Henselianizations over countable index sets
Let $A$ be a ring, $Isubset A$ a finitely generated ideal.
The henselianization $A^h$ of $A$ along $I$ is the universal $A$-algebra that is henselian along $I$ and can be presented as a direct limit of étale ring maps that are the identity on mod $I$ fibers:
$$A^h = varinjlim_{sin S} A_s$$
where $Ato A_s$ is étale and such that $A/Ito A_s/I$ is the identity, and $S$ is an index set.
When $A$ is smooth over the Noetherian henselian valuation ring $R = mathbf{Z}_{(p)}^h$ and $I = pA$ is principal, can $S$ be arranged to be a countable set?
ag.algebraic-geometry ac.commutative-algebra arithmetic-geometry etale-cohomology
asked Dec 9 '18 at 18:14
Ben
575
add a comment |
Let $A$ be a ring, $Isubset A$ a finitely generated ideal.
The henselianization $A^h$ of $A$ along $I$ is the universal $A$-algebra that is henselian along $I$ and can be presented as a direct limit of étale ring maps that are the identity on mod $I$ fibers:
$$A^h = varinjlim_{sin S} A_s$$
where $Ato A_s$ is étale and such that $A/Ito A_s/I$ is the identity, and $S$ is an index set.
When $A$ is smooth over the Noetherian henselian valuation ring $R = mathbf{Z}_{(p)}^h$ and $I = pA$ is principal, can $S$ be arranged to be a countable set?
ag.algebraic-geometry ac.commutative-algebra arithmetic-geometry etale-cohomology
asked Dec 9 '18 at 18:14
Ben
575
add a comment |
Let $A$ be a ring, $Isubset A$ a finitely generated ideal.
The henselianization $A^h$ of $A$ along $I$ is the universal $A$-algebra that is henselian along $I$ and can be presented as a direct limit of étale ring maps that are the identity on mod $I$ fibers:
$$A^h = varinjlim_{sin S} A_s$$
where $Ato A_s$ is étale and such that $A/Ito A_s/I$ is the identity, and $S$ is an index set.
When $A$ is smooth over the Noetherian henselian valuation ring $R = mathbf{Z}_{(p)}^h$ and $I = pA$ is principal, can $S$ be arranged to be a countable set?
ag.algebraic-geometry ac.commutative-algebra arithmetic-geometry etale-cohomology
asked Dec 9 '18 at 18:14
Ben
575
Let $A$ be a ring, $Isubset A$ a finitely generated ideal.
The henselianization $A^h$ of $A$ along $I$ is the universal $A$-algebra that is henselian along $I$ and can be presented as a direct limit of étale ring maps that are the identity on mod $I$ fibers:
$$A^h = varinjlim_{sin S} A_s$$
where $Ato A_s$ is étale and such that $A/Ito A_s/I$ is the identity, and $S$ is an index set.
When $A$ is smooth over the Noetherian henselian valuation ring $R = mathbf{Z}_{(p)}^h$ and $I = pA$ is principal, can $S$ be arranged to be a countable set?
ag.algebraic-geometry ac.commutative-algebra arithmetic-geometry etale-cohomology
ag.algebraic-geometry ac.commutative-algebra arithmetic-geometry etale-cohomology
asked Dec 9 '18 at 18:14
Ben
575
asked Dec 9 '18 at 18:14
Ben
575
edited Dec 9 '18 at 19:38
asked Dec 9 '18 at 18:14
Ben
575
asked Dec 9 '18 at 18:14
Ben
575
asked Dec 9 '18 at 18:14
Ben
575
575
add a comment |
add a comment |
1 Answer
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No.
Take $A = mathbb{C}[x]$ and $I=(x)$. Suppose that $A^h$ is the direct limit of a system of etale algebras $A_i$ such that $A_i/(x) cong mathbb{C}$. We can assume that each $A_i$ is finitely presented.
For $ain mathbb{C}setminus {0}$, consider the algebra $A[1/(x-a)]$. Then:
(1) for every nonzero $ain mathbb{C}$, the map $Ato A^h$ factors through $A[1/(x-a)]$ (by universal property of henselization)
(2) for every index $i$, there are only finitely many nonzero $ain mathbb{C}$ for which $Ato A_i$ factors through $A[1/(x-a)]$ (because the image of ${rm Spec}(A_i) to {rm Spec}(A)$ is open and dense).
