Under what conditions does a non-empty intersection over $mathbb{Z}$ remain non-empty after a change of base...
Let $V$ be a Zariski open subset of an affine variety over $mathbb{Z}$, and let $f: V to P$ be a morphism into a projective space $P$ over $mathbb{Z}[i]$. Assume there given a projective hypersurface $Z subset P$ given by the vanishing of a homogeneous polynomial with coefficients in $mathbb{Z}[i]$, and that
$f(V) cap Z = phi$
We would like to consider these objects over a commutative ring $A$ with $1$, in which $-1$ is not a square. If we now let $V_A$ be the variety $V$ defined now over $A$, let $P_A$ and $Z_A$ be the projective varieties $P$ and $Z$ respectively defined now over $A[i]$, and denote by
$f_A: V_A to P_A$
the map corresponding to $f$. Under what conditions on $V$, $P$, $f$ and $A$, do we have
$f_A(V_A) cap Z_A = phi$ ?
algebraic-geometry commutative-algebra
asked Nov 20 at 15:17
Malkoun
1,8291612
add a comment |
Let $V$ be a Zariski open subset of an affine variety over $mathbb{Z}$, and let $f: V to P$ be a morphism into a projective space $P$ over $mathbb{Z}[i]$. Assume there given a projective hypersurface $Z subset P$ given by the vanishing of a homogeneous polynomial with coefficients in $mathbb{Z}[i]$, and that
$f(V) cap Z = phi$
We would like to consider these objects over a commutative ring $A$ with $1$, in which $-1$ is not a square. If we now let $V_A$ be the variety $V$ defined now over $A$, let $P_A$ and $Z_A$ be the projective varieties $P$ and $Z$ respectively defined now over $A[i]$, and denote by
$f_A: V_A to P_A$
the map corresponding to $f$. Under what conditions on $V$, $P$, $f$ and $A$, do we have
$f_A(V_A) cap Z_A = phi$ ?
algebraic-geometry commutative-algebra
asked Nov 20 at 15:17
Malkoun
1,8291612
Always? If $Bcap C = varnothing$ then $Atimes B cap Atimes C = varnothing$. Am I missing something?
– Ben
Nov 20 at 18:35
@Ben, I have thought about what you wrote. Consider the following. Let $f$ be the map from the affine line $V$ over $mathbb{Z}$ into $P^1$, which maps any point to the point $[1:3]$. Assume that $Z = {[1:0]}$. We have that $f(V) cap Z = phi$, but after reducing mod $3$, i.e. taking $A$ to be the integers modulo $3$, we get that the intersection is no longer empty.
– Malkoun
Nov 22 at 6:43
I have to admit that I have used the word "extension" which was probably confusing. I meant a change of base ring, which is not necessarily an extension of $mathbb{Z}$. I will fix that.
– Malkoun
Nov 22 at 6:55
add a comment |
Let $V$ be a Zariski open subset of an affine variety over $mathbb{Z}$, and let $f: V to P$ be a morphism into a projective space $P$ over $mathbb{Z}[i]$. Assume there given a projective hypersurface $Z subset P$ given by the vanishing of a homogeneous polynomial with coefficients in $mathbb{Z}[i]$, and that
$f(V) cap Z = phi$
We would like to consider these objects over a commutative ring $A$ with $1$, in which $-1$ is not a square. If we now let $V_A$ be the variety $V$ defined now over $A$, let $P_A$ and $Z_A$ be the projective varieties $P$ and $Z$ respectively defined now over $A[i]$, and denote by
$f_A: V_A to P_A$
the map corresponding to $f$. Under what conditions on $V$, $P$, $f$ and $A$, do we have
$f_A(V_A) cap Z_A = phi$ ?
algebraic-geometry commutative-algebra
asked Nov 20 at 15:17
Malkoun
1,8291612
Let $V$ be a Zariski open subset of an affine variety over $mathbb{Z}$, and let $f: V to P$ be a morphism into a projective space $P$ over $mathbb{Z}[i]$. Assume there given a projective hypersurface $Z subset P$ given by the vanishing of a homogeneous polynomial with coefficients in $mathbb{Z}[i]$, and that
$f(V) cap Z = phi$
We would like to consider these objects over a commutative ring $A$ with $1$, in which $-1$ is not a square. If we now let $V_A$ be the variety $V$ defined now over $A$, let $P_A$ and $Z_A$ be the projective varieties $P$ and $Z$ respectively defined now over $A[i]$, and denote by
$f_A: V_A to P_A$
the map corresponding to $f$. Under what conditions on $V$, $P$, $f$ and $A$, do we have
$f_A(V_A) cap Z_A = phi$ ?
algebraic-geometry commutative-algebra
algebraic-geometry commutative-algebra
asked Nov 20 at 15:17
Malkoun
1,8291612
asked Nov 20 at 15:17
Malkoun
1,8291612
edited Nov 22 at 6:55
asked Nov 20 at 15:17
Malkoun
1,8291612
asked Nov 20 at 15:17
Malkoun
1,8291612
asked Nov 20 at 15:17
Malkoun
1,8291612
1,8291612
Always? If $Bcap C = varnothing$ then $Atimes B cap Atimes C = varnothing$. Am I missing something?
