How to show the canonical homomorphism $A_{mathfrak p}to B_{mathfrak q}$ is injective?
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Let $A,B$ be commutative rings with identity and $f:Ato B$ an injective homomorphism of rings. For any prime ideal $mathfrak q$ of $B$, denote $f^{-1}(mathfrak q)$ by $mathfrak p$, how to show the canonical homomorphism $A_{mathfrak p}to B_{mathfrak q}$ is injective?
abstract-algebra ring-theory commutative-algebra
asked Nov 14 at 5:13
Born to be proud
770510
add a comment |
up vote
0
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Let $A,B$ be commutative rings with identity and $f:Ato B$ an injective homomorphism of rings. For any prime ideal $mathfrak q$ of $B$, denote $f^{-1}(mathfrak q)$ by $mathfrak p$, how to show the canonical homomorphism $A_{mathfrak p}to B_{mathfrak q}$ is injective?
abstract-algebra ring-theory commutative-algebra
asked Nov 14 at 5:13
Born to be proud
770510
This seems like the kind of thing that might follow from the universal property. Did you try using that?
– Alfred Yerger
Nov 14 at 5:50
@AlfredYerger Yes, I factor it as $A_{mathfrak p}to B_{mathfrak p}to B_{mathfrak q}$, the first is injective, but I wonder if the second is injective.
– Born to be proud
Nov 14 at 5:57
@AlfredYerger I wonder if jgon's answer is correct. math.stackexchange.com/questions/2997043/…
– Born to be proud
Nov 14 at 6:02
Related
– André 3000
Nov 14 at 7:41
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $A,B$ be commutative rings with identity and $f:Ato B$ an injective homomorphism of rings. For any prime ideal $mathfrak q$ of $B$, denote $f^{-1}(mathfrak q)$ by $mathfrak p$, how to show the canonical homomorphism $A_{mathfrak p}to B_{mathfrak q}$ is injective?
abstract-algebra ring-theory commutative-algebra
asked Nov 14 at 5:13
Born to be proud
770510
Let $A,B$ be commutative rings with identity and $f:Ato B$ an injective homomorphism of rings. For any prime ideal $mathfrak q$ of $B$, denote $f^{-1}(mathfrak q)$ by $mathfrak p$, how to show the canonical homomorphism $A_{mathfrak p}to B_{mathfrak q}$ is injective?
abstract-algebra ring-theory commutative-algebra
abstract-algebra ring-theory commutative-algebra
asked Nov 14 at 5:13
Born to be proud
770510
asked Nov 14 at 5:13
Born to be proud
770510
asked Nov 14 at 5:13
Born to be proud
770510
asked Nov 14 at 5:13
Born to be proud
770510
asked Nov 14 at 5:13
Born to be proud
770510
770510
This seems like the kind of thing that might follow from the universal property. Did you try using that?
– Alfred Yerger
Nov 14 at 5:50
@AlfredYerger Yes, I factor it as $A_{mathfrak p}to B_{mathfrak p}to B_{mathfrak q}$, the first is injective, but I wonder if the second is injective.
– Born to be proud
Nov 14 at 5:57
@AlfredYerger I wonder if jgon's answer is correct. math.stackexchange.com/questions/2997043/…
– Born to be proud
Nov 14 at 6:02
Related
– André 3000
Nov 14 at 7:41
add a comment |
This seems like the kind of thing that might follow from the universal property. Did you try using that?
– Alfred Yerger
Nov 14 at 5:50
@AlfredYerger Yes, I factor it as $A_{mathfrak p}to B_{mathfrak p}to B_{mathfrak q}$, the first is injective, but I wonder if the second is injective.
– Born to be proud
Nov 14 at 5:57
@AlfredYerger I wonder if jgon's answer is correct. math.stackexchange.com/questions/2997043/…
– Born to be proud
Nov 14 at 6:02
Related
– André 3000
Nov 14 at 7:41
This seems like the kind of thing that might follow from the universal property. Did you try using that?
– Alfred Yerger
Nov 14 at 5:50
This seems like the kind of thing that might follow from the universal property. Did you try using that?
– Alfred Yerger
Nov 14 at 5:50
@AlfredYerger Yes, I factor it as $A_{mathfrak p}to B_{mathfrak p}to B_{mathfrak q}$, the first is injective, but I wonder if the second is injective.
– Born to be proud
Nov 14 at 5:57
@AlfredYerger Yes, I factor it as $A_{mathfrak p}to B_{mathfrak p}to B_{mathfrak q}$, the first is injective, but I wonder if the second is injective.
– Born to be proud
Nov 14 at 5:57
@AlfredYerger I wonder if jgon's answer is correct. math.stackexchange.com/questions/2997043/…
– Born to be proud
Nov 14 at 6:02
@AlfredYerger I wonder if jgon's answer is correct. math.stackexchange.com/questions/2997043/…
– Born to be proud
Nov 14 at 6:02
Related
– André 3000
Nov 14 at 7:41
Related
– André 3000
Nov 14 at 7:41
add a comment |
1 Answer
1
active
oldest
votes
up vote
6
down vote
accepted
This is false. For instance, let $k$ be a field, let $A=k[x]$ and $B=k[x,y]/(xy)$, and let $f:Ato B$ be the obvious inclusion. Then the ideal $mathfrak{q}=xBsubset B$ is prime, and the induced map $A_{f^{-1}(mathfrak{q})}to B_mathfrak{q}$ is not injective: $x$ is nonzero in $A_{f^{-1}(mathfrak{q})}$ but is $0$ in $B_mathfrak{q}$ since $xy=0$ and $ynotinmathfrak{q}$.
answered Nov 14 at 6:06
Eric Wofsey
175k12202326
Nice! I was typing this exact example, but you beat me to it.