This shows that the index set has to have cardinality at least equal to the cardinality of $mathbb{C}$.
answered Dec 9 '18 at 19:24
Piotr Achinger
8,17312752
I see your point. However, in my question $A$ is smooth over some Noetherian henselian valuation ring with uniformizer $x$ (I should have made it clear the valuation ring is non-archimedean: at least I always mean that), and the ideal $I$ should be the extension of the principal maximal ideal of the valuation ring, ie. $xA$. In my question, I am thinking about $R = mathbf{Z}_{(p)}^h$, $A$ smooth over $R$, $I = pA$. I have edited my final question to only include this case
– Ben
Dec 9 '18 at 19:37
1
Sorry, I misunderstood the question. But I think you can make a variant of the above counterexample work with $A = mathbb{Z}_p[t]$ and $a in mathbb{Q}_psetminus mathbb{Z}_p$. However, in your particular case ($mathbb{Z}_{(p)}$ rather than the $p$-adics), it seems that the rings are countable, so there are only countably many f.p. etale algebras over them, no?
– Piotr Achinger
Dec 9 '18 at 19:51
1
More precisely, the variant with $p$-adics would be: $A=mathbb{Z}_p[x]$, $I=(p)$ and for $a = c/p^n in mathbb{Q}_p$ with $cinmathbb{Z}^times_p$ and $n>0$, the algebra $A[1/(p^n x-c)]$.
– Piotr Achinger
Dec 9 '18 at 20:03
add a comment |
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No.
Take $A = mathbb{C}[x]$ and $I=(x)$. Suppose that $A^h$ is the direct limit of a system of etale algebras $A_i$ such that $A_i/(x) cong mathbb{C}$. We can assume that each $A_i$ is finitely presented.
For $ain mathbb{C}setminus {0}$, consider the algebra $A[1/(x-a)]$. Then:
(1) for every nonzero $ain mathbb{C}$, the map $Ato A^h$ factors through $A[1/(x-a)]$ (by universal property of henselization)
(2) for every index $i$, there are only finitely many nonzero $ain mathbb{C}$ for which $Ato A_i$ factors through $A[1/(x-a)]$ (because the image of ${rm Spec}(A_i) to {rm Spec}(A)$ is open and dense).
This shows that the index set has to have cardinality at least equal to the cardinality of $mathbb{C}$.
answered Dec 9 '18 at 19:24
Piotr Achinger
8,17312752
I see your point. However, in my question $A$ is smooth over some Noetherian henselian valuation ring with uniformizer $x$ (I should have made it clear the valuation ring is non-archimedean: at least I always mean that), and the ideal $I$ should be the extension of the principal maximal ideal of the valuation ring, ie. $xA$. In my question, I am thinking about $R = mathbf{Z}_{(p)}^h$, $A$ smooth over $R$, $I = pA$. I have edited my final question to only include this case
– Ben
Dec 9 '18 at 19:37
1
Sorry, I misunderstood the question. But I think you can make a variant of the above counterexample work with $A = mathbb{Z}_p[t]$ and $a in mathbb{Q}_psetminus mathbb{Z}_p$. However, in your particular case ($mathbb{Z}_{(p)}$ rather than the $p$-adics), it seems that the rings are countable, so there are only countably many f.p. etale algebras over them, no?
– Piotr Achinger
Dec 9 '18 at 19:51
1
More precisely, the variant with $p$-adics would be: $A=mathbb{Z}_p[x]$, $I=(p)$ and for $a = c/p^n in mathbb{Q}_p$ with $cinmathbb{Z}^times_p$ and $n>0$, the algebra $A[1/(p^n x-c)]$.
– Piotr Achinger
Dec 9 '18 at 20:03
add a comment |
No.
Take $A = mathbb{C}[x]$ and $I=(x)$. Suppose that $A^h$ is the direct limit of a system of etale algebras $A_i$ such that $A_i/(x) cong mathbb{C}$. We can assume that each $A_i$ is finitely presented.
For $ain mathbb{C}setminus {0}$, consider the algebra $A[1/(x-a)]$. Then:
(1) for every nonzero $ain mathbb{C}$, the map $Ato A^h$ factors through $A[1/(x-a)]$ (by universal property of henselization)
(2) for every index $i$, there are only finitely many nonzero $ain mathbb{C}$ for which $Ato A_i$ factors through $A[1/(x-a)]$ (because the image of ${rm Spec}(A_i) to {rm Spec}(A)$ is open and dense).
