When is a map of local rings finite?












0














Let $A,B$ be Noetherian local rings, and let $A to B$ be a ring homomorphism such that the induced map $operatorname{Spec} B to operatorname{Spec} A$ is surjective and quasifinite (of finite type and has finite fibers). Is it necessarily true that $B$ is finite over $A$?



What I know so far:




  • The usual counterexample to finiteness of quasifinite morphisms is given by taking $A = k[[x]]$, $B = k((x))$, and $A to B$ to be the obvious inclusion map, but this counterexample fails to satisfy the surjectivity requirement.

  • One approach is to try to show that the map $operatorname{Spec} B to operatorname{Spec} A$ is universally closed, because universally closed affine maps are integral. By https://stacks.math.columbia.edu/tag/0205, it suffices to show that the pulled-back map $mathbb{A}_B^n to mathbb{A}_A^n$ is closed for every $n$. The surjectivity assumption implies that this is true for $n = 0$, but I'm not sure how to prove closedness for $n > 0$.






algebraic-geometry commutative-algebra affine-schemes local-rings







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    Let $A,B$ be Noetherian local rings, and let $A to B$ be a ring homomorphism such that the induced map $operatorname{Spec} B to operatorname{Spec} A$ is surjective and quasifinite (of finite type and has finite fibers). Is it necessarily true that $B$ is finite over $A$?



    What I know so far:




    • The usual counterexample to finiteness of quasifinite morphisms is given by taking $A = k[[x]]$, $B = k((x))$, and $A to B$ to be the obvious inclusion map, but this counterexample fails to satisfy the surjectivity requirement.

    • One approach is to try to show that the map $operatorname{Spec} B to operatorname{Spec} A$ is universally closed, because universally closed affine maps are integral. By https://stacks.math.columbia.edu/tag/0205, it suffices to show that the pulled-back map $mathbb{A}_B^n to mathbb{A}_A^n$ is closed for every $n$. The surjectivity assumption implies that this is true for $n = 0$, but I'm not sure how to prove closedness for $n > 0$.






    algebraic-geometry commutative-algebra affine-schemes local-rings







    share|cite|improve this question

























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      Let $A,B$ be Noetherian local rings, and let $A to B$ be a ring homomorphism such that the induced map $operatorname{Spec} B to operatorname{Spec} A$ is surjective and quasifinite (of finite type and has finite fibers). Is it necessarily true that $B$ is finite over $A$?



      What I know so far:




      • The usual counterexample to finiteness of quasifinite morphisms is given by taking $A = k[[x]]$, $B = k((x))$, and $A to B$ to be the obvious inclusion map, but this counterexample fails to satisfy the surjectivity requirement.

      • One approach is to try to show that the map $operatorname{Spec} B to operatorname{Spec} A$ is universally closed, because universally closed affine maps are integral. By https://stacks.math.columbia.edu/tag/0205, it suffices to show that the pulled-back map $mathbb{A}_B^n to mathbb{A}_A^n$ is closed for every $n$. The surjectivity assumption implies that this is true for $n = 0$, but I'm not sure how to prove closedness for $n > 0$.






      algebraic-geometry commutative-algebra affine-schemes local-rings







      share|cite|improve this question













      Let $A,B$ be Noetherian local rings, and let $A to B$ be a ring homomorphism such that the induced map $operatorname{Spec} B to operatorname{Spec} A$ is surjective and quasifinite (of finite type and has finite fibers). Is it necessarily true that $B$ is finite over $A$?



      What I know so far:




      • The usual counterexample to finiteness of quasifinite morphisms is given by taking $A = k[[x]]$, $B = k((x))$, and $A to B$ to be the obvious inclusion map, but this counterexample fails to satisfy the surjectivity requirement.

