Mixing
Projective and flat vs. faithfully flat
Clash Royale CLAN TAG#URR8PPP
up vote
3
down vote
favorite
Let $R$ be a commutative ring with unity and let $M$ be a projective and faithful $R$-module. Then is $M$ faithfully flat ? Is it true at least if $M$ is finitely generated, or say Noetherian ?
I know that I only have to show that $Motimes_R Nne 0$ for every non-zero $R$-module $N$. Now if $M$ is finitely generated, then by faithful ness of $M$, I can show that $Motimes_R Nne 0$ for every non-zero , finitely generated $R$-module $N$, because for finitely generated $R$-modules $M$ and $N$, $ operatornameSupp(M otimes_R N)=V( operatornameAnn_R(M) + operatornameAnn_R(N))$.
I am unable to proceed further.
Please help.
commutative-algebra modules tensor-products projective-module flatness
edited Aug 3 at 19:49
Pedro Tamaroff♦
93.6k10143290
asked Aug 3 at 18:40
user521337
606
add a comment |Â
up vote
3
down vote
favorite
Let $R$ be a commutative ring with unity and let $M$ be a projective and faithful $R$-module. Then is $M$ faithfully flat ? Is it true at least if $M$ is finitely generated, or say Noetherian ?
I know that I only have to show that $Motimes_R Nne 0$ for every non-zero $R$-module $N$. Now if $M$ is finitely generated, then by faithful ness of $M$, I can show that $Motimes_R Nne 0$ for every non-zero , finitely generated $R$-module $N$, because for finitely generated $R$-modules $M$ and $N$, $ operatornameSupp(M otimes_R N)=V( operatornameAnn_R(M) + operatornameAnn_R(N))$.
I am unable to proceed further.
Please help.
commutative-algebra modules tensor-products projective-module flatness
edited Aug 3 at 19:49
Pedro Tamaroff♦
93.6k10143290
asked Aug 3 at 18:40
user521337
606
1
As Pedro's beautiful answer shows, this is false in general. However, I believe it does hold if we add the extra condition that $r_M(P) > 0$ for all $P in Spec(R)$. (Recall that every projective module over a commutative local ring is free; here, $r_M(P)$ denotes the rank of $M_P$.) Indeed, suppose this is the case, and let $N$ be a nonzero $R$-module. Then for any $P in Spec(R)$, $(M otimes_R N)_P cong M_P otimes_R_P N_P cong N_P^r_M(P)$. Since $r_M(P) > 0$ for all $P$ and $N_P$ must be nonzero for some $P$, it follows that $(M otimes_R N)_P$ must be...
– Alex Wertheim
Aug 3 at 19:37
1
...nonzero for this same $P$, whence $M otimes_R N$ is nonzero. I don't know how strong or artificial the requirement that $r_M(P) > 0$ for all $P in Spec(R)$ is, however.
– Alex Wertheim
Aug 3 at 19:39
@AlexWertheim Yes, I had that proof in mind too! Well, if $R$ is connected then $r_M(P)$ is constant, for example. Of course in Bourbaki's counterexample this fails miserably. :)
– Pedro Tamaroff♦
Aug 3 at 19:46
See this.
– Pedro Tamaroff♦
Aug 3 at 19:57
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Let $R$ be a commutative ring with unity and let $M$ be a projective and faithful $R$-module. Then is $M$ faithfully flat ? Is it true at least if $M$ is finitely generated, or say Noetherian ?
I know that I only have to show that $Motimes_R Nne 0$ for every non-zero $R$-module $N$. Now if $M$ is finitely generated, then by faithful ness of $M$, I can show that $Motimes_R Nne 0$ for every non-zero , finitely generated $R$-module $N$, because for finitely generated $R$-modules $M$ and $N$, $ operatornameSupp(M otimes_R N)=V( operatornameAnn_R(M) + operatornameAnn_R(N))$.
I am unable to proceed further.
Please help.
commutative-algebra modules tensor-products projective-module flatness
edited Aug 3 at 19:49
Pedro Tamaroff♦
93.6k10143290
asked Aug 3 at 18:40
user521337
606
Let $R$ be a commutative ring with unity and let $M$ be a projective and faithful $R$-module. Then is $M$ faithfully flat ? Is it true at least if $M$ is finitely generated, or say Noetherian ?
I know that I only have to show that $Motimes_R Nne 0$ for every non-zero $R$-module $N$. Now if $M$ is finitely generated, then by faithful ness of $M$, I can show that $Motimes_R Nne 0$ for every non-zero , finitely generated $R$-module $N$, because for finitely generated $R$-modules $M$ and $N$, $ operatornameSupp(M otimes_R N)=V( operatornameAnn_R(M) + operatornameAnn_R(N))$.
