A short proof for $dim(R[T])=dim(R)+1$?
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If $R$ is a commutative ring, it is easy to prove $dim(R[T]) geq dim(R)+1$. For noetherian $R$, we have equality. Every proof I'm aware of uses quite a bit of commutative algebra and non-trivial theorems such as Krull's intersection theorem.
Recently T. Coquand and H. Lombardi have found a surprisingly elementary "almost" first-order characterization of the Krull dimension, see here. It states:
For $l in mathbb{N}$ we have $dim(R) leq l$ if and only if for all $x_0,dotsc,x_l in R$ there are $a_0,dotsc,a_l in R$ and $m_0,ldots,m_l in mathbb{N}$ such that $x_0^{m_0} (cdots ( x_l^{m_l} (1+a_l x_l)+cdots)+a_0 x_0)=0$.
A consequence of this is a new short proof of $dim(K[x_1,dotsc,x_n])=n$, where $K$ is a field. Using Noether normalization and the fact that integral extensions don't change the dimension, it follows that $dim(Rotimes_K S)=dim(R)+dim(S)$ if $R,S$ are finitely generated $K$- algebras. In particular $dim(R[T])=dim(R)+1$. This could be useful for introductory courses on algebraic geometry which don't want to waste too much time with dimension theory.
Can we use this elementary characterization of the Krull dimension to give a new short proof of $dim(R[T])=dim(R)+1$ for noetherian commutative rings $R$?
Maybe this question is a bit naïve. I suspect that this can only work if we find a first-order property of rings which is equivalent or even weaker than Noetherian and prove the formula for these rings. Notice that in contrast to that the Gelfand-Kirillov dimension satisfies $mathrm{GK}dim(R[T])=mathrm{GK}dim(R)+1$ (see here) for every $K$-algebra $R$.
commutative-algebra ring-theory dimension-theory krull-dimension noetherian
edited Jan 15 at 9:35
El borito
664216
asked Apr 11 '13 at 15:19
Martin BrandenburgMartin Brandenburg
109k13166335
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show 9 more comments
$begingroup$
If $R$ is a commutative ring, it is easy to prove $dim(R[T]) geq dim(R)+1$. For noetherian $R$, we have equality. Every proof I'm aware of uses quite a bit of commutative algebra and non-trivial theorems such as Krull's intersection theorem.
Recently T. Coquand and H. Lombardi have found a surprisingly elementary "almost" first-order characterization of the Krull dimension, see here. It states:
For $l in mathbb{N}$ we have $dim(R) leq l$ if and only if for all $x_0,dotsc,x_l in R$ there are $a_0,dotsc,a_l in R$ and $m_0,ldots,m_l in mathbb{N}$ such that $x_0^{m_0} (cdots ( x_l^{m_l} (1+a_l x_l)+cdots)+a_0 x_0)=0$.
A consequence of this is a new short proof of $dim(K[x_1,dotsc,x_n])=n$, where $K$ is a field. Using Noether normalization and the fact that integral extensions don't change the dimension, it follows that $dim(Rotimes_K S)=dim(R)+dim(S)$ if $R,S$ are finitely generated $K$- algebras. In particular $dim(R[T])=dim(R)+1$. This could be useful for introductory courses on algebraic geometry which don't want to waste too much time with dimension theory.
Can we use this elementary characterization of the Krull dimension to give a new short proof of $dim(R[T])=dim(R)+1$ for noetherian commutative rings $R$?
Maybe this question is a bit naïve. I suspect that this can only work if we find a first-order property of rings which is equivalent or even weaker than Noetherian and prove the formula for these rings. Notice that in contrast to that the Gelfand-Kirillov dimension satisfies $mathrm{GK}dim(R[T])=mathrm{GK}dim(R)+1$ (see here) for every $K$-algebra $R$.
commutative-algebra ring-theory dimension-theory krull-dimension noetherian
edited Jan 15 at 9:35
El borito
664216
asked Apr 11 '13 at 15:19
Martin BrandenburgMartin Brandenburg
109k13166335
$endgroup$
10
$begingroup$
At first sight I can't see many hopes to do this as long as the characterization of the Krull dimension in that paper deals with elements instead of ideals, while the property of being noetherian relies on some properties of ideals. But who knows?
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– user26857
Apr 12 '13 at 8:11
4
$begingroup$
Yes. But maybe there is a first-order characterization for noetherian rings, like the one for Krull dimension? Or maybe some weaker property already suffices? I don't know.
$endgroup$
– Martin Brandenburg
Apr 12 '13 at 9:11
88
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Shall I ask this on mathoverflow?
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– Martin Brandenburg
Aug 26 '13 at 12:43
27
$begingroup$
MO copy of the question: mathoverflow.net/questions/172350/a-short-proof-for-dimrt-dimr1
$endgroup$
– Martin Sleziak
Jun 21 '14 at 19:55
5
$begingroup$
Since the question is on MO now, should it be closed here?