This shows that the index set has to have cardinality at least equal to the cardinality of $mathbb{C}$.
answered Dec 9 '18 at 19:24
Piotr Achinger
8,17312752
I see your point. However, in my question $A$ is smooth over some Noetherian henselian valuation ring with uniformizer $x$ (I should have made it clear the valuation ring is non-archimedean: at least I always mean that), and the ideal $I$ should be the extension of the principal maximal ideal of the valuation ring, ie. $xA$. In my question, I am thinking about $R = mathbf{Z}_{(p)}^h$, $A$ smooth over $R$, $I = pA$. I have edited my final question to only include this case
– Ben
Dec 9 '18 at 19:37
1
Sorry, I misunderstood the question. But I think you can make a variant of the above counterexample work with $A = mathbb{Z}_p[t]$ and $a in mathbb{Q}_psetminus mathbb{Z}_p$. However, in your particular case ($mathbb{Z}_{(p)}$ rather than the $p$-adics), it seems that the rings are countable, so there are only countably many f.p. etale algebras over them, no?
– Piotr Achinger
Dec 9 '18 at 19:51
1
More precisely, the variant with $p$-adics would be: $A=mathbb{Z}_p[x]$, $I=(p)$ and for $a = c/p^n in mathbb{Q}_p$ with $cinmathbb{Z}^times_p$ and $n>0$, the algebra $A[1/(p^n x-c)]$.
– Piotr Achinger
Dec 9 '18 at 20:03
add a comment |
No.
Take $A = mathbb{C}[x]$ and $I=(x)$. Suppose that $A^h$ is the direct limit of a system of etale algebras $A_i$ such that $A_i/(x) cong mathbb{C}$. We can assume that each $A_i$ is finitely presented.
For $ain mathbb{C}setminus {0}$, consider the algebra $A[1/(x-a)]$. Then:
(1) for every nonzero $ain mathbb{C}$, the map $Ato A^h$ factors through $A[1/(x-a)]$ (by universal property of henselization)
(2) for every index $i$, there are only finitely many nonzero $ain mathbb{C}$ for which $Ato A_i$ factors through $A[1/(x-a)]$ (because the image of ${rm Spec}(A_i) to {rm Spec}(A)$ is open and dense).
This shows that the index set has to have cardinality at least equal to the cardinality of $mathbb{C}$.
answered Dec 9 '18 at 19:24
Piotr Achinger
8,17312752
No.
Take $A = mathbb{C}[x]$ and $I=(x)$. Suppose that $A^h$ is the direct limit of a system of etale algebras $A_i$ such that $A_i/(x) cong mathbb{C}$. We can assume that each $A_i$ is finitely presented.
For $ain mathbb{C}setminus {0}$, consider the algebra $A[1/(x-a)]$. Then:
(1) for every nonzero $ain mathbb{C}$, the map $Ato A^h$ factors through $A[1/(x-a)]$ (by universal property of henselization)
(2) for every index $i$, there are only finitely many nonzero $ain mathbb{C}$ for which $Ato A_i$ factors through $A[1/(x-a)]$ (because the image of ${rm Spec}(A_i) to {rm Spec}(A)$ is open and dense).
This shows that the index set has to have cardinality at least equal to the cardinality of $mathbb{C}$.
answered Dec 9 '18 at 19:24
Piotr Achinger
8,17312752
answered Dec 9 '18 at 19:24
Piotr Achinger
8,17312752
answered Dec 9 '18 at 19:24
Piotr Achinger
8,17312752
answered Dec 9 '18 at 19:24
Piotr Achinger
8,17312752
8,17312752
I see your point. However, in my question $A$ is smooth over some Noetherian henselian valuation ring with uniformizer $x$ (I should have made it clear the valuation ring is non-archimedean: at least I always mean that), and the ideal $I$ should be the extension of the principal maximal ideal of the valuation ring, ie. $xA$. In my question, I am thinking about $R = mathbf{Z}_{(p)}^h$, $A$ smooth over $R$, $I = pA$. I have edited my final question to only include this case
– Ben
Dec 9 '18 at 19:37
1
Sorry, I misunderstood the question. But I think you can make a variant of the above counterexample work with $A = mathbb{Z}_p[t]$ and $a in mathbb{Q}_psetminus mathbb{Z}_p$. However, in your particular case ($mathbb{Z}_{(p)}$ rather than the $p$-adics), it seems that the rings are countable, so there are only countably many f.p. etale algebras over them, no?
