Show $A_{f(b)} cong A otimes _B B_g $
$begingroup$
Let $f: B to A$ a ring morphism and $g in B$.
My goal is to show that $A_{f(g)} cong A otimes _B B_g $
($B_g$ means localization at $g$)
Firstly using universal property of tensor product the canonical maps $A to A_{f(g)}, B_g to A_{f(g)}$ induce a map $h:A otimes _B B_g to A_{f(g)}$.
Obviously it's surjective since $A$ and the multiplicative system ${ f(g)^n vert n in mathbb{N}}$ are contained in the image.
Does anybody have an argument for injectivity?
ring-theory commutative-algebra tensor-products
edited Dec 27 '18 at 10:09
user26857
39.3k124183
asked Dec 26 '18 at 23:50
KarlPeterKarlPeter
6101315
$endgroup$
add a comment |
$begingroup$
Let $f: B to A$ a ring morphism and $g in B$.
My goal is to show that $A_{f(g)} cong A otimes _B B_g $
($B_g$ means localization at $g$)
Firstly using universal property of tensor product the canonical maps $A to A_{f(g)}, B_g to A_{f(g)}$ induce a map $h:A otimes _B B_g to A_{f(g)}$.
Obviously it's surjective since $A$ and the multiplicative system ${ f(g)^n vert n in mathbb{N}}$ are contained in the image.
Does anybody have an argument for injectivity?
ring-theory commutative-algebra tensor-products
edited Dec 27 '18 at 10:09
user26857
39.3k124183
asked Dec 26 '18 at 23:50
KarlPeterKarlPeter
6101315
$endgroup$
add a comment |
$begingroup$
Let $f: B to A$ a ring morphism and $g in B$.
My goal is to show that $A_{f(g)} cong A otimes _B B_g $
($B_g$ means localization at $g$)
Firstly using universal property of tensor product the canonical maps $A to A_{f(g)}, B_g to A_{f(g)}$ induce a map $h:A otimes _B B_g to A_{f(g)}$.
Obviously it's surjective since $A$ and the multiplicative system ${ f(g)^n vert n in mathbb{N}}$ are contained in the image.
Does anybody have an argument for injectivity?
ring-theory commutative-algebra tensor-products
edited Dec 27 '18 at 10:09
user26857
39.3k124183
asked Dec 26 '18 at 23:50
KarlPeterKarlPeter
6101315
$endgroup$
Let $f: B to A$ a ring morphism and $g in B$.
My goal is to show that $A_{f(g)} cong A otimes _B B_g $
($B_g$ means localization at $g$)
Firstly using universal property of tensor product the canonical maps $A to A_{f(g)}, B_g to A_{f(g)}$ induce a map $h:A otimes _B B_g to A_{f(g)}$.
Obviously it's surjective since $A$ and the multiplicative system ${ f(g)^n vert n in mathbb{N}}$ are contained in the image.
Does anybody have an argument for injectivity?
ring-theory commutative-algebra tensor-products
ring-theory commutative-algebra tensor-products
edited Dec 27 '18 at 10:09
user26857
39.3k124183
asked Dec 26 '18 at 23:50
KarlPeterKarlPeter
6101315
edited Dec 27 '18 at 10:09
user26857
39.3k124183
asked Dec 26 '18 at 23:50
KarlPeterKarlPeter
6101315
edited Dec 27 '18 at 10:09
user26857
39.3k124183
edited Dec 27 '18 at 10:09
user26857
39.3k124183
edited Dec 27 '18 at 10:09
user26857
39.3k124183
39.3k124183
asked Dec 26 '18 at 23:50
KarlPeterKarlPeter
6101315
asked Dec 26 '18 at 23:50
KarlPeterKarlPeter
6101315
asked Dec 26 '18 at 23:50
KarlPeterKarlPeter
6101315
6101315
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
This should follow from the following more general fact: Let $M$ be a $B$-module. Then, given any multiplicative system $S$ in $B$, we have
$$
S^{-1}Mcong Motimes_B S^{-1}B
$$
as $B$-modules.
In your particular case we take $S={1,g,g^2,dots}$ and $M=A$ as a $B$-module. All that'd remain to show is that this map is also a(n iso)morphism in whatever category you have in mind ($B$-algebras, commutative rings?).