      • One approach is to try to show that the map $operatorname{Spec} B to operatorname{Spec} A$ is universally closed, because universally closed affine maps are integral. By https://stacks.math.columbia.edu/tag/0205, it suffices to show that the pulled-back map $mathbb{A}_B^n to mathbb{A}_A^n$ is closed for every $n$. The surjectivity assumption implies that this is true for $n = 0$, but I'm not sure how to prove closedness for $n > 0$.






      algebraic-geometry commutative-algebra affine-schemes local-rings




      algebraic-geometry commutative-algebra affine-schemes local-rings






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      asked Jan 4 at 10:28









      Ashvin SwaminathanAshvin Swaminathan

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          Consider $k[t]subset k[u]$, $tmapsto u^2$. Then, $t-1mapsto u^2-1=(u-1)(u+1)$. Let $A$ be the localization of $k[t]$ at the maximal ideal $(t-1)$ and $B$ be the localization of $k[u]$ at the maximal ideal $(u-1)$. Then, $Ato B$ satisfies all your requirements, but not finite.






          share|cite|improve this answer





















          • Thanks for the helpful response, but I did your example and got that it was finite. Your claim seems to be that, for example, the projection of a parabola $x=y^2-y$ onto the $x$-axis fails to be finite if one localizes the parabola at the origin $x=y=0$. But clearly $k[[x,y]]/(x-y^2+y)$ is finite over $k[[x]]$ with basis $(1,y)$.
            – Ashvin Swaminathan
            Jan 4 at 15:39












          • No, it is not finite. The integral closure of $A$ in the fraction field of $B$ is semilocal and not local, since there are two points above $t=1$.
            – Mohan
            Jan 4 at 17:04










          • Thanks, just to confirm, are you saying that in my comment, $k[[x,y]]/(x-y^2+y)$ is not finite over $k[[x]]$, or are you saying my comment is actually different from your counterexample?
            – Ashvin Swaminathan
            Jan 4 at 17:53






          • 1




            No, your case is different from mine. First, an application of Weirstrass preparation theorem will tell you that if $A$ is complete (for example $A=k[[x]]$), then quasifinite is finite. If you took your example over $k[x]$ localized at say $(x-1)$ (not powerseries) and localize $k[x,y]/(x-y^2+y)$ at one of the roots of $y^2-y-1$, you will have a quasifinite non-finite extension as you want.
            – Mohan
            Jan 4 at 17:59











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          Consider $k[t]subset k[u]$, $tmapsto u^2$. Then, $t-1mapsto u^2-1=(u-1)(u+1)$. Let $A$ be the localization of $k[t]$ at the maximal ideal $(t-1)$ and $B$ be the localization of $k[u]$ at the maximal ideal $(u-1)$. Then, $Ato B$ satisfies all your requirements, but not finite.






          share|cite|improve this answer





















          • Thanks for the helpful response, but I did your example and got that it was finite. Your claim seems to be that, for example, the projection of a parabola $x=y^2-y$ onto the $x$-axis fails to be finite if one localizes the parabola at the origin $x=y=0$. But clearly $k[[x,y]]/(x-y^2+y)$ is finite over $k[[x]]$ with basis $(1,y)$.
            – Ashvin Swaminathan
            Jan 4 at 15:39












          • No, it is not finite. The integral closure of $A$ in the fraction field of $B$ is semilocal and not local, since there are two points above $t=1$.
            – Mohan
            Jan 4 at 17:04










          • Thanks, just to confirm, are you saying that in my comment, $k[[x,y]]/(x-y^2+y)$ is not finite over $k[[x]]$, or are you saying my comment is actually different from your counterexample?
            – Ashvin Swaminathan
            Jan 4 at 17:53






          • 1




            No, your case is different from mine. First, an application of Weirstrass preparation theorem will tell you that if $A$ is complete (for example $A=k[[x]]$), then quasifinite is finite. If you took your example over $k[x]$ localized at say $(x-1)$ (not powerseries) and localize $k[x,y]/(x-y^2+y)$ at one of the roots of $y^2-y-1$, you will have a quasifinite non-finite extension as you want.
            – Mohan
            Jan 4 at 17:59
















