Henselianizations over countable index sets












4














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







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    4














    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







    share|cite|improve this question



























      4












      4








      4







      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







      share|cite|improve this question















      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






      share|cite|improve this question















      share|cite|improve this question













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      edited Dec 9 '18 at 19:38

























      asked Dec 9 '18 at 18:14









      Ben

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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 $mathbb{C}$.






          share|cite|improve this answer





















          • 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











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          1 Answer
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          1 Answer
          1






          active

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          active

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          active

          oldest

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          4














          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 $mathbb{C}$.






          share|cite|improve this answer





















          • 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
















          4














          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 $mathbb{C}$.






          share|cite|improve this answer





















          • 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














          4












          4








          4






          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 $mathbb{C}$.






          share|cite|improve this answer












          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 $mathbb{C}$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          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


















          • 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


















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