I am unable to proceed further.
Please help.
commutative-algebra modules tensor-products projective-module flatness
edited Aug 3 at 19:49
Pedro Tamaroff♦
93.6k10143290
asked Aug 3 at 18:40
user521337
606
edited Aug 3 at 19:49
Pedro Tamaroff♦
93.6k10143290
edited Aug 3 at 19:49
Pedro Tamaroff♦
93.6k10143290
edited Aug 3 at 19:49
Pedro Tamaroff♦
93.6k10143290
93.6k10143290
asked Aug 3 at 18:40
user521337
606
asked Aug 3 at 18:40
user521337
606
asked Aug 3 at 18:40
user521337
606
606
1
As Pedro's beautiful answer shows, this is false in general. However, I believe it does hold if we add the extra condition that $r_M(P) > 0$ for all $P in Spec(R)$. (Recall that every projective module over a commutative local ring is free; here, $r_M(P)$ denotes the rank of $M_P$.) Indeed, suppose this is the case, and let $N$ be a nonzero $R$-module. Then for any $P in Spec(R)$, $(M otimes_R N)_P cong M_P otimes_R_P N_P cong N_P^r_M(P)$. Since $r_M(P) > 0$ for all $P$ and $N_P$ must be nonzero for some $P$, it follows that $(M otimes_R N)_P$ must be...
– Alex Wertheim
Aug 3 at 19:37
1
...nonzero for this same $P$, whence $M otimes_R N$ is nonzero. I don't know how strong or artificial the requirement that $r_M(P) > 0$ for all $P in Spec(R)$ is, however.
– Alex Wertheim
Aug 3 at 19:39
@AlexWertheim Yes, I had that proof in mind too! Well, if $R$ is connected then $r_M(P)$ is constant, for example. Of course in Bourbaki's counterexample this fails miserably. :)
– Pedro Tamaroff♦
Aug 3 at 19:46
See this.
– Pedro Tamaroff♦
Aug 3 at 19:57
add a comment |Â
1
As Pedro's beautiful answer shows, this is false in general. However, I believe it does hold if we add the extra condition that $r_M(P) > 0$ for all $P in Spec(R)$. (Recall that every projective module over a commutative local ring is free; here, $r_M(P)$ denotes the rank of $M_P$.) Indeed, suppose this is the case, and let $N$ be a nonzero $R$-module. Then for any $P in Spec(R)$, $(M otimes_R N)_P cong M_P otimes_R_P N_P cong N_P^r_M(P)$. Since $r_M(P) > 0$ for all $P$ and $N_P$ must be nonzero for some $P$, it follows that $(M otimes_R N)_P$ must be...
– Alex Wertheim
Aug 3 at 19:37
1
...nonzero for this same $P$, whence $M otimes_R N$ is nonzero. I don't know how strong or artificial the requirement that $r_M(P) > 0$ for all $P in Spec(R)$ is, however.
– Alex Wertheim
Aug 3 at 19:39
@AlexWertheim Yes, I had that proof in mind too! Well, if $R$ is connected then $r_M(P)$ is constant, for example. Of course in Bourbaki's counterexample this fails miserably. :)
– Pedro Tamaroff♦
Aug 3 at 19:46
See this.
– Pedro Tamaroff♦
Aug 3 at 19:57
1
1
As Pedro's beautiful answer shows, this is false in general. However, I believe it does hold if we add the extra condition that $r_M(P) > 0$ for all $P in Spec(R)$. (Recall that every projective module over a commutative local ring is free; here, $r_M(P)$ denotes the rank of $M_P$.) Indeed, suppose this is the case, and let $N$ be a nonzero $R$-module. Then for any $P in Spec(R)$, $(M otimes_R N)_P cong M_P otimes_R_P N_P cong N_P^r_M(P)$. Since $r_M(P) > 0$ for all $P$ and $N_P$ must be nonzero for some $P$, it follows that $(M otimes_R N)_P$ must be...
– Alex Wertheim
Aug 3 at 19:37
As Pedro's beautiful answer shows, this is false in general. However, I believe it does hold if we add the extra condition that $r_M(P) > 0$ for all $P in Spec(R)$. (Recall that every projective module over a commutative local ring is free; here, $r_M(P)$ denotes the rank of $M_P$.) Indeed, suppose this is the case, and let $N$ be a nonzero $R$-module. Then for any $P in Spec(R)$, $(M otimes_R N)_P cong M_P otimes_R_P N_P cong N_P^r_M(P)$. Since $r_M(P) > 0$ for all $P$ and $N_P$ must be nonzero for some $P$, it follows that $(M otimes_R N)_P$ must be...