For a proof of this standard result see Lemma 6.19 in these notes by Gathmann: https://www.mathematik.uni-kl.de/~gathmann/class/commalg-2013/commalg-2013-c6.pdf
answered Dec 27 '18 at 0:40
user347489user347489
1,188617
$endgroup$
add a comment |
$begingroup$
Consider the ring homomorphism $varphicolon Ato Aotimes_BB_g$, $amapsto aotimes(1/1)$.
Suppose $psicolon Ato C$ is a ring homomorphism, where $psi(f(g))$ is invertible.
By the property of $B_g$, there exists a unique homomorphism $taucolon B_gto C$ such that $tau(b/g^n)=psi(f(b))psi(f(g))^{-n}$.
Therefore there exists a unique homomorphism $hat{psi}colon Aotimes_BB_gto C$ such that
$$
hat{psi}(aotimes(b/g^n))=psi(a)tau(b/g^n)=psi(a)psi(f(b))psi(f(g))^{-n}
$$
and
$$
hat{psi}circvarphi(a)=psi(a)
$$
Therefore $Aotimes_BB_g$ satisfies the property required for $A_{f(g)}$.
answered Dec 27 '18 at 10:21
egregegreg
182k1485203
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
This should follow from the following more general fact: Let $M$ be a $B$-module. Then, given any multiplicative system $S$ in $B$, we have
$$
S^{-1}Mcong Motimes_B S^{-1}B
$$
as $B$-modules.
In your particular case we take $S={1,g,g^2,dots}$ and $M=A$ as a $B$-module. All that'd remain to show is that this map is also a(n iso)morphism in whatever category you have in mind ($B$-algebras, commutative rings?).
For a proof of this standard result see Lemma 6.19 in these notes by Gathmann: https://www.mathematik.uni-kl.de/~gathmann/class/commalg-2013/commalg-2013-c6.pdf
answered Dec 27 '18 at 0:40
user347489user347489
1,188617
$endgroup$
add a comment |
$begingroup$
This should follow from the following more general fact: Let $M$ be a $B$-module. Then, given any multiplicative system $S$ in $B$, we have
$$
S^{-1}Mcong Motimes_B S^{-1}B
$$
as $B$-modules.
In your particular case we take $S={1,g,g^2,dots}$ and $M=A$ as a $B$-module. All that'd remain to show is that this map is also a(n iso)morphism in whatever category you have in mind ($B$-algebras, commutative rings?).
For a proof of this standard result see Lemma 6.19 in these notes by Gathmann: https://www.mathematik.uni-kl.de/~gathmann/class/commalg-2013/commalg-2013-c6.pdf
answered Dec 27 '18 at 0:40
user347489user347489
1,188617
$endgroup$
add a comment |
$begingroup$
This should follow from the following more general fact: Let $M$ be a $B$-module. Then, given any multiplicative system $S$ in $B$, we have
$$
S^{-1}Mcong Motimes_B S^{-1}B
$$
as $B$-modules.
In your particular case we take $S={1,g,g^2,dots}$ and $M=A$ as a $B$-module. All that'd remain to show is that this map is also a(n iso)morphism in whatever category you have in mind ($B$-algebras, commutative rings?).
For a proof of this standard result see Lemma 6.19 in these notes by Gathmann: https://www.mathematik.uni-kl.de/~gathmann/class/commalg-2013/commalg-2013-c6.pdf
answered Dec 27 '18 at 0:40
user347489user347489
1,188617
$endgroup$
This should follow from the following more general fact: Let $M$ be a $B$-module. Then, given any multiplicative system $S$ in $B$, we have
$$
S^{-1}Mcong Motimes_B S^{-1}B
$$
as $B$-modules.
In your particular case we take $S={1,g,g^2,dots}$ and $M=A$ as a $B$-module. All that'd remain to show is that this map is also a(n iso)morphism in whatever category you have in mind ($B$-algebras, commutative rings?).
