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Birational and faithfully flat $implies$ isomorphism
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up vote
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Let $A subseteq B$ be integral domains with the same field of fractions. Assume that $A to B$ is faithfully flat. Why do we have $A=B$?
This is an exercise in Matsumura's book. Here is my idea: If $b in B$, consider $I = a in A : ab in A$. This is an ideal of $A$. By asumption $I neq 0$, and our goal is to show that $I=A$. It suffices to prove $IB=B$. But how can we achieve this?
commutative-algebra
asked Sep 3 '13 at 10:51
Martin Brandenburg
105k13150316
 |Â
show 2 more comments
up vote
11
down vote
favorite
Let $A subseteq B$ be integral domains with the same field of fractions. Assume that $A to B$ is faithfully flat. Why do we have $A=B$?
This is an exercise in Matsumura's book. Here is my idea: If $b in B$, consider $I = a in A : ab in A$. This is an ideal of $A$. By asumption $I neq 0$, and our goal is to show that $I=A$. It suffices to prove $IB=B$. But how can we achieve this?
commutative-algebra
asked Sep 3 '13 at 10:51
Martin Brandenburg
105k13150316
1
This is just a hunch, but if they are integral domains with the same field of fractions, wouldn't $B$ be a localization of $A$?
– Arthur
Sep 3 '13 at 11:28
@Arthur: No. This holds in some special cases (for example when $A$ is a PID), but it fails in general. Birational morphisms can be quite complicated.
– Martin Brandenburg
Sep 3 '13 at 13:12
Perhaps Arthur means localisation in the broad sense of $A[S^-1]$ for some multiplicatively-closed set $S$. Surely this is true, by taking $S$ to be the set of units in $B$?
– Zhen Lin
Sep 3 '13 at 13:18
@ZhenLin No: math.stackexchange.com/a/287259/38268
– user38268
Sep 3 '13 at 13:21
1
Hmmm, pity. Otherwise we could just base change to $B$ and use faithful-flatness to deduce that $operatornamecoker f = 0$.
– Zhen Lin
Sep 3 '13 at 13:22
 |Â
show 2 more comments
up vote
11
down vote
favorite
up vote
11
down vote
favorite
Let $A subseteq B$ be integral domains with the same field of fractions. Assume that $A to B$ is faithfully flat. Why do we have $A=B$?
This is an exercise in Matsumura's book. Here is my idea: If $b in B$, consider $I = a in A : ab in A$. This is an ideal of $A$. By asumption $I neq 0$, and our goal is to show that $I=A$. It suffices to prove $IB=B$. But how can we achieve this?
commutative-algebra
asked Sep 3 '13 at 10:51
Martin Brandenburg
105k13150316
Let $A subseteq B$ be integral domains with the same field of fractions. Assume that $A to B$ is faithfully flat. Why do we have $A=B$?
This is an exercise in Matsumura's book. Here is my idea: If $b in B$, consider $I = a in A : ab in A$. This is an ideal of $A$. By asumption $I neq 0$, and our goal is to show that $I=A$. It suffices to prove $IB=B$. But how can we achieve this?
commutative-algebra
asked Sep 3 '13 at 10:51
Martin Brandenburg
105k13150316
edited Sep 3 '13 at 13:01
user38268
asked Sep 3 '13 at 10:51
Martin Brandenburg
105k13150316
asked Sep 3 '13 at 10:51
Martin Brandenburg
105k13150316
asked Sep 3 '13 at 10:51
Martin Brandenburg
105k13150316
105k13150316
1
This is just a hunch, but if they are integral domains with the same field of fractions, wouldn't $B$ be a localization of $A$?
– Arthur
Sep 3 '13 at 11:28
@Arthur: No. This holds in some special cases (for example when $A$ is a PID), but it fails in general. Birational morphisms can be quite complicated.
– Martin Brandenburg
Sep 3 '13 at 13:12
Perhaps Arthur means localisation in the broad sense of $A[S^-1]$ for some multiplicatively-closed set $S$. Surely this is true, by taking $S$ to be the set of units in $B$?
– Zhen Lin
Sep 3 '13 at 13:18
@ZhenLin No: math.stackexchange.com/a/287259/38268
– user38268
Sep 3 '13 at 13:21
1
Hmmm, pity. Otherwise we could just base change to $B$ and use faithful-flatness to deduce that $operatornamecoker f = 0$.
