b_0 in p'$.
$endgroup$
– Lei Feima
Jan 23 at 14:05
add a comment |
$begingroup$
Suppose $mathfrak{p}$ is a nonzero prime ideal of $A := K[X^{2}, X^{3}]$; we want to show that $mathfrak{p}$ is maximal. Note that $K[X^{2}]$ is a subring of $A$, and $K[X^{2}]$ is a principal ideal domain, since it is isomorphic to $K[X]$ via the morphism of $K$-algebras $K[X] to K[X^{2}], X mapsto X^{2}$. Note that $mathfrak{m} := mathfrak{p} cap K[X^{2}]$ is a prime ideal of $K[X^{2}]$, since it is the preimage of $mathfrak{p}$ under the inclusion morphism $K[X^{2}] hookrightarrow A$. If $mathfrak{m}$ is nonzero, then $mathfrak{m}$ is maximal, since $K[X^{2}]$ is a PID. Moreover, since the inclusion $K[X^{2}] hookrightarrow A$ is integral, so too is the induced (injective) morphism $K[X^{2}]/mathfrak{m} hookrightarrow A/mathfrak{p}$. Since $K[X^{2}]/mathfrak{m}$ is a field and $A$ is a domain, $A$ must be a field as well (this is, e.g., Proposition 5.7 in Atiyah Macdonald).
It therefore suffices to show that $mathfrak{p} cap K[X^{2}]$ is nonzero for any nonzero prime ideal $mathfrak{p}$ of $A$. This amounts to showing that any nonzero $mathfrak{p}$ contains a polynomial whose monomial terms all have even degree. Take $f(X) in mathfrak{p}$ nonzero, and write $f(X) = g(X) + h(X)$, where $g$ has monomial terms of even degree only, and $h$ has monomial terms of odd degree only. Then $f(-X) = g(X) - h(X)$, and $f(-X) in A$, so $f(-X)f(X) = g(X)^{2} - h(X)^{2} in mathfrak{p}$, which clearly has monomial terms of even degree only.
answered Jan 23 at 2:58
Alex WertheimAlex Wertheim
16.1k22848
$endgroup$
$begingroup$
Thanks a lot! This is exactly the proof that I had in mind when I posed the question, even if I showed that $pcap K[X^2]$ is nonempty in a slightly different way...
$endgroup$
– Lei Feima
Jan 23 at 9:00
1
$begingroup$
@LeiFeima: great minds think alike. :) I suppose it's good to have it recorded here, if nothing else.
$endgroup$
– Alex Wertheim
Jan 23 at 9:04
$begingroup$
Just for the sake of completeness and clarity: I considered the $p$ prime ideal of $R$ and $p':=pcap K[X^3]$. Since $R$ is integral over $K[X^3]$, I consider a nonzero $cin p$ such that, for $b_i in K[X^3]$ with $b_0 neq 0$, $c^n+c^{n-1}b_{n-1}+...+b_0 =0$. Then $c(c^{n-1}+b_{n-1}c^{n-2}+...+b_1)=-b_0 in p'$.
$endgroup$
– Lei Feima
Jan 23 at 14:05
add a comment |
$begingroup$
Suppose $mathfrak{p}$ is a nonzero prime ideal of $A := K[X^{2}, X^{3}]$; we want to show that $mathfrak{p}$ is maximal. Note that $K[X^{2}]$ is a subring of $A$, and $K[X^{2}]$ is a principal ideal domain, since it is isomorphic to $K[X]$ via the morphism of $K$-algebras $K[X] to K[X^{2}], X mapsto X^{2}$. Note that $mathfrak{m} := mathfrak{p} cap K[X^{2}]$ is a prime ideal of $K[X^{2}]$, since it is the preimage of $mathfrak{p}$ under the inclusion morphism $K[X^{2}] hookrightarrow A$. If $mathfrak{m}$ is nonzero, then $mathfrak{m}$ is maximal, since $K[X^{2}]$ is a PID. Moreover, since the inclusion $K[X^{2}] hookrightarrow A$ is integral, so too is the induced (injective) morphism $K[X^{2}]/mathfrak{m} hookrightarrow A/mathfrak{p}$. Since $K[X^{2}]/mathfrak{m}$ is a field and $A$ is a domain, $A$ must be a field as well (this is, e.g., Proposition 5.7 in Atiyah Macdonald).