– Alex Wertheim
Nov 14 at 6:07
1
I wonder if jgon's answer is correct, please help me check it, thank you. math.stackexchange.com/questions/2997043/…
– Born to be proud
Nov 14 at 6:15
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
6
down vote
accepted
This is false. For instance, let $k$ be a field, let $A=k[x]$ and $B=k[x,y]/(xy)$, and let $f:Ato B$ be the obvious inclusion. Then the ideal $mathfrak{q}=xBsubset B$ is prime, and the induced map $A_{f^{-1}(mathfrak{q})}to B_mathfrak{q}$ is not injective: $x$ is nonzero in $A_{f^{-1}(mathfrak{q})}$ but is $0$ in $B_mathfrak{q}$ since $xy=0$ and $ynotinmathfrak{q}$.
answered Nov 14 at 6:06
Eric Wofsey
175k12202326
Nice! I was typing this exact example, but you beat me to it.
– Alex Wertheim
Nov 14 at 6:07
1
I wonder if jgon's answer is correct, please help me check it, thank you. math.stackexchange.com/questions/2997043/…
– Born to be proud
Nov 14 at 6:15
add a comment |
up vote
6
down vote
accepted
This is false. For instance, let $k$ be a field, let $A=k[x]$ and $B=k[x,y]/(xy)$, and let $f:Ato B$ be the obvious inclusion. Then the ideal $mathfrak{q}=xBsubset B$ is prime, and the induced map $A_{f^{-1}(mathfrak{q})}to B_mathfrak{q}$ is not injective: $x$ is nonzero in $A_{f^{-1}(mathfrak{q})}$ but is $0$ in $B_mathfrak{q}$ since $xy=0$ and $ynotinmathfrak{q}$.
answered Nov 14 at 6:06
Eric Wofsey
175k12202326
Nice! I was typing this exact example, but you beat me to it.
– Alex Wertheim
Nov 14 at 6:07
1
I wonder if jgon's answer is correct, please help me check it, thank you. math.stackexchange.com/questions/2997043/…
– Born to be proud
Nov 14 at 6:15
add a comment |
up vote
6
down vote
accepted
up vote
6
down vote
accepted
This is false. For instance, let $k$ be a field, let $A=k[x]$ and $B=k[x,y]/(xy)$, and let $f:Ato B$ be the obvious inclusion. Then the ideal $mathfrak{q}=xBsubset B$ is prime, and the induced map $A_{f^{-1}(mathfrak{q})}to B_mathfrak{q}$ is not injective: $x$ is nonzero in $A_{f^{-1}(mathfrak{q})}$ but is $0$ in $B_mathfrak{q}$ since $xy=0$ and $ynotinmathfrak{q}$.
answered Nov 14 at 6:06
Eric Wofsey
175k12202326
This is false. For instance, let $k$ be a field, let $A=k[x]$ and $B=k[x,y]/(xy)$, and let $f:Ato B$ be the obvious inclusion. Then the ideal $mathfrak{q}=xBsubset B$ is prime, and the induced map $A_{f^{-1}(mathfrak{q})}to B_mathfrak{q}$ is not injective: $x$ is nonzero in $A_{f^{-1}(mathfrak{q})}$ but is $0$ in $B_mathfrak{q}$ since $xy=0$ and $ynotinmathfrak{q}$.
answered Nov 14 at 6:06
Eric Wofsey
175k12202326
answered Nov 14 at 6:06
Eric Wofsey
175k12202326
answered Nov 14 at 6:06
Eric Wofsey
175k12202326
answered Nov 14 at 6:06
Eric Wofsey
175k12202326
175k12202326
Nice! I was typing this exact example, but you beat me to it.
– Alex Wertheim
Nov 14 at 6:07
1
I wonder if jgon's answer is correct, please help me check it, thank you. math.stackexchange.com/questions/2997043/…
– Born to be proud
Nov 14 at 6:15
add a comment |
Nice! I was typing this exact example, but you beat me to it.
– Alex Wertheim
Nov 14 at 6:07
1
I wonder if jgon's answer is correct, please help me check it, thank you. math.stackexchange.com/questions/2997043/…
– Born to be proud
Nov 14 at 6:15
Nice! I was typing this exact example, but you beat me to it.
– Alex Wertheim
Nov 14 at 6:07
Nice! I was typing this exact example, but you beat me to it.
– Alex Wertheim
Nov 14 at 6:07
1
1
I wonder if jgon's answer is correct, please help me check it, thank you. math.stackexchange.com/questions/2997043/…
– Born to be proud
Nov 14 at 6:15
I wonder if jgon's answer is correct, please help me check it, thank you. math.stackexchange.com/questions/2997043/…
– Born to be proud
Nov 14 at 6:15
add a comment |
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This seems like the kind of thing that might follow from the universal property. Did you try using that?
– Alfred Yerger
Nov 14 at 5:50
@AlfredYerger Yes, I factor it as $A_{mathfrak p}to B_{mathfrak p}to B_{mathfrak q}$, the first is injective, but I wonder if the second is injective.
– Born to be proud
Nov 14 at 5:57
@AlfredYerger I wonder if jgon's answer is correct. math.stackexchange.com/questions/2997043/…
– Born to be proud
Nov 14 at 6:02
Related
– André 3000
Nov 14 at 7:41