For a proof of this standard result see Lemma 6.19 in these notes by Gathmann: https://www.mathematik.uni-kl.de/~gathmann/class/commalg-2013/commalg-2013-c6.pdf
answered Dec 27 '18 at 0:40
user347489user347489
1,188617
answered Dec 27 '18 at 0:40
user347489user347489
1,188617
answered Dec 27 '18 at 0:40
user347489user347489
1,188617
answered Dec 27 '18 at 0:40
user347489user347489
1,188617
1,188617
add a comment |
add a comment |
$begingroup$
Consider the ring homomorphism $varphicolon Ato Aotimes_BB_g$, $amapsto aotimes(1/1)$.
Suppose $psicolon Ato C$ is a ring homomorphism, where $psi(f(g))$ is invertible.
By the property of $B_g$, there exists a unique homomorphism $taucolon B_gto C$ such that $tau(b/g^n)=psi(f(b))psi(f(g))^{-n}$.
Therefore there exists a unique homomorphism $hat{psi}colon Aotimes_BB_gto C$ such that
$$
hat{psi}(aotimes(b/g^n))=psi(a)tau(b/g^n)=psi(a)psi(f(b))psi(f(g))^{-n}
$$
and
$$
hat{psi}circvarphi(a)=psi(a)
$$
Therefore $Aotimes_BB_g$ satisfies the property required for $A_{f(g)}$.
answered Dec 27 '18 at 10:21
egregegreg
182k1485203
$endgroup$
add a comment |
$begingroup$
Consider the ring homomorphism $varphicolon Ato Aotimes_BB_g$, $amapsto aotimes(1/1)$.
Suppose $psicolon Ato C$ is a ring homomorphism, where $psi(f(g))$ is invertible.
By the property of $B_g$, there exists a unique homomorphism $taucolon B_gto C$ such that $tau(b/g^n)=psi(f(b))psi(f(g))^{-n}$.
Therefore there exists a unique homomorphism $hat{psi}colon Aotimes_BB_gto C$ such that
$$
hat{psi}(aotimes(b/g^n))=psi(a)tau(b/g^n)=psi(a)psi(f(b))psi(f(g))^{-n}
$$
and
$$
hat{psi}circvarphi(a)=psi(a)
$$
Therefore $Aotimes_BB_g$ satisfies the property required for $A_{f(g)}$.
answered Dec 27 '18 at 10:21
egregegreg
182k1485203
$endgroup$
add a comment |
$begingroup$
Consider the ring homomorphism $varphicolon Ato Aotimes_BB_g$, $amapsto aotimes(1/1)$.
Suppose $psicolon Ato C$ is a ring homomorphism, where $psi(f(g))$ is invertible.
By the property of $B_g$, there exists a unique homomorphism $taucolon B_gto C$ such that $tau(b/g^n)=psi(f(b))psi(f(g))^{-n}$.
Therefore there exists a unique homomorphism $hat{psi}colon Aotimes_BB_gto C$ such that
$$
hat{psi}(aotimes(b/g^n))=psi(a)tau(b/g^n)=psi(a)psi(f(b))psi(f(g))^{-n}
$$
and
$$
hat{psi}circvarphi(a)=psi(a)
$$
Therefore $Aotimes_BB_g$ satisfies the property required for $A_{f(g)}$.
answered Dec 27 '18 at 10:21
egregegreg
182k1485203
$endgroup$
Consider the ring homomorphism $varphicolon Ato Aotimes_BB_g$, $amapsto aotimes(1/1)$.
Suppose $psicolon Ato C$ is a ring homomorphism, where $psi(f(g))$ is invertible.
By the property of $B_g$, there exists a unique homomorphism $taucolon B_gto C$ such that $tau(b/g^n)=psi(f(b))psi(f(g))^{-n}$.
Therefore there exists a unique homomorphism $hat{psi}colon Aotimes_BB_gto C$ such that
$$
hat{psi}(aotimes(b/g^n))=psi(a)tau(b/g^n)=psi(a)psi(f(b))psi(f(g))^{-n}
$$
and
$$
hat{psi}circvarphi(a)=psi(a)
$$
Therefore $Aotimes_BB_g$ satisfies the property required for $A_{f(g)}$.
answered Dec 27 '18 at 10:21
egregegreg
182k1485203
answered Dec 27 '18 at 10:21
egregegreg
182k1485203
answered Dec 27 '18 at 10:21
egregegreg
182k1485203
answered Dec 27 '18 at 10:21
egregegreg
182k1485203
182k1485203
add a comment |
add a comment |
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