It therefore suffices to show that $mathfrak{p} cap K[X^{2}]$ is nonzero for any nonzero prime ideal $mathfrak{p}$ of $A$. This amounts to showing that any nonzero $mathfrak{p}$ contains a polynomial whose monomial terms all have even degree. Take $f(X) in mathfrak{p}$ nonzero, and write $f(X) = g(X) + h(X)$, where $g$ has monomial terms of even degree only, and $h$ has monomial terms of odd degree only. Then $f(-X) = g(X) - h(X)$, and $f(-X) in A$, so $f(-X)f(X) = g(X)^{2} - h(X)^{2} in mathfrak{p}$, which clearly has monomial terms of even degree only.
answered Jan 23 at 2:58
Alex WertheimAlex Wertheim
16.1k22848
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Suppose $mathfrak{p}$ is a nonzero prime ideal of $A := K[X^{2}, X^{3}]$; we want to show that $mathfrak{p}$ is maximal. Note that $K[X^{2}]$ is a subring of $A$, and $K[X^{2}]$ is a principal ideal domain, since it is isomorphic to $K[X]$ via the morphism of $K$-algebras $K[X] to K[X^{2}], X mapsto X^{2}$. Note that $mathfrak{m} := mathfrak{p} cap K[X^{2}]$ is a prime ideal of $K[X^{2}]$, since it is the preimage of $mathfrak{p}$ under the inclusion morphism $K[X^{2}] hookrightarrow A$. If $mathfrak{m}$ is nonzero, then $mathfrak{m}$ is maximal, since $K[X^{2}]$ is a PID. Moreover, since the inclusion $K[X^{2}] hookrightarrow A$ is integral, so too is the induced (injective) morphism $K[X^{2}]/mathfrak{m} hookrightarrow A/mathfrak{p}$. Since $K[X^{2}]/mathfrak{m}$ is a field and $A$ is a domain, $A$ must be a field as well (this is, e.g., Proposition 5.7 in Atiyah Macdonald).
It therefore suffices to show that $mathfrak{p} cap K[X^{2}]$ is nonzero for any nonzero prime ideal $mathfrak{p}$ of $A$. This amounts to showing that any nonzero $mathfrak{p}$ contains a polynomial whose monomial terms all have even degree. Take $f(X) in mathfrak{p}$ nonzero, and write $f(X) = g(X) + h(X)$, where $g$ has monomial terms of even degree only, and $h$ has monomial terms of odd degree only. Then $f(-X) = g(X) - h(X)$, and $f(-X) in A$, so $f(-X)f(X) = g(X)^{2} - h(X)^{2} in mathfrak{p}$, which clearly has monomial terms of even degree only.
answered Jan 23 at 2:58
Alex WertheimAlex Wertheim
16.1k22848
answered Jan 23 at 2:58
Alex WertheimAlex Wertheim
16.1k22848
answered Jan 23 at 2:58
Alex WertheimAlex Wertheim
16.1k22848
answered Jan 23 at 2:58
Alex WertheimAlex Wertheim
16.1k22848
16.1k22848
$begingroup$
Thanks a lot! This is exactly the proof that I had in mind when I posed the question, even if I showed that $pcap K[X^2]$ is nonempty in a slightly different way...
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– Lei Feima
Jan 23 at 9:00
1
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@LeiFeima: great minds think alike. :) I suppose it's good to have it recorded here, if nothing else.