– Zhen Lin
Sep 3 '13 at 13:22
 |Â
show 2 more comments
1
This is just a hunch, but if they are integral domains with the same field of fractions, wouldn't $B$ be a localization of $A$?
– Arthur
Sep 3 '13 at 11:28
@Arthur: No. This holds in some special cases (for example when $A$ is a PID), but it fails in general. Birational morphisms can be quite complicated.
– Martin Brandenburg
Sep 3 '13 at 13:12
Perhaps Arthur means localisation in the broad sense of $A[S^-1]$ for some multiplicatively-closed set $S$. Surely this is true, by taking $S$ to be the set of units in $B$?
– Zhen Lin
Sep 3 '13 at 13:18
@ZhenLin No: math.stackexchange.com/a/287259/38268
– user38268
Sep 3 '13 at 13:21
1
Hmmm, pity. Otherwise we could just base change to $B$ and use faithful-flatness to deduce that $operatornamecoker f = 0$.
– Zhen Lin
Sep 3 '13 at 13:22
1
1
This is just a hunch, but if they are integral domains with the same field of fractions, wouldn't $B$ be a localization of $A$?
– Arthur
Sep 3 '13 at 11:28
This is just a hunch, but if they are integral domains with the same field of fractions, wouldn't $B$ be a localization of $A$?
– Arthur
Sep 3 '13 at 11:28
@Arthur: No. This holds in some special cases (for example when $A$ is a PID), but it fails in general. Birational morphisms can be quite complicated.
– Martin Brandenburg
Sep 3 '13 at 13:12
@Arthur: No. This holds in some special cases (for example when $A$ is a PID), but it fails in general. Birational morphisms can be quite complicated.
– Martin Brandenburg
Sep 3 '13 at 13:12
Perhaps Arthur means localisation in the broad sense of $A[S^-1]$ for some multiplicatively-closed set $S$. Surely this is true, by taking $S$ to be the set of units in $B$?
– Zhen Lin
Sep 3 '13 at 13:18
Perhaps Arthur means localisation in the broad sense of $A[S^-1]$ for some multiplicatively-closed set $S$. Surely this is true, by taking $S$ to be the set of units in $B$?
– Zhen Lin
Sep 3 '13 at 13:18
@ZhenLin No: math.stackexchange.com/a/287259/38268
– user38268
Sep 3 '13 at 13:21
@ZhenLin No: math.stackexchange.com/a/287259/38268
– user38268
Sep 3 '13 at 13:21
1
1
Hmmm, pity. Otherwise we could just base change to $B$ and use faithful-flatness to deduce that $operatornamecoker f = 0$.
– Zhen Lin
Sep 3 '13 at 13:22
Hmmm, pity. Otherwise we could just base change to $B$ and use faithful-flatness to deduce that $operatornamecoker f = 0$.
– Zhen Lin
Sep 3 '13 at 13:22
 |Â
show 2 more comments
2 Answers
2
active
oldest
votes
up vote
7
down vote
accepted
We can show $I = A$ as follows, the inclusion $I subseteq A$ being clear. Choose $a in A$ and write $b = a_1/a_2$. Then $aba_2 = acdot a_1 in A$ and so $acdot a_1 in a a_2 B cap A = aa_2 A$ where the last equality comes from faithful flatness. It follows that $aba_2 in aa_2A$ and so $aba_2 = a a_2 a'$ for some $a' in A$. Thus $ab = a a' in A$ and so indeed $a in I$. Thus $I = A$.
edited Sep 3 '13 at 15:07
Martin Brandenburg
105k13150316
2
Of course it suffices to take $a=1$. So the proof is just: If $b=a_1/a_2$, then $a_1 in a_2 B cap A = a_2 A$, hence $b in A$. I should have seen this!
– Martin Brandenburg
Sep 3 '13 at 21:13
add a comment |Â
up vote
1
down vote
As pointed out in the comments, the problem is easy if we know that $B$ is a localisation of $A$, because then the codiagonal $nabla : B otimes_A B to B$ is an isomorphism, and so we can (co)base change $A to B$ along itself to deduce that $A to B$ is an isomorphism. (This is where we use the fact that $A to B$ is faithfully flat.) But in fact $nabla : B otimes_A B to B$ is an isomorphism if and only if $A to B$ is an epimorphism, and this is certainly true if $B$ is flat and embeds in $operatornameFrac A$ as an $A$-algebra. (See comments below.)
answered Sep 3 '13 at 14:16
Zhen Lin
59k4100206
Ok. But for the last claim we also need that $B$ is flat over $A$, right?