$endgroup$
– Martin Brandenburg
Jun 27 '14 at 18:59
|
show 9 more comments
$begingroup$
If $R$ is a commutative ring, it is easy to prove $dim(R[T]) geq dim(R)+1$. For noetherian $R$, we have equality. Every proof I'm aware of uses quite a bit of commutative algebra and non-trivial theorems such as Krull's intersection theorem.
Recently T. Coquand and H. Lombardi have found a surprisingly elementary "almost" first-order characterization of the Krull dimension, see here. It states:
For $l in mathbb{N}$ we have $dim(R) leq l$ if and only if for all $x_0,dotsc,x_l in R$ there are $a_0,dotsc,a_l in R$ and $m_0,ldots,m_l in mathbb{N}$ such that $x_0^{m_0} (cdots ( x_l^{m_l} (1+a_l x_l)+cdots)+a_0 x_0)=0$.
A consequence of this is a new short proof of $dim(K[x_1,dotsc,x_n])=n$, where $K$ is a field. Using Noether normalization and the fact that integral extensions don't change the dimension, it follows that $dim(Rotimes_K S)=dim(R)+dim(S)$ if $R,S$ are finitely generated $K$- algebras. In particular $dim(R[T])=dim(R)+1$. This could be useful for introductory courses on algebraic geometry which don't want to waste too much time with dimension theory.
Can we use this elementary characterization of the Krull dimension to give a new short proof of $dim(R[T])=dim(R)+1$ for noetherian commutative rings $R$?
Maybe this question is a bit naïve. I suspect that this can only work if we find a first-order property of rings which is equivalent or even weaker than Noetherian and prove the formula for these rings. Notice that in contrast to that the Gelfand-Kirillov dimension satisfies $mathrm{GK}dim(R[T])=mathrm{GK}dim(R)+1$ (see here) for every $K$-algebra $R$.
commutative-algebra ring-theory dimension-theory krull-dimension noetherian
edited Jan 15 at 9:35
El borito
664216
asked Apr 11 '13 at 15:19
Martin BrandenburgMartin Brandenburg
109k13166335
$endgroup$
If $R$ is a commutative ring, it is easy to prove $dim(R[T]) geq dim(R)+1$. For noetherian $R$, we have equality. Every proof I'm aware of uses quite a bit of commutative algebra and non-trivial theorems such as Krull's intersection theorem.
Recently T. Coquand and H. Lombardi have found a surprisingly elementary "almost" first-order characterization of the Krull dimension, see here. It states:
For $l in mathbb{N}$ we have $dim(R) leq l$ if and only if for all $x_0,dotsc,x_l in R$ there are $a_0,dotsc,a_l in R$ and $m_0,ldots,m_l in mathbb{N}$ such that $x_0^{m_0} (cdots ( x_l^{m_l} (1+a_l x_l)+cdots)+a_0 x_0)=0$.
A consequence of this is a new short proof of $dim(K[x_1,dotsc,x_n])=n$, where $K$ is a field. Using Noether normalization and the fact that integral extensions don't change the dimension, it follows that $dim(Rotimes_K S)=dim(R)+dim(S)$ if $R,S$ are finitely generated $K$- algebras. In particular $dim(R[T])=dim(R)+1$. This could be useful for introductory courses on algebraic geometry which don't want to waste too much time with dimension theory.
Can we use this elementary characterization of the Krull dimension to give a new short proof of $dim(R[T])=dim(R)+1$ for noetherian commutative rings $R$?
Maybe this question is a bit naïve. I suspect that this can only work if we find a first-order property of rings which is equivalent or even weaker than Noetherian and prove the formula for these rings. Notice that in contrast to that the Gelfand-Kirillov dimension satisfies $mathrm{GK}dim(R[T])=mathrm{GK}dim(R)+1$ (see here) for every $K$-algebra $R$.
commutative-algebra ring-theory dimension-theory krull-dimension noetherian
commutative-algebra ring-theory dimension-theory krull-dimension noetherian
edited Jan 15 at 9:35
El borito
664216
asked Apr 11 '13 at 15:19
Martin BrandenburgMartin Brandenburg
109k13166335
edited Jan 15 at 9:35
El borito
664216
asked Apr 11 '13 at 15:19
Martin BrandenburgMartin Brandenburg
109k13166335
edited Jan 15 at 9:35
El borito
664216
edited Jan 15 at 9:35
El borito
664216
edited Jan 15 at 9:35
El borito
664216
664216
asked Apr 11 '13 at 15:19
Martin BrandenburgMartin Brandenburg
109k13166335
asked Apr 11 '13 at 15:19
Martin BrandenburgMartin Brandenburg
109k13166335
asked Apr 11 '13 at 15:19
Martin BrandenburgMartin Brandenburg
109k13166335
109k13166335
10
$begingroup$
At first sight I can't see many hopes to do this as long as the characterization of the Krull dimension in that paper deals with elements instead of ideals, while the property of being noetherian relies on some properties of ideals. But who knows?
$endgroup$
– user26857
Apr 12 '13 at 8:11
4
$begingroup$
Yes. But maybe there is a first-order characterization for noetherian rings, like the one for Krull dimension? Or maybe some weaker property already suffices? I don't know.
