When can $A$ an abelian group be made into a vector space over $BbbF_p$?
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Let $BbbF_p$ be the finite field of integers modulo $p, p$ a prime, let $A$ be an abelian group. Precisely when can $A$ be made into a vector space over $BbbF_p$?
abstract-algebra group-theory finite-fields
asked Sep 21 '15 at 3:25
rakhil11
1439
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up vote
3
down vote
favorite
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Let $BbbF_p$ be the finite field of integers modulo $p, p$ a prime, let $A$ be an abelian group. Precisely when can $A$ be made into a vector space over $BbbF_p$?
abstract-algebra group-theory finite-fields
asked Sep 21 '15 at 3:25
rakhil11
1439
How much linear algebra do you know? Like, for example, can there be two non-isomorphic vector spaces of dimension $2$ over any given field? Or is your focus on the infinite-dimensional case?
– pjs36
Sep 21 '15 at 3:30
1
If dim(U) = dim(V), they're isomorphic, so no. I don't think this is referring to the infinite-dimensional case, although I guess that's not specified in the problem
– rakhil11
Sep 21 '15 at 3:36
1
Can you find a necessary condition? What needs to be checked then to make it a sufficient condition as well? Can you do the finitely generated case? Go step by step.
– guest
Sep 21 '15 at 3:41
2
@rakhil11 Given that comment, I would just think (in the finite-dimensional case) about all the vector spaces $V$ over $Bbb F_q$ you can think of, paying particular attention to the additive group $(V, +)$. Before trying to characterize everything, just characterize what you know.
– pjs36
Sep 21 '15 at 3:53
Taking pjs36's hint one step further. In $BbbF_p$ we have $0=1+1+cdots+1$ ($p$ summands on the right, all equal to $1$). What does this imply about the scalar multiplication (or addition) in $A$?
– Jyrki Lahtonen
Sep 21 '15 at 5:12
 |Â
show 2 more comments
up vote
3
down vote
favorite
1
up vote
3
down vote
favorite
1
1
Let $BbbF_p$ be the finite field of integers modulo $p, p$ a prime, let $A$ be an abelian group. Precisely when can $A$ be made into a vector space over $BbbF_p$?
abstract-algebra group-theory finite-fields
asked Sep 21 '15 at 3:25
rakhil11
1439
Let $BbbF_p$ be the finite field of integers modulo $p, p$ a prime, let $A$ be an abelian group. Precisely when can $A$ be made into a vector space over $BbbF_p$?
abstract-algebra group-theory finite-fields
abstract-algebra group-theory finite-fields
asked Sep 21 '15 at 3:25
rakhil11
1439
asked Sep 21 '15 at 3:25
rakhil11
1439
asked Sep 21 '15 at 3:25
rakhil11
1439
asked Sep 21 '15 at 3:25
rakhil11
1439
asked Sep 21 '15 at 3:25
rakhil11
1439
1439
How much linear algebra do you know? Like, for example, can there be two non-isomorphic vector spaces of dimension $2$ over any given field? Or is your focus on the infinite-dimensional case?
– pjs36
Sep 21 '15 at 3:30
1
If dim(U) = dim(V), they're isomorphic, so no. I don't think this is referring to the infinite-dimensional case, although I guess that's not specified in the problem
– rakhil11
Sep 21 '15 at 3:36
1
Can you find a necessary condition? What needs to be checked then to make it a sufficient condition as well? Can you do the finitely generated case? Go step by step.
– guest
Sep 21 '15 at 3:41
2
@rakhil11 Given that comment, I would just think (in the finite-dimensional case) about all the vector spaces $V$ over $Bbb F_q$ you can think of, paying particular attention to the additive group $(V, +)$. Before trying to characterize everything, just characterize what you know.
– pjs36
Sep 21 '15 at 3:53
Taking pjs36's hint one step further. In $BbbF_p$ we have $0=1+1+cdots+1$ ($p$ summands on the right, all equal to $1$). What does this imply about the scalar multiplication (or addition) in $A$?