– Martin Brandenburg
Sep 3 '13 at 14:34
1
Ah, sorry, I was thinking that the usual proof that $A to operatornameFrac A$ is an epimorphism would go through here as well, but it seems a more subtle approach is needed. Hmmm...
– Zhen Lin
Sep 3 '13 at 14:40
2
Yes. I meant that $B otimes_A B to Q(A) otimes_A Q(A)$ is injective (flatness), hence $b otimes 1 = 1 otimes b$ holds in $B otimes_A B$.
– Martin Brandenburg
Sep 3 '13 at 15:05
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
7
down vote
accepted
We can show $I = A$ as follows, the inclusion $I subseteq A$ being clear. Choose $a in A$ and write $b = a_1/a_2$. Then $aba_2 = acdot a_1 in A$ and so $acdot a_1 in a a_2 B cap A = aa_2 A$ where the last equality comes from faithful flatness. It follows that $aba_2 in aa_2A$ and so $aba_2 = a a_2 a'$ for some $a' in A$. Thus $ab = a a' in A$ and so indeed $a in I$. Thus $I = A$.
edited Sep 3 '13 at 15:07
Martin Brandenburg
105k13150316
2
Of course it suffices to take $a=1$. So the proof is just: If $b=a_1/a_2$, then $a_1 in a_2 B cap A = a_2 A$, hence $b in A$. I should have seen this!
– Martin Brandenburg
Sep 3 '13 at 21:13
add a comment |Â
up vote
7
down vote
accepted
We can show $I = A$ as follows, the inclusion $I subseteq A$ being clear. Choose $a in A$ and write $b = a_1/a_2$. Then $aba_2 = acdot a_1 in A$ and so $acdot a_1 in a a_2 B cap A = aa_2 A$ where the last equality comes from faithful flatness. It follows that $aba_2 in aa_2A$ and so $aba_2 = a a_2 a'$ for some $a' in A$. Thus $ab = a a' in A$ and so indeed $a in I$. Thus $I = A$.
edited Sep 3 '13 at 15:07
Martin Brandenburg
105k13150316
2
Of course it suffices to take $a=1$. So the proof is just: If $b=a_1/a_2$, then $a_1 in a_2 B cap A = a_2 A$, hence $b in A$. I should have seen this!
– Martin Brandenburg
Sep 3 '13 at 21:13
add a comment |Â
up vote
7
down vote
accepted
up vote
7
down vote
accepted
We can show $I = A$ as follows, the inclusion $I subseteq A$ being clear. Choose $a in A$ and write $b = a_1/a_2$. Then $aba_2 = acdot a_1 in A$ and so $acdot a_1 in a a_2 B cap A = aa_2 A$ where the last equality comes from faithful flatness. It follows that $aba_2 in aa_2A$ and so $aba_2 = a a_2 a'$ for some $a' in A$. Thus $ab = a a' in A$ and so indeed $a in I$. Thus $I = A$.
edited Sep 3 '13 at 15:07
Martin Brandenburg
105k13150316
We can show $I = A$ as follows, the inclusion $I subseteq A$ being clear. Choose $a in A$ and write $b = a_1/a_2$. Then $aba_2 = acdot a_1 in A$ and so $acdot a_1 in a a_2 B cap A = aa_2 A$ where the last equality comes from faithful flatness. It follows that $aba_2 in aa_2A$ and so $aba_2 = a a_2 a'$ for some $a' in A$. Thus $ab = a a' in A$ and so indeed $a in I$. Thus $I = A$.
edited Sep 3 '13 at 15:07
Martin Brandenburg
105k13150316
edited Sep 3 '13 at 15:07
Martin Brandenburg
105k13150316
edited Sep 3 '13 at 15:07
Martin Brandenburg
105k13150316
edited Sep 3 '13 at 15:07
Martin Brandenburg
105k13150316
105k13150316
answered Sep 3 '13 at 14:49
user38268
2
Of course it suffices to take $a=1$. So the proof is just: If $b=a_1/a_2$, then $a_1 in a_2 B cap A = a_2 A$, hence $b in A$. I should have seen this!
