Is extending an ideal through a ring homomorphism the same as through extension of scalars?
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Suppose we have commutative rings $A$ and $B$, a (maybe injective) ring homomorphism $f: A rightarrow B$ and and ideal $I subseteq A$. Is it true that $I^e cong I otimes_A B$, where $I^e$ denotes the extension of $I$ into $B$?
In other words, does the extended ideal have the same module structure as the module obtained through extension of scalars?
commutative-algebra
asked Nov 17 at 4:13
beanshadow
138212
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up vote
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Suppose we have commutative rings $A$ and $B$, a (maybe injective) ring homomorphism $f: A rightarrow B$ and and ideal $I subseteq A$. Is it true that $I^e cong I otimes_A B$, where $I^e$ denotes the extension of $I$ into $B$?
In other words, does the extended ideal have the same module structure as the module obtained through extension of scalars?
commutative-algebra
asked Nov 17 at 4:13
beanshadow
138212
Is it possible that this provides a counter-example? math.stackexchange.com/questions/2220392/… The correspondence between your terminology and from that example: your A is their R, your B is their N, your I is their (t). It seems to me like that example is saying $I otimes_A B rightarrow B$ is not injective, so it shouldn't be isomorphic to $I^e subseteq B$, or at least the isomorphism shouldn't be induced by the natural map from $I otimes_A B rightarrow B$. Does that seem right? (PS: If $A rightarrow B$ is flat, see Matsumura CRT Theorem 7.7.)
– CJD
Nov 17 at 4:39
1
For $A=k[x^2,x^3]$, $B=k[x]$, and $I=(x^2,x^3)$ we have $I^e=x^2Bsimeq B$ (as $B$-modules). If $Iotimes_ABsimeq B$ then the canonical morphism $Iotimes_ABto I^e$ is an isomorphism, a contradiction.
– user26857
Nov 18 at 23:25
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Suppose we have commutative rings $A$ and $B$, a (maybe injective) ring homomorphism $f: A rightarrow B$ and and ideal $I subseteq A$. Is it true that $I^e cong I otimes_A B$, where $I^e$ denotes the extension of $I$ into $B$?
In other words, does the extended ideal have the same module structure as the module obtained through extension of scalars?
commutative-algebra
asked Nov 17 at 4:13
beanshadow
138212
Suppose we have commutative rings $A$ and $B$, a (maybe injective) ring homomorphism $f: A rightarrow B$ and and ideal $I subseteq A$. Is it true that $I^e cong I otimes_A B$, where $I^e$ denotes the extension of $I$ into $B$?
In other words, does the extended ideal have the same module structure as the module obtained through extension of scalars?
commutative-algebra
commutative-algebra
asked Nov 17 at 4:13
beanshadow
138212
asked Nov 17 at 4:13
beanshadow
138212
asked Nov 17 at 4:13
beanshadow
138212
asked Nov 17 at 4:13
beanshadow
138212
asked Nov 17 at 4:13
beanshadow
138212
138212
Is it possible that this provides a counter-example? math.stackexchange.com/questions/2220392/… The correspondence between your terminology and from that example: your A is their R, your B is their N, your I is their (t). It seems to me like that example is saying $I otimes_A B rightarrow B$ is not injective, so it shouldn't be isomorphic to $I^e subseteq B$, or at least the isomorphism shouldn't be induced by the natural map from $I otimes_A B rightarrow B$. Does that seem right? (PS: If $A rightarrow B$ is flat, see Matsumura CRT Theorem 7.7.)
– CJD
Nov 17 at 4:39
1
For $A=k[x^2,x^3]$, $B=k[x]$, and $I=(x^2,x^3)$ we have $I^e=x^2Bsimeq B$ (as $B$-modules). If $Iotimes_ABsimeq B$ then the canonical morphism $Iotimes_ABto I^e$ is an isomorphism, a contradiction.
– user26857
Nov 18 at 23:25
add a comment |
Is it possible that this provides a counter-example? math.stackexchange.com/questions/2220392/… The correspondence between your terminology and from that example: your A is their R, your B is their N, your I is their (t). It seems to me like that example is saying $I otimes_A B rightarrow B$ is not injective, so it shouldn't be isomorphic to $I^e subseteq B$, or at least the isomorphism shouldn't be induced by the natural map from $I otimes_A B rightarrow B$. Does that seem right? (PS: If $A rightarrow B$ is flat, see Matsumura CRT Theorem 7.7.)
– CJD
Nov 17 at 4:39
1
For $A=k[x^2,x^3]$, $B=k[x]$, and $I=(x^2,x^3)$ we have $I^e=x^2Bsimeq B$ (as $B$-modules). If $Iotimes_ABsimeq B$ then the canonical morphism $Iotimes_ABto I^e$ is an isomorphism, a contradiction.
– user26857
Nov 18 at 23:25
Is it possible that this provides a counter-example? math.stackexchange.com/questions/2220392/… The correspondence between your terminology and from that example: your A is their R, your B is their N, your I is their (t). It seems to me like that example is saying $I otimes_A B rightarrow B$ is not injective, so it shouldn't be isomorphic to $I^e subseteq B$, or at least the isomorphism shouldn't be induced by the natural map from $I otimes_A B rightarrow B$. Does that seem right? (PS: If $A rightarrow B$ is flat, see Matsumura CRT Theorem 7.7.)
– CJD
Nov 17 at 4:39
Is it possible that this provides a counter-example? math.stackexchange.com/questions/2220392/… The correspondence between your terminology and from that example: your A is their R, your B is their N, your I is their (t). It seems to me like that example is saying $I otimes_A B rightarrow B$ is not injective, so it shouldn't be isomorphic to $I^e subseteq B$, or at least the isomorphism shouldn't be induced by the natural map from $I otimes_A B rightarrow B$. Does that seem right? (PS: If $A rightarrow B$ is flat, see Matsumura CRT Theorem 7.7.)
– CJD
Nov 17 at 4:39
1
1
For $A=k[x^2,x^3]$, $B=k[x]$, and $I=(x^2,x^3)$ we have $I^e=x^2Bsimeq B$ (as $B$-modules). If $Iotimes_ABsimeq B$ then the canonical morphism $Iotimes_ABto I^e$ is an isomorphism, a contradiction.
– user26857
Nov 18 at 23:25
For $A=k[x^2,x^3]$, $B=k[x]$, and $I=(x^2,x^3)$ we have $I^e=x^2Bsimeq B$ (as $B$-modules). If $Iotimes_ABsimeq B$ then the canonical morphism $Iotimes_ABto I^e$ is an isomorphism, a contradiction.
– user26857
Nov 18 at 23:25
add a comment |
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Is it possible that this provides a counter-example? math.stackexchange.com/questions/2220392/… The correspondence between your terminology and from that example: your A is their R, your B is their N, your I is their (t). It seems to me like that example is saying $I otimes_A B rightarrow B$ is not injective, so it shouldn't be isomorphic to $I^e subseteq B$, or at least the isomorphism shouldn't be induced by the natural map from $I otimes_A B rightarrow B$. Does that seem right? (PS: If $A rightarrow B$ is flat, see Matsumura CRT Theorem 7.7.)
– CJD
Nov 17 at 4:39
1
For $A=k[x^2,x^3]$, $B=k[x]$, and $I=(x^2,x^3)$ we have $I^e=x^2Bsimeq B$ (as $B$-modules). If $Iotimes_ABsimeq B$ then the canonical morphism $Iotimes_ABto I^e$ is an isomorphism, a contradiction.
– user26857
Nov 18 at 23:25