– Jyrki Lahtonen
Sep 21 '15 at 5:12
 |Â
show 2 more comments
How much linear algebra do you know? Like, for example, can there be two non-isomorphic vector spaces of dimension $2$ over any given field? Or is your focus on the infinite-dimensional case?
– pjs36
Sep 21 '15 at 3:30
1
If dim(U) = dim(V), they're isomorphic, so no. I don't think this is referring to the infinite-dimensional case, although I guess that's not specified in the problem
– rakhil11
Sep 21 '15 at 3:36
1
Can you find a necessary condition? What needs to be checked then to make it a sufficient condition as well? Can you do the finitely generated case? Go step by step.
– guest
Sep 21 '15 at 3:41
2
@rakhil11 Given that comment, I would just think (in the finite-dimensional case) about all the vector spaces $V$ over $Bbb F_q$ you can think of, paying particular attention to the additive group $(V, +)$. Before trying to characterize everything, just characterize what you know.
– pjs36
Sep 21 '15 at 3:53
Taking pjs36's hint one step further. In $BbbF_p$ we have $0=1+1+cdots+1$ ($p$ summands on the right, all equal to $1$). What does this imply about the scalar multiplication (or addition) in $A$?
– Jyrki Lahtonen
Sep 21 '15 at 5:12
How much linear algebra do you know? Like, for example, can there be two non-isomorphic vector spaces of dimension $2$ over any given field? Or is your focus on the infinite-dimensional case?
– pjs36
Sep 21 '15 at 3:30
How much linear algebra do you know? Like, for example, can there be two non-isomorphic vector spaces of dimension $2$ over any given field? Or is your focus on the infinite-dimensional case?
– pjs36
Sep 21 '15 at 3:30
1
1
If dim(U) = dim(V), they're isomorphic, so no. I don't think this is referring to the infinite-dimensional case, although I guess that's not specified in the problem
– rakhil11
Sep 21 '15 at 3:36
If dim(U) = dim(V), they're isomorphic, so no. I don't think this is referring to the infinite-dimensional case, although I guess that's not specified in the problem
– rakhil11
Sep 21 '15 at 3:36
1
1
Can you find a necessary condition? What needs to be checked then to make it a sufficient condition as well? Can you do the finitely generated case? Go step by step.
– guest
Sep 21 '15 at 3:41
Can you find a necessary condition? What needs to be checked then to make it a sufficient condition as well? Can you do the finitely generated case? Go step by step.
– guest
Sep 21 '15 at 3:41
2
2
@rakhil11 Given that comment, I would just think (in the finite-dimensional case) about all the vector spaces $V$ over $Bbb F_q$ you can think of, paying particular attention to the additive group $(V, +)$. Before trying to characterize everything, just characterize what you know.
– pjs36
Sep 21 '15 at 3:53
@rakhil11 Given that comment, I would just think (in the finite-dimensional case) about all the vector spaces $V$ over $Bbb F_q$ you can think of, paying particular attention to the additive group $(V, +)$. Before trying to characterize everything, just characterize what you know.
– pjs36
Sep 21 '15 at 3:53
Taking pjs36's hint one step further. In $BbbF_p$ we have $0=1+1+cdots+1$ ($p$ summands on the right, all equal to $1$). What does this imply about the scalar multiplication (or addition) in $A$?
– Jyrki Lahtonen
Sep 21 '15 at 5:12
Taking pjs36's hint one step further. In $BbbF_p$ we have $0=1+1+cdots+1$ ($p$ summands on the right, all equal to $1$). What does this imply about the scalar multiplication (or addition) in $A$?
– Jyrki Lahtonen
Sep 21 '15 at 5:12
 |Â
show 2 more comments
1 Answer
1
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I'll try to give a full answer based on the comments
An abelian group $A$ is a $mathbbZ-$module and $ann_mathbbZA=(d)$ for some $dinmathbbZ$.
