If possible, construct a simple module that is not projective.
$begingroup$
Suppose that $M$ is a simple (or irreducible) $R$-module. Does this imply that
$M$ is projective?
Vector spaces are all projective since they are free, so no counter example can be gotten from there.
I tried to find a counter-example with the $mathbb{Z}$-modules $mathbb{Z}/{pmathbb Z}$
where $p$ is prime (since they are irreducible) but could not get anywhere.
At this point I'm not even sure that the result is false.
Please give only hints and not the answer.
modules projective-module
asked Dec 5 '18 at 12:16
UserAUserA
542216
$endgroup$
add a comment |
$begingroup$
Suppose that $M$ is a simple (or irreducible) $R$-module. Does this imply that
$M$ is projective?
Vector spaces are all projective since they are free, so no counter example can be gotten from there.
I tried to find a counter-example with the $mathbb{Z}$-modules $mathbb{Z}/{pmathbb Z}$
where $p$ is prime (since they are irreducible) but could not get anywhere.
At this point I'm not even sure that the result is false.
Please give only hints and not the answer.
modules projective-module
asked Dec 5 '18 at 12:16
UserAUserA
542216
$endgroup$
$begingroup$
Try $R=Bbb Z/nBbb Z$ for some non-prime $n$.
$endgroup$
– Lord Shark the Unknown
Dec 5 '18 at 12:21
$begingroup$
You can obtain counterexamples to this with quivers, by choosing a vertex and the simple module associated to it.
$endgroup$
– Pedro Tamaroff♦
Dec 5 '18 at 12:22
add a comment |
$begingroup$
Suppose that $M$ is a simple (or irreducible) $R$-module. Does this imply that
$M$ is projective?
Vector spaces are all projective since they are free, so no counter example can be gotten from there.
I tried to find a counter-example with the $mathbb{Z}$-modules $mathbb{Z}/{pmathbb Z}$
where $p$ is prime (since they are irreducible) but could not get anywhere.
At this point I'm not even sure that the result is false.
Please give only hints and not the answer.
modules projective-module
asked Dec 5 '18 at 12:16
UserAUserA
542216
$endgroup$
Suppose that $M$ is a simple (or irreducible) $R$-module. Does this imply that
$M$ is projective?
Vector spaces are all projective since they are free, so no counter example can be gotten from there.
I tried to find a counter-example with the $mathbb{Z}$-modules $mathbb{Z}/{pmathbb Z}$
where $p$ is prime (since they are irreducible) but could not get anywhere.
At this point I'm not even sure that the result is false.
Please give only hints and not the answer.
modules projective-module
modules projective-module
asked Dec 5 '18 at 12:16
UserAUserA
542216
asked Dec 5 '18 at 12:16
UserAUserA
542216
asked Dec 5 '18 at 12:16
UserAUserA
542216
asked Dec 5 '18 at 12:16
UserAUserA
542216
asked Dec 5 '18 at 12:16
UserAUserA
542216
542216
$begingroup$
Try $R=Bbb Z/nBbb Z$ for some non-prime $n$.
$endgroup$
– Lord Shark the Unknown
Dec 5 '18 at 12:21
$begingroup$
You can obtain counterexamples to this with quivers, by choosing a vertex and the simple module associated to it.
$endgroup$
– Pedro Tamaroff♦
Dec 5 '18 at 12:22
add a comment |
$begingroup$
Try $R=Bbb Z/nBbb Z$ for some non-prime $n$.
$endgroup$
– Lord Shark the Unknown
Dec 5 '18 at 12:21
$begingroup$
You can obtain counterexamples to this with quivers, by choosing a vertex and the simple module associated to it.
$endgroup$
– Pedro Tamaroff♦
Dec 5 '18 at 12:22
$begingroup$
Try $R=Bbb Z/nBbb Z$ for some non-prime $n$.
$endgroup$
– Lord Shark the Unknown
Dec 5 '18 at 12:21
$begingroup$
Try $R=Bbb Z/nBbb Z$ for some non-prime $n$.
$endgroup$
– Lord Shark the Unknown
Dec 5 '18 at 12:21
$begingroup$
You can obtain counterexamples to this with quivers, by choosing a vertex and the simple module associated to it.
$endgroup$
– Pedro Tamaroff♦
Dec 5 '18 at 12:22
$begingroup$
You can obtain counterexamples to this with quivers, by choosing a vertex and the simple module associated to it.
$endgroup$
– Pedro Tamaroff♦
Dec 5 '18 at 12:22
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Your attempt was really good! Since $mathbb{Z}/pmathbb{Z}$ are very good examples of such objects. You already know that they are irreducible. Now we just need to show that they are not projective.