– Martin Brandenburg
Sep 3 '13 at 21:13
add a comment |Â
2
Of course it suffices to take $a=1$. So the proof is just: If $b=a_1/a_2$, then $a_1 in a_2 B cap A = a_2 A$, hence $b in A$. I should have seen this!
– Martin Brandenburg
Sep 3 '13 at 21:13
2
2
Of course it suffices to take $a=1$. So the proof is just: If $b=a_1/a_2$, then $a_1 in a_2 B cap A = a_2 A$, hence $b in A$. I should have seen this!
– Martin Brandenburg
Sep 3 '13 at 21:13
Of course it suffices to take $a=1$. So the proof is just: If $b=a_1/a_2$, then $a_1 in a_2 B cap A = a_2 A$, hence $b in A$. I should have seen this!
– Martin Brandenburg
Sep 3 '13 at 21:13
add a comment |Â
up vote
1
down vote
As pointed out in the comments, the problem is easy if we know that $B$ is a localisation of $A$, because then the codiagonal $nabla : B otimes_A B to B$ is an isomorphism, and so we can (co)base change $A to B$ along itself to deduce that $A to B$ is an isomorphism. (This is where we use the fact that $A to B$ is faithfully flat.) But in fact $nabla : B otimes_A B to B$ is an isomorphism if and only if $A to B$ is an epimorphism, and this is certainly true if $B$ is flat and embeds in $operatornameFrac A$ as an $A$-algebra. (See comments below.)
answered Sep 3 '13 at 14:16
Zhen Lin
59k4100206
Ok. But for the last claim we also need that $B$ is flat over $A$, right?
– Martin Brandenburg
Sep 3 '13 at 14:34
1
Ah, sorry, I was thinking that the usual proof that $A to operatornameFrac A$ is an epimorphism would go through here as well, but it seems a more subtle approach is needed. Hmmm...
– Zhen Lin
Sep 3 '13 at 14:40
2
Yes. I meant that $B otimes_A B to Q(A) otimes_A Q(A)$ is injective (flatness), hence $b otimes 1 = 1 otimes b$ holds in $B otimes_A B$.
– Martin Brandenburg
Sep 3 '13 at 15:05
add a comment |Â
up vote
1
down vote
As pointed out in the comments, the problem is easy if we know that $B$ is a localisation of $A$, because then the codiagonal $nabla : B otimes_A B to B$ is an isomorphism, and so we can (co)base change $A to B$ along itself to deduce that $A to B$ is an isomorphism. (This is where we use the fact that $A to B$ is faithfully flat.) But in fact $nabla : B otimes_A B to B$ is an isomorphism if and only if $A to B$ is an epimorphism, and this is certainly true if $B$ is flat and embeds in $operatornameFrac A$ as an $A$-algebra. (See comments below.)
answered Sep 3 '13 at 14:16
Zhen Lin
59k4100206
Ok. But for the last claim we also need that $B$ is flat over $A$, right?
– Martin Brandenburg
Sep 3 '13 at 14:34
1
Ah, sorry, I was thinking that the usual proof that $A to operatornameFrac A$ is an epimorphism would go through here as well, but it seems a more subtle approach is needed. Hmmm...
– Zhen Lin
Sep 3 '13 at 14:40
2
Yes. I meant that $B otimes_A B to Q(A) otimes_A Q(A)$ is injective (flatness), hence $b otimes 1 = 1 otimes b$ holds in $B otimes_A B$.
– Martin Brandenburg
Sep 3 '13 at 15:05
add a comment |Â
up vote
1
down vote
up vote
1
down vote
As pointed out in the comments, the problem is easy if we know that $B$ is a localisation of $A$, because then the codiagonal $nabla : B otimes_A B to B$ is an isomorphism, and so we can (co)base change $A to B$ along itself to deduce that $A to B$ is an isomorphism. (This is where we use the fact that $A to B$ is faithfully flat.) But in fact $nabla : B otimes_A B to B$ is an isomorphism if and only if $A to B$ is an epimorphism, and this is certainly true if $B$ is flat and embeds in $operatornameFrac A$ as an $A$-algebra. (See comments below.)
