Existence of mod $m^n$-points for every $n$
Let A be a complete noetherian local ring and $m$ be its maximal ideal. If I have some polynomials $f_i$ with coeffecients in $A$, and they have a common zero $x_n$ in $A/m^n$ for every $n$, then must they have a common zero in $A$?
Note if the corresponding scheme $X$ is smooth over $A$, then we can use Hensel lemma to conclude. The main problem is compatibility for $x_n$, if $A/m$ is finite then it's true because inverse limit of finite set is non-empty.
number-theory algebraic-geometry arithmetic-geometry
asked Dec 9 at 17:16
zzy
2,3571419
add a comment |
Let A be a complete noetherian local ring and $m$ be its maximal ideal. If I have some polynomials $f_i$ with coeffecients in $A$, and they have a common zero $x_n$ in $A/m^n$ for every $n$, then must they have a common zero in $A$?
Note if the corresponding scheme $X$ is smooth over $A$, then we can use Hensel lemma to conclude. The main problem is compatibility for $x_n$, if $A/m$ is finite then it's true because inverse limit of finite set is non-empty.
number-theory algebraic-geometry arithmetic-geometry
asked Dec 9 at 17:16
zzy
2,3571419
If $A,B$ are complete local rings with maximal ideals $M,N$ containing a coefficient field $k$ and if $A/M^ncong B/N^n$ for all $n$, then you can find a compatible isomorphism, so that $Acong B$. I think this should be enough to get what you want, at least for one polynomial.
– Mohan
Dec 10 at 14:11
@Mohan Thank you, I am also interested in higher dimensional case.
– zzy
Dec 10 at 20:56
add a comment |
Let A be a complete noetherian local ring and $m$ be its maximal ideal. If I have some polynomials $f_i$ with coeffecients in $A$, and they have a common zero $x_n$ in $A/m^n$ for every $n$, then must they have a common zero in $A$?
Note if the corresponding scheme $X$ is smooth over $A$, then we can use Hensel lemma to conclude. The main problem is compatibility for $x_n$, if $A/m$ is finite then it's true because inverse limit of finite set is non-empty.
number-theory algebraic-geometry arithmetic-geometry
asked Dec 9 at 17:16
zzy
2,3571419
Let A be a complete noetherian local ring and $m$ be its maximal ideal. If I have some polynomials $f_i$ with coeffecients in $A$, and they have a common zero $x_n$ in $A/m^n$ for every $n$, then must they have a common zero in $A$?
Note if the corresponding scheme $X$ is smooth over $A$, then we can use Hensel lemma to conclude. The main problem is compatibility for $x_n$, if $A/m$ is finite then it's true because inverse limit of finite set is non-empty.
number-theory algebraic-geometry arithmetic-geometry
number-theory algebraic-geometry arithmetic-geometry
asked Dec 9 at 17:16
zzy
2,3571419
asked Dec 9 at 17:16
zzy
2,3571419
asked Dec 9 at 17:16
zzy
2,3571419
asked Dec 9 at 17:16
zzy
2,3571419
asked Dec 9 at 17:16
zzy
2,3571419
2,3571419
If $A,B$ are complete local rings with maximal ideals $M,N$ containing a coefficient field $k$ and if $A/M^ncong B/N^n$ for all $n$, then you can find a compatible isomorphism, so that $Acong B$. I think this should be enough to get what you want, at least for one polynomial.
– Mohan
Dec 10 at 14:11
@Mohan Thank you, I am also interested in higher dimensional case.
– zzy
Dec 10 at 20:56
add a comment |
If $A,B$ are complete local rings with maximal ideals $M,N$ containing a coefficient field $k$ and if $A/M^ncong B/N^n$ for all $n$, then you can find a compatible isomorphism, so that $Acong B$. I think this should be enough to get what you want, at least for one polynomial.
– Mohan
Dec 10 at 14:11
@Mohan Thank you, I am also interested in higher dimensional case.
– zzy
Dec 10 at 20:56
If $A,B$ are complete local rings with maximal ideals $M,N$ containing a coefficient field $k$ and if $A/M^ncong B/N^n$ for all $n$, then you can find a compatible isomorphism, so that $Acong B$. I think this should be enough to get what you want, at least for one polynomial.
– Mohan
Dec 10 at 14:11
If $A,B$ are complete local rings with maximal ideals $M,N$ containing a coefficient field $k$ and if $A/M^ncong B/N^n$ for all $n$, then you can find a compatible isomorphism, so that $Acong B$. I think this should be enough to get what you want, at least for one polynomial.
– Mohan
Dec 10 at 14:11
@Mohan Thank you, I am also interested in higher dimensional case.
– zzy
Dec 10 at 20:56
@Mohan Thank you, I am also interested in higher dimensional case.
– zzy
Dec 10 at 20:56
add a comment |
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If $A,B$ are complete local rings with maximal ideals $M,N$ containing a coefficient field $k$ and if $A/M^ncong B/N^n$ for all $n$, then you can find a compatible isomorphism, so that $Acong B$. I think this should be enough to get what you want, at least for one polynomial.
– Mohan
Dec 10 at 14:11
@Mohan Thank you, I am also interested in higher dimensional case.
– zzy
Dec 10 at 20:56