I'll show that $A$ is a $mathbbF_p$ vector space iff $(p)subset (d)Leftrightarrow d|pLeftrightarrow p=d$
- Let $ann_mathbbZ=(p)$. Then we can define a $dfracmathbbZpmathbbZ-$module structure as follows $$dfracmathbbZpmathbbZtimes Ato A$$
with $(z+pmathbbZ,x)mapsto zx$. This is well-defined : if $(z+pmathbbZ,x)=(z'+pmathbbZ,x')Rightarrow x=x',z-z'in pmathbbZRightarrow (z-z')x=0Rightarrow zx=z'x'.$ This way $A$ becomes a $mathbbF_p-$module (=$mathbbF_p$-vector space) - Let $A$ an $mathbbF_p-$vector space (=$dfracmathbbZpmathbbZ$-module). Then consider the ring homomorphism $$f:mathbbZto dfracmathbbZpmathbbZ$$ with $f(z)=[z]_p$. Then we can see $A$ as a $mathbbZ-$module with multiplication $$zcdot a=f(z)a,quad forall ain A,zin mathbbZ$$. Then $$zcdot a=0Leftrightarrow f(z)a=0Leftrightarrow f(z)=[0]_pLeftrightarrow p|z$$ hence $ann_mathbbZ=(p)$
answered Sep 6 at 7:56
giannispapav
1,340323
Don't you mean $(p) = (d)$? (Which is equivalent to $d = pm p$.)
– Torsten Schoeneberg
Sep 6 at 14:58
Yes thanks, my omission for the $pm$. In fact since $(p)$ is maximal in $mathbbZ$ and $Mnot=0$, $(p)⊂(d)⇔(p)=(d)$
– giannispapav
Sep 7 at 12:49
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
I'll try to give a full answer based on the comments
An abelian group $A$ is a $mathbbZ-$module and $ann_mathbbZA=(d)$ for some $dinmathbbZ$.
I'll show that $A$ is a $mathbbF_p$ vector space iff $(p)subset (d)Leftrightarrow d|pLeftrightarrow p=d$
- Let $ann_mathbbZ=(p)$. Then we can define a $dfracmathbbZpmathbbZ-$module structure as follows $$dfracmathbbZpmathbbZtimes Ato A$$
with $(z+pmathbbZ,x)mapsto zx$. This is well-defined : if $(z+pmathbbZ,x)=(z'+pmathbbZ,x')Rightarrow x=x',z-z'in pmathbbZRightarrow (z-z')x=0Rightarrow zx=z'x'.$ This way $A$ becomes a $mathbbF_p-$module (=$mathbbF_p$-vector space) - Let $A$ an $mathbbF_p-$vector space (=$dfracmathbbZpmathbbZ$-module). Then consider the ring homomorphism $$f:mathbbZto dfracmathbbZpmathbbZ$$ with $f(z)=[z]_p$. Then we can see $A$ as a $mathbbZ-$module with multiplication $$zcdot a=f(z)a,quad forall ain A,zin mathbbZ$$. Then $$zcdot a=0Leftrightarrow f(z)a=0Leftrightarrow f(z)=[0]_pLeftrightarrow p|z$$ hence $ann_mathbbZ=(p)$
answered Sep 6 at 7:56
giannispapav
1,340323
Don't you mean $(p) = (d)$? (Which is equivalent to $d = pm p$.)
– Torsten Schoeneberg
Sep 6 at 14:58
Yes thanks, my omission for the $pm$. In fact since $(p)$ is maximal in $mathbbZ$ and $Mnot=0$, $(p)⊂(d)⇔(p)=(d)$
– giannispapav
Sep 7 at 12:49
add a comment |Â
up vote
2
down vote
I'll try to give a full answer based on the comments
An abelian group $A$ is a $mathbbZ-$module and $ann_mathbbZA=(d)$ for some $dinmathbbZ$.