– Ben
Nov 20 at 18:35
@Ben, I have thought about what you wrote. Consider the following. Let $f$ be the map from the affine line $V$ over $mathbb{Z}$ into $P^1$, which maps any point to the point $[1:3]$. Assume that $Z = {[1:0]}$. We have that $f(V) cap Z = phi$, but after reducing mod $3$, i.e. taking $A$ to be the integers modulo $3$, we get that the intersection is no longer empty.
– Malkoun
Nov 22 at 6:43
I have to admit that I have used the word "extension" which was probably confusing. I meant a change of base ring, which is not necessarily an extension of $mathbb{Z}$. I will fix that.
– Malkoun
Nov 22 at 6:55
add a comment |
Always? If $Bcap C = varnothing$ then $Atimes B cap Atimes C = varnothing$. Am I missing something?
– Ben
Nov 20 at 18:35
@Ben, I have thought about what you wrote. Consider the following. Let $f$ be the map from the affine line $V$ over $mathbb{Z}$ into $P^1$, which maps any point to the point $[1:3]$. Assume that $Z = {[1:0]}$. We have that $f(V) cap Z = phi$, but after reducing mod $3$, i.e. taking $A$ to be the integers modulo $3$, we get that the intersection is no longer empty.
– Malkoun
Nov 22 at 6:43
I have to admit that I have used the word "extension" which was probably confusing. I meant a change of base ring, which is not necessarily an extension of $mathbb{Z}$. I will fix that.
– Malkoun
Nov 22 at 6:55
Always? If $Bcap C = varnothing$ then $Atimes B cap Atimes C = varnothing$. Am I missing something?
– Ben
Nov 20 at 18:35
Always? If $Bcap C = varnothing$ then $Atimes B cap Atimes C = varnothing$. Am I missing something?
– Ben
Nov 20 at 18:35
@Ben, I have thought about what you wrote. Consider the following. Let $f$ be the map from the affine line $V$ over $mathbb{Z}$ into $P^1$, which maps any point to the point $[1:3]$. Assume that $Z = {[1:0]}$. We have that $f(V) cap Z = phi$, but after reducing mod $3$, i.e. taking $A$ to be the integers modulo $3$, we get that the intersection is no longer empty.
– Malkoun
Nov 22 at 6:43
@Ben, I have thought about what you wrote. Consider the following. Let $f$ be the map from the affine line $V$ over $mathbb{Z}$ into $P^1$, which maps any point to the point $[1:3]$. Assume that $Z = {[1:0]}$. We have that $f(V) cap Z = phi$, but after reducing mod $3$, i.e. taking $A$ to be the integers modulo $3$, we get that the intersection is no longer empty.
– Malkoun
Nov 22 at 6:43
I have to admit that I have used the word "extension" which was probably confusing. I meant a change of base ring, which is not necessarily an extension of $mathbb{Z}$. I will fix that.
– Malkoun
Nov 22 at 6:55
I have to admit that I have used the word "extension" which was probably confusing. I meant a change of base ring, which is not necessarily an extension of $mathbb{Z}$. I will fix that.
– Malkoun
Nov 22 at 6:55
add a comment |
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Always? If $Bcap C = varnothing$ then $Atimes B cap Atimes C = varnothing$. Am I missing something?
– Ben
Nov 20 at 18:35
@Ben, I have thought about what you wrote. Consider the following. Let $f$ be the map from the affine line $V$ over $mathbb{Z}$ into $P^1$, which maps any point to the point $[1:3]$. Assume that $Z = {[1:0]}$. We have that $f(V) cap Z = phi$, but after reducing mod $3$, i.e. taking $A$ to be the integers modulo $3$, we get that the intersection is no longer empty.
– Malkoun
Nov 22 at 6:43
I have to admit that I have used the word "extension" which was probably confusing. I meant a change of base ring, which is not necessarily an extension of $mathbb{Z}$. I will fix that.
– Malkoun
Nov 22 at 6:55