– Piotr Achinger
Dec 9 '18 at 19:51
1
More precisely, the variant with $p$-adics would be: $A=mathbb{Z}_p[x]$, $I=(p)$ and for $a = c/p^n in mathbb{Q}_p$ with $cinmathbb{Z}^times_p$ and $n>0$, the algebra $A[1/(p^n x-c)]$.
– Piotr Achinger
Dec 9 '18 at 20:03
add a comment |
I see your point. However, in my question $A$ is smooth over some Noetherian henselian valuation ring with uniformizer $x$ (I should have made it clear the valuation ring is non-archimedean: at least I always mean that), and the ideal $I$ should be the extension of the principal maximal ideal of the valuation ring, ie. $xA$. In my question, I am thinking about $R = mathbf{Z}_{(p)}^h$, $A$ smooth over $R$, $I = pA$. I have edited my final question to only include this case
– Ben
Dec 9 '18 at 19:37
1
Sorry, I misunderstood the question. But I think you can make a variant of the above counterexample work with $A = mathbb{Z}_p[t]$ and $a in mathbb{Q}_psetminus mathbb{Z}_p$. However, in your particular case ($mathbb{Z}_{(p)}$ rather than the $p$-adics), it seems that the rings are countable, so there are only countably many f.p. etale algebras over them, no?
– Piotr Achinger
Dec 9 '18 at 19:51
1
More precisely, the variant with $p$-adics would be: $A=mathbb{Z}_p[x]$, $I=(p)$ and for $a = c/p^n in mathbb{Q}_p$ with $cinmathbb{Z}^times_p$ and $n>0$, the algebra $A[1/(p^n x-c)]$.
– Piotr Achinger
Dec 9 '18 at 20:03
I see your point. However, in my question $A$ is smooth over some Noetherian henselian valuation ring with uniformizer $x$ (I should have made it clear the valuation ring is non-archimedean: at least I always mean that), and the ideal $I$ should be the extension of the principal maximal ideal of the valuation ring, ie. $xA$. In my question, I am thinking about $R = mathbf{Z}_{(p)}^h$, $A$ smooth over $R$, $I = pA$. I have edited my final question to only include this case
– Ben
Dec 9 '18 at 19:37
I see your point. However, in my question $A$ is smooth over some Noetherian henselian valuation ring with uniformizer $x$ (I should have made it clear the valuation ring is non-archimedean: at least I always mean that), and the ideal $I$ should be the extension of the principal maximal ideal of the valuation ring, ie. $xA$. In my question, I am thinking about $R = mathbf{Z}_{(p)}^h$, $A$ smooth over $R$, $I = pA$. I have edited my final question to only include this case
– Ben
Dec 9 '18 at 19:37
1
1
Sorry, I misunderstood the question. But I think you can make a variant of the above counterexample work with $A = mathbb{Z}_p[t]$ and $a in mathbb{Q}_psetminus mathbb{Z}_p$. However, in your particular case ($mathbb{Z}_{(p)}$ rather than the $p$-adics), it seems that the rings are countable, so there are only countably many f.p. etale algebras over them, no?
– Piotr Achinger
Dec 9 '18 at 19:51
Sorry, I misunderstood the question. But I think you can make a variant of the above counterexample work with $A = mathbb{Z}_p[t]$ and $a in mathbb{Q}_psetminus mathbb{Z}_p$. However, in your particular case ($mathbb{Z}_{(p)}$ rather than the $p$-adics), it seems that the rings are countable, so there are only countably many f.p. etale algebras over them, no?
– Piotr Achinger
Dec 9 '18 at 19:51
1
1
More precisely, the variant with $p$-adics would be: $A=mathbb{Z}_p[x]$, $I=(p)$ and for $a = c/p^n in mathbb{Q}_p$ with $cinmathbb{Z}^times_p$ and $n>0$, the algebra $A[1/(p^n x-c)]$.
– Piotr Achinger
Dec 9 '18 at 20:03
More precisely, the variant with $p$-adics would be: $A=mathbb{Z}_p[x]$, $I=(p)$ and for $a = c/p^n in mathbb{Q}_p$ with $cinmathbb{Z}^times_p$ and $n>0$, the algebra $A[1/(p^n x-c)]$.
– Piotr Achinger
Dec 9 '18 at 20:03
add a comment |
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