– Alex Wertheim
Aug 3 at 19:37
1
1
...nonzero for this same $P$, whence $M otimes_R N$ is nonzero. I don't know how strong or artificial the requirement that $r_M(P) > 0$ for all $P in Spec(R)$ is, however.
– Alex Wertheim
Aug 3 at 19:39
...nonzero for this same $P$, whence $M otimes_R N$ is nonzero. I don't know how strong or artificial the requirement that $r_M(P) > 0$ for all $P in Spec(R)$ is, however.
– Alex Wertheim
Aug 3 at 19:39
@AlexWertheim Yes, I had that proof in mind too! Well, if $R$ is connected then $r_M(P)$ is constant, for example. Of course in Bourbaki's counterexample this fails miserably. :)
– Pedro Tamaroff♦
Aug 3 at 19:46
@AlexWertheim Yes, I had that proof in mind too! Well, if $R$ is connected then $r_M(P)$ is constant, for example. Of course in Bourbaki's counterexample this fails miserably. :)
– Pedro Tamaroff♦
Aug 3 at 19:46
See this.
– Pedro Tamaroff♦
Aug 3 at 19:57
See this.
– Pedro Tamaroff♦
Aug 3 at 19:57
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
According to Bourbaki, Commutative Algebra, this is false for general rings (page 29, Example (2)). See here.
The counterexample is the following: take the direct product $A$ of the rings $mathbb Z/p$ over all primes, and consider the ideal $I$ obtained from taking the direct sum. Then $I$ is projective and faithful, but not faithfully flat.
To see that $I$ is not faithfully flat, recall that a module $M$ is faithfully flat iff it is flat and $JMneq M$ for every (maximal) ideal. But note that $I^2=I$, so it follows that $I$ is not faithfully flat.
On the other hand, if we take $e_pin I$ the $p$-th coordinate element, we see that for $ain A$, $ae_p = a_p$, so if $a$ annihilates $M$ we deduce that $a=0$, and $M$ is faithful.
Finally, $I$ is projective: I will exhibit a dual basis for it. Consider the coordinate elements $e_p$ and the $A$-linear maps $e_p^*$ that are the composition of the projection to the $p$th component with the inclusion $mathbb Z/pto A$. It is immediate from the definitions that if $xin I$, then $sum_p e_p^*(x)e_p =x$
answered Aug 3 at 19:05
Pedro Tamaroff♦
93.6k10143290
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
According to Bourbaki, Commutative Algebra, this is false for general rings (page 29, Example (2)). See here.
The counterexample is the following: take the direct product $A$ of the rings $mathbb Z/p$ over all primes, and consider the ideal $I$ obtained from taking the direct sum. Then $I$ is projective and faithful, but not faithfully flat.
To see that $I$ is not faithfully flat, recall that a module $M$ is faithfully flat iff it is flat and $JMneq M$ for every (maximal) ideal. But note that $I^2=I$, so it follows that $I$ is not faithfully flat.
On the other hand, if we take $e_pin I$ the $p$-th coordinate element, we see that for $ain A$, $ae_p = a_p$, so if $a$ annihilates $M$ we deduce that $a=0$, and $M$ is faithful.
Finally, $I$ is projective: I will exhibit a dual basis for it. Consider the coordinate elements $e_p$ and the $A$-linear maps $e_p^*$ that are the composition of the projection to the $p$th component with the inclusion $mathbb Z/pto A$. It is immediate from the definitions that if $xin I$, then $sum_p e_p^*(x)e_p =x$
answered Aug 3 at 19:05
Pedro Tamaroff♦
93.6k10143290
add a comment |Â
up vote
2
down vote
accepted
According to Bourbaki, Commutative Algebra, this is false for general rings (page 29, Example (2)). See here.
The counterexample is the following: take the direct product $A$ of the rings $mathbb Z/p$ over all primes, and consider the ideal $I$ obtained from taking the direct sum. Then $I$ is projective and faithful, but not faithfully flat.
To see that $I$ is not faithfully flat, recall that a module $M$ is faithfully flat iff it is flat and $JMneq M$ for every (maximal) ideal. But note that $I^2=I$, so it follows that $I$ is not faithfully flat.
On the other hand, if we take $e_pin I$ the $p$-th coordinate element, we see that for $ain A$, $ae_p = a_p$, so if $a$ annihilates $M$ we deduce that $a=0$, and $M$ is faithful.