answered Sep 3 '13 at 14:16
Zhen Lin
59k4100206
As pointed out in the comments, the problem is easy if we know that $B$ is a localisation of $A$, because then the codiagonal $nabla : B otimes_A B to B$ is an isomorphism, and so we can (co)base change $A to B$ along itself to deduce that $A to B$ is an isomorphism. (This is where we use the fact that $A to B$ is faithfully flat.) But in fact $nabla : B otimes_A B to B$ is an isomorphism if and only if $A to B$ is an epimorphism, and this is certainly true if $B$ is flat and embeds in $operatornameFrac A$ as an $A$-algebra. (See comments below.)
answered Sep 3 '13 at 14:16
Zhen Lin
59k4100206
edited Sep 3 '13 at 15:47
answered Sep 3 '13 at 14:16
Zhen Lin
59k4100206
answered Sep 3 '13 at 14:16
Zhen Lin
59k4100206
answered Sep 3 '13 at 14:16
Zhen Lin
59k4100206
59k4100206
Ok. But for the last claim we also need that $B$ is flat over $A$, right?
– Martin Brandenburg
Sep 3 '13 at 14:34
1
Ah, sorry, I was thinking that the usual proof that $A to operatornameFrac A$ is an epimorphism would go through here as well, but it seems a more subtle approach is needed. Hmmm...
– Zhen Lin
Sep 3 '13 at 14:40
2
Yes. I meant that $B otimes_A B to Q(A) otimes_A Q(A)$ is injective (flatness), hence $b otimes 1 = 1 otimes b$ holds in $B otimes_A B$.
– Martin Brandenburg
Sep 3 '13 at 15:05
add a comment |Â
Ok. But for the last claim we also need that $B$ is flat over $A$, right?
– Martin Brandenburg
Sep 3 '13 at 14:34
1
Ah, sorry, I was thinking that the usual proof that $A to operatornameFrac A$ is an epimorphism would go through here as well, but it seems a more subtle approach is needed. Hmmm...
– Zhen Lin
Sep 3 '13 at 14:40
2
Yes. I meant that $B otimes_A B to Q(A) otimes_A Q(A)$ is injective (flatness), hence $b otimes 1 = 1 otimes b$ holds in $B otimes_A B$.
– Martin Brandenburg
Sep 3 '13 at 15:05
Ok. But for the last claim we also need that $B$ is flat over $A$, right?
– Martin Brandenburg
Sep 3 '13 at 14:34
Ok. But for the last claim we also need that $B$ is flat over $A$, right?
– Martin Brandenburg
Sep 3 '13 at 14:34
1
1
Ah, sorry, I was thinking that the usual proof that $A to operatornameFrac A$ is an epimorphism would go through here as well, but it seems a more subtle approach is needed. Hmmm...
– Zhen Lin
Sep 3 '13 at 14:40
Ah, sorry, I was thinking that the usual proof that $A to operatornameFrac A$ is an epimorphism would go through here as well, but it seems a more subtle approach is needed. Hmmm...
– Zhen Lin
Sep 3 '13 at 14:40
2
2
Yes. I meant that $B otimes_A B to Q(A) otimes_A Q(A)$ is injective (flatness), hence $b otimes 1 = 1 otimes b$ holds in $B otimes_A B$.
– Martin Brandenburg
Sep 3 '13 at 15:05
Yes. I meant that $B otimes_A B to Q(A) otimes_A Q(A)$ is injective (flatness), hence $b otimes 1 = 1 otimes b$ holds in $B otimes_A B$.
– Martin Brandenburg
Sep 3 '13 at 15:05
add a comment |Â
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1
This is just a hunch, but if they are integral domains with the same field of fractions, wouldn't $B$ be a localization of $A$?
– Arthur
Sep 3 '13 at 11:28
@Arthur: No. This holds in some special cases (for example when $A$ is a PID), but it fails in general. Birational morphisms can be quite complicated.
– Martin Brandenburg
Sep 3 '13 at 13:12
Perhaps Arthur means localisation in the broad sense of $A[S^-1]$ for some multiplicatively-closed set $S$. Surely this is true, by taking $S$ to be the set of units in $B$?
– Zhen Lin
Sep 3 '13 at 13:18
@ZhenLin No: math.stackexchange.com/a/287259/38268
– user38268
Sep 3 '13 at 13:21
1
Hmmm, pity. Otherwise we could just base change to $B$ and use faithful-flatness to deduce that $operatornamecoker f = 0$.
– Zhen Lin
Sep 3 '13 at 13:22