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– Alex Wertheim
Jan 23 at 9:04
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Just for the sake of completeness and clarity: I considered the $p$ prime ideal of $R$ and $p':=pcap K[X^3]$. Since $R$ is integral over $K[X^3]$, I consider a nonzero $cin p$ such that, for $b_i in K[X^3]$ with $b_0 neq 0$, $c^n+c^{n-1}b_{n-1}+...+b_0 =0$. Then $c(c^{n-1}+b_{n-1}c^{n-2}+...+b_1)=-b_0 in p'$.
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– Lei Feima
Jan 23 at 14:05
add a comment |
$begingroup$
Thanks a lot! This is exactly the proof that I had in mind when I posed the question, even if I showed that $pcap K[X^2]$ is nonempty in a slightly different way...
$endgroup$
– Lei Feima
Jan 23 at 9:00
1
$begingroup$
@LeiFeima: great minds think alike. :) I suppose it's good to have it recorded here, if nothing else.
$endgroup$
– Alex Wertheim
Jan 23 at 9:04
$begingroup$
Just for the sake of completeness and clarity: I considered the $p$ prime ideal of $R$ and $p':=pcap K[X^3]$. Since $R$ is integral over $K[X^3]$, I consider a nonzero $cin p$ such that, for $b_i in K[X^3]$ with $b_0 neq 0$, $c^n+c^{n-1}b_{n-1}+...+b_0 =0$. Then $c(c^{n-1}+b_{n-1}c^{n-2}+...+b_1)=-b_0 in p'$.
$endgroup$
– Lei Feima
Jan 23 at 14:05
$begingroup$
Thanks a lot! This is exactly the proof that I had in mind when I posed the question, even if I showed that $pcap K[X^2]$ is nonempty in a slightly different way...
$endgroup$
– Lei Feima
Jan 23 at 9:00
$begingroup$
Thanks a lot! This is exactly the proof that I had in mind when I posed the question, even if I showed that $pcap K[X^2]$ is nonempty in a slightly different way...
$endgroup$
– Lei Feima
Jan 23 at 9:00
1
1
$begingroup$
@LeiFeima: great minds think alike. :) I suppose it's good to have it recorded here, if nothing else.
$endgroup$
– Alex Wertheim
Jan 23 at 9:04
$begingroup$
@LeiFeima: great minds think alike. :) I suppose it's good to have it recorded here, if nothing else.
$endgroup$
– Alex Wertheim
Jan 23 at 9:04
$begingroup$
Just for the sake of completeness and clarity: I considered the $p$ prime ideal of $R$ and $p':=pcap K[X^3]$. Since $R$ is integral over $K[X^3]$, I consider a nonzero $cin p$ such that, for $b_i in K[X^3]$ with $b_0 neq 0$, $c^n+c^{n-1}b_{n-1}+...+b_0 =0$. Then $c(c^{n-1}+b_{n-1}c^{n-2}+...+b_1)=-b_0 in p'$.
$endgroup$
– Lei Feima
Jan 23 at 14:05
$begingroup$
Just for the sake of completeness and clarity: I considered the $p$ prime ideal of $R$ and $p':=pcap K[X^3]$. Since $R$ is integral over $K[X^3]$, I consider a nonzero $cin p$ such that, for $b_i in K[X^3]$ with $b_0 neq 0$, $c^n+c^{n-1}b_{n-1}+...+b_0 =0$. Then $c(c^{n-1}+b_{n-1}c^{n-2}+...+b_1)=-b_0 in p'$.
$endgroup$
– Lei Feima
Jan 23 at 14:05
add a comment |
$begingroup$
Merely because $K[X]$ is an integral extension of $K[X^2,X^3]$.
More precisely, the going up theorem ensures any prime of $K[X^2,X^3]$ is a contraction of $K[X]$, and you must know that an ideal $mathfrak{m}subseteq K[X]$ is maximal if and only if $mathfrak{m}cap K[X^2,X^3]subseteq K[X^2,X^3]$ is maximal.
answered Jan 23 at 3:52
Cube BearCube Bear
597211
$endgroup$
add a comment |
$begingroup$
Merely because $K[X]$ is an integral extension of $K[X^2,X^3]$.