I'll show that $A$ is a $mathbbF_p$ vector space iff $(p)subset (d)Leftrightarrow d|pLeftrightarrow p=d$
- Let $ann_mathbbZ=(p)$. Then we can define a $dfracmathbbZpmathbbZ-$module structure as follows $$dfracmathbbZpmathbbZtimes Ato A$$
with $(z+pmathbbZ,x)mapsto zx$. This is well-defined : if $(z+pmathbbZ,x)=(z'+pmathbbZ,x')Rightarrow x=x',z-z'in pmathbbZRightarrow (z-z')x=0Rightarrow zx=z'x'.$ This way $A$ becomes a $mathbbF_p-$module (=$mathbbF_p$-vector space) - Let $A$ an $mathbbF_p-$vector space (=$dfracmathbbZpmathbbZ$-module). Then consider the ring homomorphism $$f:mathbbZto dfracmathbbZpmathbbZ$$ with $f(z)=[z]_p$. Then we can see $A$ as a $mathbbZ-$module with multiplication $$zcdot a=f(z)a,quad forall ain A,zin mathbbZ$$. Then $$zcdot a=0Leftrightarrow f(z)a=0Leftrightarrow f(z)=[0]_pLeftrightarrow p|z$$ hence $ann_mathbbZ=(p)$
answered Sep 6 at 7:56
giannispapav
1,340323
Don't you mean $(p) = (d)$? (Which is equivalent to $d = pm p$.)
– Torsten Schoeneberg
Sep 6 at 14:58
Yes thanks, my omission for the $pm$. In fact since $(p)$ is maximal in $mathbbZ$ and $Mnot=0$, $(p)⊂(d)⇔(p)=(d)$
– giannispapav
Sep 7 at 12:49
add a comment |Â
up vote
2
down vote
up vote
2
down vote
I'll try to give a full answer based on the comments
An abelian group $A$ is a $mathbbZ-$module and $ann_mathbbZA=(d)$ for some $dinmathbbZ$.
I'll show that $A$ is a $mathbbF_p$ vector space iff $(p)subset (d)Leftrightarrow d|pLeftrightarrow p=d$
- Let $ann_mathbbZ=(p)$. Then we can define a $dfracmathbbZpmathbbZ-$module structure as follows $$dfracmathbbZpmathbbZtimes Ato A$$
with $(z+pmathbbZ,x)mapsto zx$. This is well-defined : if $(z+pmathbbZ,x)=(z'+pmathbbZ,x')Rightarrow x=x',z-z'in pmathbbZRightarrow (z-z')x=0Rightarrow zx=z'x'.$ This way $A$ becomes a $mathbbF_p-$module (=$mathbbF_p$-vector space) - Let $A$ an $mathbbF_p-$vector space (=$dfracmathbbZpmathbbZ$-module). Then consider the ring homomorphism $$f:mathbbZto dfracmathbbZpmathbbZ$$ with $f(z)=[z]_p$. Then we can see $A$ as a $mathbbZ-$module with multiplication $$zcdot a=f(z)a,quad forall ain A,zin mathbbZ$$. Then $$zcdot a=0Leftrightarrow f(z)a=0Leftrightarrow f(z)=[0]_pLeftrightarrow p|z$$ hence $ann_mathbbZ=(p)$
answered Sep 6 at 7:56
giannispapav
1,340323
I'll try to give a full answer based on the comments
An abelian group $A$ is a $mathbbZ-$module and $ann_mathbbZA=(d)$ for some $dinmathbbZ$.