Finally, $I$ is projective: I will exhibit a dual basis for it. Consider the coordinate elements $e_p$ and the $A$-linear maps $e_p^*$ that are the composition of the projection to the $p$th component with the inclusion $mathbb Z/pto A$. It is immediate from the definitions that if $xin I$, then $sum_p e_p^*(x)e_p =x$
answered Aug 3 at 19:05
Pedro Tamaroff♦
93.6k10143290
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
According to Bourbaki, Commutative Algebra, this is false for general rings (page 29, Example (2)). See here.
The counterexample is the following: take the direct product $A$ of the rings $mathbb Z/p$ over all primes, and consider the ideal $I$ obtained from taking the direct sum. Then $I$ is projective and faithful, but not faithfully flat.
To see that $I$ is not faithfully flat, recall that a module $M$ is faithfully flat iff it is flat and $JMneq M$ for every (maximal) ideal. But note that $I^2=I$, so it follows that $I$ is not faithfully flat.
On the other hand, if we take $e_pin I$ the $p$-th coordinate element, we see that for $ain A$, $ae_p = a_p$, so if $a$ annihilates $M$ we deduce that $a=0$, and $M$ is faithful.
Finally, $I$ is projective: I will exhibit a dual basis for it. Consider the coordinate elements $e_p$ and the $A$-linear maps $e_p^*$ that are the composition of the projection to the $p$th component with the inclusion $mathbb Z/pto A$. It is immediate from the definitions that if $xin I$, then $sum_p e_p^*(x)e_p =x$
answered Aug 3 at 19:05
Pedro Tamaroff♦
93.6k10143290
According to Bourbaki, Commutative Algebra, this is false for general rings (page 29, Example (2)). See here.
The counterexample is the following: take the direct product $A$ of the rings $mathbb Z/p$ over all primes, and consider the ideal $I$ obtained from taking the direct sum. Then $I$ is projective and faithful, but not faithfully flat.
To see that $I$ is not faithfully flat, recall that a module $M$ is faithfully flat iff it is flat and $JMneq M$ for every (maximal) ideal. But note that $I^2=I$, so it follows that $I$ is not faithfully flat.
On the other hand, if we take $e_pin I$ the $p$-th coordinate element, we see that for $ain A$, $ae_p = a_p$, so if $a$ annihilates $M$ we deduce that $a=0$, and $M$ is faithful.
Finally, $I$ is projective: I will exhibit a dual basis for it. Consider the coordinate elements $e_p$ and the $A$-linear maps $e_p^*$ that are the composition of the projection to the $p$th component with the inclusion $mathbb Z/pto A$. It is immediate from the definitions that if $xin I$, then $sum_p e_p^*(x)e_p =x$
answered Aug 3 at 19:05
Pedro Tamaroff♦
93.6k10143290
edited Aug 3 at 19:29
answered Aug 3 at 19:05
Pedro Tamaroff♦
93.6k10143290
answered Aug 3 at 19:05
Pedro Tamaroff♦
93.6k10143290
answered Aug 3 at 19:05
Pedro Tamaroff♦
93.6k10143290
93.6k10143290
add a comment |Â
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2871379%2fprojective-and-flat-vs-faithfully-flat%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
1
As Pedro's beautiful answer shows, this is false in general. However, I believe it does hold if we add the extra condition that $r_M(P) > 0$ for all $P in Spec(R)$. (Recall that every projective module over a commutative local ring is free; here, $r_M(P)$ denotes the rank of $M_P$.) Indeed, suppose this is the case, and let $N$ be a nonzero $R$-module. Then for any $P in Spec(R)$, $(M otimes_R N)_P cong M_P otimes_R_P N_P cong N_P^r_M(P)$. Since $r_M(P) > 0$ for all $P$ and $N_P$ must be nonzero for some $P$, it follows that $(M otimes_R N)_P$ must be...
– Alex Wertheim
Aug 3 at 19:37
1
...nonzero for this same $P$, whence $M otimes_R N$ is nonzero. I don't know how strong or artificial the requirement that $r_M(P) > 0$ for all $P in Spec(R)$ is, however.
– Alex Wertheim
Aug 3 at 19:39
@AlexWertheim Yes, I had that proof in mind too! Well, if $R$ is connected then $r_M(P)$ is constant, for example. Of course in Bourbaki's counterexample this fails miserably. :)
– Pedro Tamaroff♦
Aug 3 at 19:46
See this.
– Pedro Tamaroff♦
Aug 3 at 19:57