          1














          Consider $k[t]subset k[u]$, $tmapsto u^2$. Then, $t-1mapsto u^2-1=(u-1)(u+1)$. Let $A$ be the localization of $k[t]$ at the maximal ideal $(t-1)$ and $B$ be the localization of $k[u]$ at the maximal ideal $(u-1)$. Then, $Ato B$ satisfies all your requirements, but not finite.






          share|cite|improve this answer





















          • Thanks for the helpful response, but I did your example and got that it was finite. Your claim seems to be that, for example, the projection of a parabola $x=y^2-y$ onto the $x$-axis fails to be finite if one localizes the parabola at the origin $x=y=0$. But clearly $k[[x,y]]/(x-y^2+y)$ is finite over $k[[x]]$ with basis $(1,y)$.
            – Ashvin Swaminathan
            Jan 4 at 15:39












          • No, it is not finite. The integral closure of $A$ in the fraction field of $B$ is semilocal and not local, since there are two points above $t=1$.
            – Mohan
            Jan 4 at 17:04










          • Thanks, just to confirm, are you saying that in my comment, $k[[x,y]]/(x-y^2+y)$ is not finite over $k[[x]]$, or are you saying my comment is actually different from your counterexample?
            – Ashvin Swaminathan
            Jan 4 at 17:53






          • 1




            No, your case is different from mine. First, an application of Weirstrass preparation theorem will tell you that if $A$ is complete (for example $A=k[[x]]$), then quasifinite is finite. If you took your example over $k[x]$ localized at say $(x-1)$ (not powerseries) and localize $k[x,y]/(x-y^2+y)$ at one of the roots of $y^2-y-1$, you will have a quasifinite non-finite extension as you want.
            – Mohan
            Jan 4 at 17:59














          1












          1








          1






          Consider $k[t]subset k[u]$, $tmapsto u^2$. Then, $t-1mapsto u^2-1=(u-1)(u+1)$. Let $A$ be the localization of $k[t]$ at the maximal ideal $(t-1)$ and $B$ be the localization of $k[u]$ at the maximal ideal $(u-1)$. Then, $Ato B$ satisfies all your requirements, but not finite.






          share|cite|improve this answer












          Consider $k[t]subset k[u]$, $tmapsto u^2$. Then, $t-1mapsto u^2-1=(u-1)(u+1)$. Let $A$ be the localization of $k[t]$ at the maximal ideal $(t-1)$ and $B$ be the localization of $k[u]$ at the maximal ideal $(u-1)$. Then, $Ato B$ satisfies all your requirements, but not finite.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 4 at 14:57









          MohanMohan

          11.6k1817




          11.6k1817












          • Thanks for the helpful response, but I did your example and got that it was finite. Your claim seems to be that, for example, the projection of a parabola $x=y^2-y$ onto the $x$-axis fails to be finite if one localizes the parabola at the origin $x=y=0$. But clearly $k[[x,y]]/(x-y^2+y)$ is finite over $k[[x]]$ with basis $(1,y)$.
            – Ashvin Swaminathan
            Jan 4 at 15:39












          • No, it is not finite. The integral closure of $A$ in the fraction field of $B$ is semilocal and not local, since there are two points above $t=1$.
            – Mohan
            Jan 4 at 17:04










          • Thanks, just to confirm, are you saying that in my comment, $k[[x,y]]/(x-y^2+y)$ is not finite over $k[[x]]$, or are you saying my comment is actually different from your counterexample?
            – Ashvin Swaminathan
            Jan 4 at 17:53






          • 1




            No, your case is different from mine. First, an application of Weirstrass preparation theorem will tell you that if $A$ is complete (for example $A=k[[x]]$), then quasifinite is finite. If you took your example over $k[x]$ localized at say $(x-1)$ (not powerseries) and localize $k[x,y]/(x-y^2+y)$ at one of the roots of $y^2-y-1$, you will have a quasifinite non-finite extension as you want.
            – Mohan
            Jan 4 at 17:59


