More precisely, the going up theorem ensures any prime of $K[X^2,X^3]$ is a contraction of $K[X]$, and you must know that an ideal $mathfrak{m}subseteq K[X]$ is maximal if and only if $mathfrak{m}cap K[X^2,X^3]subseteq K[X^2,X^3]$ is maximal.
answered Jan 23 at 3:52
Cube BearCube Bear
597211
$endgroup$
add a comment |
$begingroup$
Merely because $K[X]$ is an integral extension of $K[X^2,X^3]$.
More precisely, the going up theorem ensures any prime of $K[X^2,X^3]$ is a contraction of $K[X]$, and you must know that an ideal $mathfrak{m}subseteq K[X]$ is maximal if and only if $mathfrak{m}cap K[X^2,X^3]subseteq K[X^2,X^3]$ is maximal.
answered Jan 23 at 3:52
Cube BearCube Bear
597211
$endgroup$
Merely because $K[X]$ is an integral extension of $K[X^2,X^3]$.
More precisely, the going up theorem ensures any prime of $K[X^2,X^3]$ is a contraction of $K[X]$, and you must know that an ideal $mathfrak{m}subseteq K[X]$ is maximal if and only if $mathfrak{m}cap K[X^2,X^3]subseteq K[X^2,X^3]$ is maximal.
answered Jan 23 at 3:52
Cube BearCube Bear
597211
answered Jan 23 at 3:52
Cube BearCube Bear
597211
answered Jan 23 at 3:52
Cube BearCube Bear
597211
answered Jan 23 at 3:52
Cube BearCube Bear
597211
597211
add a comment |
add a comment |
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$begingroup$
Have you tried considering $R = K[X^2, X^3] cong frac{K[Y, Z]}{(Y^3 - Z^2)}$?
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– André 3000
Jan 22 at 16:25
1
$begingroup$
@André3000 so the polynomial is irreducible and the ideal generated by it is prime. Moreover, $K[Y,Z]$ should be Noetherian by Hilbert's Basissatz, so the quotient is Noetherian. Right? The dimension $1$ (i.e., all its prime ideal are maximal) is not so evident for me...
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– Lei Feima
Jan 22 at 16:35
2
$begingroup$
@LeiFeima: $langle Y^{3}-Z^{2} rangle$ is a prime ideal of $K[Y, Z]$, and any proper chain of prime ideals of $K[Y, Z]/langle Y^{3}-Z^{2} rangle$ lifts to a chain of prime ideals of $K[Y, Z]$ which begins $0 subset langle Y^{3}-Z^{2} rangle subset cdots$. Since $K[Y, Z]$ has Krull dimension $2$, it follows that $K[Y, Z]/langle Y^{3}-Z^{2} rangle$ has Krull dimension at most $1$. One can see that it has Krull dimension exactly $1$ by (e.g.) considering the projection of the maximal ideal $langle Y, Zrangle$ of $K[Y, Z]$.
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– Alex Wertheim
Jan 22 at 18:37
3
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I also have an alternative approach for showing that $K[X^{2}, X^{3}]$ has dimension $1$ which only uses basic facts on integrality, if you're interested.
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– Alex Wertheim
Jan 22 at 18:39
1
$begingroup$
@LeiFeima It depends on how elementary you'd like your proof to be. For instance, $R$ is the coordinate ring of a curve, so its dimension is $1$. Or since the ideal $(Y^3 - Z^2)$ is principal, it has codimension $1$ by Krull's Hauptidealsatz. Or as Alex Wertheim suggests, $R$ is an integral (quadratic) extension of $K[Y]$. Integral extensions preserve dimension by the Cohen-Seidenberg theorems, so $R$ has the same dimension as $K[Y]$. I suspect you may be able to reproduce this result for integral quadratic extensions by hand, if you don't want to appeal to these theorems.
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– André 3000
Jan 22 at 22:26