Given commutative rings $A, B$ and epimorphism $phi : A to B$, for $f in B$, does there exist prime ideals of...
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Let's say I have two commutative rings $A, B$ and a ring epimorphism $phi : A to B$. this induces a continuous map $phi^* : operatorname{Spec}(B) to operatorname{Spec}(A)$. Let $f in B$ and choose $x in phi^{-1}(f)$, is it true that for any $mathfrak{q} in overline{V({x})}$ there exists a $mathfrak{p} in overline{V({f})}$ such that $mathfrak{q} = phi^{-1}(mathfrak{p})$? Note that $overline{V({x})}$ and $overline{V({f})}$ denote the complements of $V({x})$ and $V({f})$ respectively.
If this statement is true, then how can I prove this? I don't immediately see any way to prove this since the only things we know about $mathfrak{q}$ is that it is a prime ideal of $A$ that doesn't contain $x$, so I don't see any way I could tell what $mathfrak{p}$ would even be given so little information.
abstract-algebra ring-theory commutative-algebra
asked Dec 13 '18 at 0:17
PerturbativePerturbative
4,33311551
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add a comment |
$begingroup$
Let's say I have two commutative rings $A, B$ and a ring epimorphism $phi : A to B$. this induces a continuous map $phi^* : operatorname{Spec}(B) to operatorname{Spec}(A)$. Let $f in B$ and choose $x in phi^{-1}(f)$, is it true that for any $mathfrak{q} in overline{V({x})}$ there exists a $mathfrak{p} in overline{V({f})}$ such that $mathfrak{q} = phi^{-1}(mathfrak{p})$? Note that $overline{V({x})}$ and $overline{V({f})}$ denote the complements of $V({x})$ and $V({f})$ respectively.
If this statement is true, then how can I prove this? I don't immediately see any way to prove this since the only things we know about $mathfrak{q}$ is that it is a prime ideal of $A$ that doesn't contain $x$, so I don't see any way I could tell what $mathfrak{p}$ would even be given so little information.
abstract-algebra ring-theory commutative-algebra
asked Dec 13 '18 at 0:17
PerturbativePerturbative
4,33311551
$endgroup$
add a comment |
$begingroup$
Let's say I have two commutative rings $A, B$ and a ring epimorphism $phi : A to B$. this induces a continuous map $phi^* : operatorname{Spec}(B) to operatorname{Spec}(A)$. Let $f in B$ and choose $x in phi^{-1}(f)$, is it true that for any $mathfrak{q} in overline{V({x})}$ there exists a $mathfrak{p} in overline{V({f})}$ such that $mathfrak{q} = phi^{-1}(mathfrak{p})$? Note that $overline{V({x})}$ and $overline{V({f})}$ denote the complements of $V({x})$ and $V({f})$ respectively.
If this statement is true, then how can I prove this? I don't immediately see any way to prove this since the only things we know about $mathfrak{q}$ is that it is a prime ideal of $A$ that doesn't contain $x$, so I don't see any way I could tell what $mathfrak{p}$ would even be given so little information.
abstract-algebra ring-theory commutative-algebra
asked Dec 13 '18 at 0:17
PerturbativePerturbative
4,33311551
$endgroup$
Let's say I have two commutative rings $A, B$ and a ring epimorphism $phi : A to B$. this induces a continuous map $phi^* : operatorname{Spec}(B) to operatorname{Spec}(A)$. Let $f in B$ and choose $x in phi^{-1}(f)$, is it true that for any $mathfrak{q} in overline{V({x})}$ there exists a $mathfrak{p} in overline{V({f})}$ such that $mathfrak{q} = phi^{-1}(mathfrak{p})$? Note that $overline{V({x})}$ and $overline{V({f})}$ denote the complements of $V({x})$ and $V({f})$ respectively.
If this statement is true, then how can I prove this? I don't immediately see any way to prove this since the only things we know about $mathfrak{q}$ is that it is a prime ideal of $A$ that doesn't contain $x$, so I don't see any way I could tell what $mathfrak{p}$ would even be given so little information.
abstract-algebra ring-theory commutative-algebra
abstract-algebra ring-theory commutative-algebra
asked Dec 13 '18 at 0:17
PerturbativePerturbative
4,33311551
asked Dec 13 '18 at 0:17
PerturbativePerturbative
4,33311551
edited Dec 13 '18 at 0:44
Perturbative
asked Dec 13 '18 at 0:17
PerturbativePerturbative
4,33311551
asked Dec 13 '18 at 0:17
PerturbativePerturbative
4,33311551
asked Dec 13 '18 at 0:17
PerturbativePerturbative
4,33311551
4,33311551
add a comment |
add a comment |
1 Answer
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This is not true, the morphism $phi^*$ is injective, but not necessarily surjective. Take $f=0$ and $x=0$, $V(0)=Spec(A)$. Then if $phi^*$ is not surjective take any $q$ which is not in its image.
Example $phi:mathbb{Z}rightarrowmathbb{Z}/p$ where $p$ is a prime.
answered Dec 13 '18 at 0:44
Tsemo AristideTsemo Aristide
57.9k11445
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1 Answer
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1 Answer
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$begingroup$
This is not true, the morphism $phi^*$ is injective, but not necessarily surjective. Take $f=0$ and $x=0$, $V(0)=Spec(A)$. Then if $phi^*$ is not surjective take any $q$ which is not in its image.
Example $phi:mathbb{Z}rightarrowmathbb{Z}/p$ where $p$ is a prime.
answered Dec 13 '18 at 0:44
Tsemo AristideTsemo Aristide
57.9k11445
$endgroup$
add a comment |
$begingroup$
This is not true, the morphism $phi^*$ is injective, but not necessarily surjective. Take $f=0$ and $x=0$, $V(0)=Spec(A)$. Then if $phi^*$ is not surjective take any $q$ which is not in its image.
Example $phi:mathbb{Z}rightarrowmathbb{Z}/p$ where $p$ is a prime.
answered Dec 13 '18 at 0:44
Tsemo AristideTsemo Aristide
57.9k11445
$endgroup$
add a comment |
$begingroup$
This is not true, the morphism $phi^*$ is injective, but not necessarily surjective. Take $f=0$ and $x=0$, $V(0)=Spec(A)$. Then if $phi^*$ is not surjective take any $q$ which is not in its image.
Example $phi:mathbb{Z}rightarrowmathbb{Z}/p$ where $p$ is a prime.
answered Dec 13 '18 at 0:44
Tsemo AristideTsemo Aristide
57.9k11445
$endgroup$
This is not true, the morphism $phi^*$ is injective, but not necessarily surjective. Take $f=0$ and $x=0$, $V(0)=Spec(A)$. Then if $phi^*$ is not surjective take any $q$ which is not in its image.
Example $phi:mathbb{Z}rightarrowmathbb{Z}/p$ where $p$ is a prime.
answered Dec 13 '18 at 0:44
Tsemo AristideTsemo Aristide
57.9k11445
answered Dec 13 '18 at 0:44
Tsemo AristideTsemo Aristide
57.9k11445
answered Dec 13 '18 at 0:44
Tsemo AristideTsemo Aristide
57.9k11445
answered Dec 13 '18 at 0:44
Tsemo AristideTsemo Aristide
57.9k11445
57.9k11445
add a comment |
add a comment |
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