          • Thanks for the helpful response, but I did your example and got that it was finite. Your claim seems to be that, for example, the projection of a parabola $x=y^2-y$ onto the $x$-axis fails to be finite if one localizes the parabola at the origin $x=y=0$. But clearly $k[[x,y]]/(x-y^2+y)$ is finite over $k[[x]]$ with basis $(1,y)$.
            – Ashvin Swaminathan
            Jan 4 at 15:39












          • No, it is not finite. The integral closure of $A$ in the fraction field of $B$ is semilocal and not local, since there are two points above $t=1$.
            – Mohan
            Jan 4 at 17:04










          • Thanks, just to confirm, are you saying that in my comment, $k[[x,y]]/(x-y^2+y)$ is not finite over $k[[x]]$, or are you saying my comment is actually different from your counterexample?
            – Ashvin Swaminathan
            Jan 4 at 17:53






          • 1




            No, your case is different from mine. First, an application of Weirstrass preparation theorem will tell you that if $A$ is complete (for example $A=k[[x]]$), then quasifinite is finite. If you took your example over $k[x]$ localized at say $(x-1)$ (not powerseries) and localize $k[x,y]/(x-y^2+y)$ at one of the roots of $y^2-y-1$, you will have a quasifinite non-finite extension as you want.
            – Mohan
            Jan 4 at 17:59
















          Thanks for the helpful response, but I did your example and got that it was finite. Your claim seems to be that, for example, the projection of a parabola $x=y^2-y$ onto the $x$-axis fails to be finite if one localizes the parabola at the origin $x=y=0$. But clearly $k[[x,y]]/(x-y^2+y)$ is finite over $k[[x]]$ with basis $(1,y)$.
          – Ashvin Swaminathan
          Jan 4 at 15:39






          Thanks for the helpful response, but I did your example and got that it was finite. Your claim seems to be that, for example, the projection of a parabola $x=y^2-y$ onto the $x$-axis fails to be finite if one localizes the parabola at the origin $x=y=0$. But clearly $k[[x,y]]/(x-y^2+y)$ is finite over $k[[x]]$ with basis $(1,y)$.
          – Ashvin Swaminathan
          Jan 4 at 15:39














          No, it is not finite. The integral closure of $A$ in the fraction field of $B$ is semilocal and not local, since there are two points above $t=1$.
          – Mohan
          Jan 4 at 17:04




          No, it is not finite. The integral closure of $A$ in the fraction field of $B$ is semilocal and not local, since there are two points above $t=1$.
          – Mohan
          Jan 4 at 17:04












          Thanks, just to confirm, are you saying that in my comment, $k[[x,y]]/(x-y^2+y)$ is not finite over $k[[x]]$, or are you saying my comment is actually different from your counterexample?
          – Ashvin Swaminathan
          Jan 4 at 17:53




          Thanks, just to confirm, are you saying that in my comment, $k[[x,y]]/(x-y^2+y)$ is not finite over $k[[x]]$, or are you saying my comment is actually different from your counterexample?
          – Ashvin Swaminathan
          Jan 4 at 17:53




          1




          1




          No, your case is different from mine. First, an application of Weirstrass preparation theorem will tell you that if $A$ is complete (for example $A=k[[x]]$), then quasifinite is finite. If you took your example over $k[x]$ localized at say $(x-1)$ (not powerseries) and localize $k[x,y]/(x-y^2+y)$ at one of the roots of $y^2-y-1$, you will have a quasifinite non-finite extension as you want.
          – Mohan
          Jan 4 at 17:59




          No, your case is different from mine. First, an application of Weirstrass preparation theorem will tell you that if $A$ is complete (for example $A=k[[x]]$), then quasifinite is finite. If you took your example over $k[x]$ localized at say $(x-1)$ (not powerseries) and localize $k[x,y]/(x-y^2+y)$ at one of the roots of $y^2-y-1$, you will have a quasifinite non-finite extension as you want.
          – Mohan
          Jan 4 at 17:59


















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