So assume that they are projective and consider the projection $$pi:mathbb{Z} twoheadrightarrow mathbb{Z}/pmathbb{Z}$$ this is a projection and together with $mathrm{id}:mathbb{Z}/pmathbb{Z} to mathbb{Z}/pmathbb{Z} $ this means that there needs to be some map $f:mathbb{Z}/pmathbb{Z} to mathbb{Z}$ such that $pi circ f=mathrm{id}$. But the only morphism $mathbb{Z}/pmathbb{Z} to mathbb{Z}$ is the zero morphism which is a contradiction.
answered Dec 5 '18 at 12:22
EnkiduEnkidu
1,36719
$endgroup$
add a comment |
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$begingroup$
Your attempt was really good! Since $mathbb{Z}/pmathbb{Z}$ are very good examples of such objects. You already know that they are irreducible. Now we just need to show that they are not projective.
So assume that they are projective and consider the projection $$pi:mathbb{Z} twoheadrightarrow mathbb{Z}/pmathbb{Z}$$ this is a projection and together with $mathrm{id}:mathbb{Z}/pmathbb{Z} to mathbb{Z}/pmathbb{Z} $ this means that there needs to be some map $f:mathbb{Z}/pmathbb{Z} to mathbb{Z}$ such that $pi circ f=mathrm{id}$. But the only morphism $mathbb{Z}/pmathbb{Z} to mathbb{Z}$ is the zero morphism which is a contradiction.
answered Dec 5 '18 at 12:22
EnkiduEnkidu
1,36719
$endgroup$
add a comment |
$begingroup$
Your attempt was really good! Since $mathbb{Z}/pmathbb{Z}$ are very good examples of such objects. You already know that they are irreducible. Now we just need to show that they are not projective.
So assume that they are projective and consider the projection $$pi:mathbb{Z} twoheadrightarrow mathbb{Z}/pmathbb{Z}$$ this is a projection and together with $mathrm{id}:mathbb{Z}/pmathbb{Z} to mathbb{Z}/pmathbb{Z} $ this means that there needs to be some map $f:mathbb{Z}/pmathbb{Z} to mathbb{Z}$ such that $pi circ f=mathrm{id}$. But the only morphism $mathbb{Z}/pmathbb{Z} to mathbb{Z}$ is the zero morphism which is a contradiction.
answered Dec 5 '18 at 12:22
EnkiduEnkidu
1,36719
$endgroup$
add a comment |
$begingroup$
Your attempt was really good! Since $mathbb{Z}/pmathbb{Z}$ are very good examples of such objects. You already know that they are irreducible. Now we just need to show that they are not projective.
So assume that they are projective and consider the projection $$pi:mathbb{Z} twoheadrightarrow mathbb{Z}/pmathbb{Z}$$ this is a projection and together with $mathrm{id}:mathbb{Z}/pmathbb{Z} to mathbb{Z}/pmathbb{Z} $ this means that there needs to be some map $f:mathbb{Z}/pmathbb{Z} to mathbb{Z}$ such that $pi circ f=mathrm{id}$. But the only morphism $mathbb{Z}/pmathbb{Z} to mathbb{Z}$ is the zero morphism which is a contradiction.
answered Dec 5 '18 at 12:22
EnkiduEnkidu
1,36719
$endgroup$
Your attempt was really good! Since $mathbb{Z}/pmathbb{Z}$ are very good examples of such objects. You already know that they are irreducible. Now we just need to show that they are not projective.
So assume that they are projective and consider the projection $$pi:mathbb{Z} twoheadrightarrow mathbb{Z}/pmathbb{Z}$$ this is a projection and together with $mathrm{id}:mathbb{Z}/pmathbb{Z} to mathbb{Z}/pmathbb{Z} $ this means that there needs to be some map $f:mathbb{Z}/pmathbb{Z} to mathbb{Z}$ such that $pi circ f=mathrm{id}$. But the only morphism $mathbb{Z}/pmathbb{Z} to mathbb{Z}$ is the zero morphism which is a contradiction.
answered Dec 5 '18 at 12:22
EnkiduEnkidu
1,36719
answered Dec 5 '18 at 12:22
EnkiduEnkidu
1,36719
answered Dec 5 '18 at 12:22
EnkiduEnkidu
1,36719
answered Dec 5 '18 at 12:22
EnkiduEnkidu
1,36719
1,36719
add a comment |
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$begingroup$
Try $R=Bbb Z/nBbb Z$ for some non-prime $n$.
$endgroup$
– Lord Shark the Unknown
Dec 5 '18 at 12:21
$begingroup$
You can obtain counterexamples to this with quivers, by choosing a vertex and the simple module associated to it.
$endgroup$
– Pedro Tamaroff♦
Dec 5 '18 at 12:22