I'll show that $A$ is a $mathbbF_p$ vector space iff $(p)subset (d)Leftrightarrow d|pLeftrightarrow p=d$
- Let $ann_mathbbZ=(p)$. Then we can define a $dfracmathbbZpmathbbZ-$module structure as follows $$dfracmathbbZpmathbbZtimes Ato A$$
with $(z+pmathbbZ,x)mapsto zx$. This is well-defined : if $(z+pmathbbZ,x)=(z'+pmathbbZ,x')Rightarrow x=x',z-z'in pmathbbZRightarrow (z-z')x=0Rightarrow zx=z'x'.$ This way $A$ becomes a $mathbbF_p-$module (=$mathbbF_p$-vector space) - Let $A$ an $mathbbF_p-$vector space (=$dfracmathbbZpmathbbZ$-module). Then consider the ring homomorphism $$f:mathbbZto dfracmathbbZpmathbbZ$$ with $f(z)=[z]_p$. Then we can see $A$ as a $mathbbZ-$module with multiplication $$zcdot a=f(z)a,quad forall ain A,zin mathbbZ$$. Then $$zcdot a=0Leftrightarrow f(z)a=0Leftrightarrow f(z)=[0]_pLeftrightarrow p|z$$ hence $ann_mathbbZ=(p)$
answered Sep 6 at 7:56
giannispapav
1,340323
answered Sep 6 at 7:56
giannispapav
1,340323
answered Sep 6 at 7:56
giannispapav
1,340323
answered Sep 6 at 7:56
giannispapav
1,340323
1,340323
Don't you mean $(p) = (d)$? (Which is equivalent to $d = pm p$.)
– Torsten Schoeneberg
Sep 6 at 14:58
Yes thanks, my omission for the $pm$. In fact since $(p)$ is maximal in $mathbbZ$ and $Mnot=0$, $(p)⊂(d)⇔(p)=(d)$
– giannispapav
Sep 7 at 12:49
add a comment |Â
Don't you mean $(p) = (d)$? (Which is equivalent to $d = pm p$.)
– Torsten Schoeneberg
Sep 6 at 14:58
Yes thanks, my omission for the $pm$. In fact since $(p)$ is maximal in $mathbbZ$ and $Mnot=0$, $(p)⊂(d)⇔(p)=(d)$
– giannispapav
Sep 7 at 12:49
Don't you mean $(p) = (d)$? (Which is equivalent to $d = pm p$.)
– Torsten Schoeneberg
Sep 6 at 14:58
Don't you mean $(p) = (d)$? (Which is equivalent to $d = pm p$.)
– Torsten Schoeneberg
Sep 6 at 14:58
Yes thanks, my omission for the $pm$. In fact since $(p)$ is maximal in $mathbbZ$ and $Mnot=0$, $(p)⊂(d)⇔(p)=(d)$
– giannispapav
Sep 7 at 12:49
Yes thanks, my omission for the $pm$. In fact since $(p)$ is maximal in $mathbbZ$ and $Mnot=0$, $(p)⊂(d)⇔(p)=(d)$
– giannispapav
Sep 7 at 12:49
add a comment |Â
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How much linear algebra do you know? Like, for example, can there be two non-isomorphic vector spaces of dimension $2$ over any given field? Or is your focus on the infinite-dimensional case?
– pjs36
Sep 21 '15 at 3:30
1
If dim(U) = dim(V), they're isomorphic, so no. I don't think this is referring to the infinite-dimensional case, although I guess that's not specified in the problem
– rakhil11
Sep 21 '15 at 3:36
1
Can you find a necessary condition? What needs to be checked then to make it a sufficient condition as well? Can you do the finitely generated case? Go step by step.
– guest
Sep 21 '15 at 3:41
2
@rakhil11 Given that comment, I would just think (in the finite-dimensional case) about all the vector spaces $V$ over $Bbb F_q$ you can think of, paying particular attention to the additive group $(V, +)$. Before trying to characterize everything, just characterize what you know.
– pjs36
Sep 21 '15 at 3:53
Taking pjs36's hint one step further. In $BbbF_p$ we have $0=1+1+cdots+1$ ($p$ summands on the right, all equal to $1$). What does this imply about the scalar multiplication (or addition) in $A$?
– Jyrki Lahtonen
Sep 21 '15 at 5:12