$endgroup$
– Martin Brandenburg
Apr 12 '13 at 9:11
88
$begingroup$
Shall I ask this on mathoverflow?
$endgroup$
– Martin Brandenburg
Aug 26 '13 at 12:43
27
$begingroup$
MO copy of the question: mathoverflow.net/questions/172350/a-short-proof-for-dimrt-dimr1
$endgroup$
– Martin Sleziak
Jun 21 '14 at 19:55
5
$begingroup$
Since the question is on MO now, should it be closed here?
$endgroup$
– Martin Brandenburg
Jun 27 '14 at 18:59
|
show 9 more comments
10
$begingroup$
At first sight I can't see many hopes to do this as long as the characterization of the Krull dimension in that paper deals with elements instead of ideals, while the property of being noetherian relies on some properties of ideals. But who knows?
$endgroup$
– user26857
Apr 12 '13 at 8:11
4
$begingroup$
Yes. But maybe there is a first-order characterization for noetherian rings, like the one for Krull dimension? Or maybe some weaker property already suffices? I don't know.
$endgroup$
– Martin Brandenburg
Apr 12 '13 at 9:11
88
$begingroup$
Shall I ask this on mathoverflow?
$endgroup$
– Martin Brandenburg
Aug 26 '13 at 12:43
27
$begingroup$
MO copy of the question: mathoverflow.net/questions/172350/a-short-proof-for-dimrt-dimr1
$endgroup$
– Martin Sleziak
Jun 21 '14 at 19:55
5
$begingroup$
Since the question is on MO now, should it be closed here?
$endgroup$
– Martin Brandenburg
Jun 27 '14 at 18:59
10
10
$begingroup$
At first sight I can't see many hopes to do this as long as the characterization of the Krull dimension in that paper deals with elements instead of ideals, while the property of being noetherian relies on some properties of ideals. But who knows?
$endgroup$
– user26857
Apr 12 '13 at 8:11
$begingroup$
At first sight I can't see many hopes to do this as long as the characterization of the Krull dimension in that paper deals with elements instead of ideals, while the property of being noetherian relies on some properties of ideals. But who knows?
$endgroup$
– user26857
Apr 12 '13 at 8:11
4
4
$begingroup$
Yes. But maybe there is a first-order characterization for noetherian rings, like the one for Krull dimension? Or maybe some weaker property already suffices? I don't know.
$endgroup$
– Martin Brandenburg
Apr 12 '13 at 9:11
$begingroup$
Yes. But maybe there is a first-order characterization for noetherian rings, like the one for Krull dimension? Or maybe some weaker property already suffices? I don't know.
$endgroup$
– Martin Brandenburg
Apr 12 '13 at 9:11
88
88
$begingroup$
Shall I ask this on mathoverflow?
$endgroup$
– Martin Brandenburg
Aug 26 '13 at 12:43
$begingroup$
Shall I ask this on mathoverflow?
$endgroup$
– Martin Brandenburg
Aug 26 '13 at 12:43
27
27
$begingroup$
MO copy of the question: mathoverflow.net/questions/172350/a-short-proof-for-dimrt-dimr1
$endgroup$
– Martin Sleziak
Jun 21 '14 at 19:55
$begingroup$
MO copy of the question: mathoverflow.net/questions/172350/a-short-proof-for-dimrt-dimr1
$endgroup$
– Martin Sleziak
Jun 21 '14 at 19:55
5
5
$begingroup$
Since the question is on MO now, should it be closed here?
$endgroup$
– Martin Brandenburg
Jun 27 '14 at 18:59
$begingroup$
Since the question is on MO now, should it be closed here?
$endgroup$
– Martin Brandenburg
Jun 27 '14 at 18:59
|
show 9 more comments
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protected by Pedro Tamaroff♦ Jan 18 '15 at 20:26
Thank you for your interest in this question.
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protected by Pedro Tamaroff♦ Jan 18 '15 at 20:26
Thank you for your interest in this question.
Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).
Would you like to answer one of these unanswered questions instead?
10
$begingroup$
At first sight I can't see many hopes to do this as long as the characterization of the Krull dimension in that paper deals with elements instead of ideals, while the property of being noetherian relies on some properties of ideals. But who knows?
$endgroup$
– user26857
Apr 12 '13 at 8:11
4
$begingroup$
Yes. But maybe there is a first-order characterization for noetherian rings, like the one for Krull dimension? Or maybe some weaker property already suffices? I don't know.
$endgroup$
– Martin Brandenburg
Apr 12 '13 at 9:11
88
$begingroup$
Shall I ask this on mathoverflow?
$endgroup$
– Martin Brandenburg
Aug 26 '13 at 12:43
27
$begingroup$
MO copy of the question: mathoverflow.net/questions/172350/a-short-proof-for-dimrt-dimr1
$endgroup$
– Martin Sleziak
Jun 21 '14 at 19:55
5
$begingroup$
Since the question is on MO now, should it be closed here?
$endgroup$
– Martin Brandenburg
Jun 27 '14 at 18:59