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Endomorphisms of $mathbb Z$-modules are the same as those of $mathbb F_p$-vector spaces
Clash Royale CLAN TAG#URR8PPP
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Let $p$ be a prime number. Consider a direct sum of $n$ copies of the cyclic group of order $p$, written $C_p$: $G=C_p^oplus n$. By a comment in this question, given an $R$-module $M$ and an ideal $Isubset R$, there is a structure of an $R/I$-module on $M$ given by $overline r x=rx$ ($rin R, xin M$) provided $IM=0$. In our case, $G$ is a $mathbb Z$-module; after setting $I=(p)$, we see that $IM=0$ is equivalent to $pG=0$, which holds. So there is a $mathbb Z/pmathbb Z$
-module structure on $G$. In other words, there is an $mathbb F_p$-vector space structure on $G$.
As I'm writing this, I came to realize that I don't really understand why is the dimension of this space $n$. (I believe this is so.) How to justify this?
Further, any group endomorphism of $G$ is by definition a $mathbb Z$-module endomorphism of $G$, and it is obvious that any such endomorphism is also an endomorphism as an endo of $mathbb F_p$-vector space. Why is the converse true? I.e., why is any endomorphism of $G$ as an $mathbb F_p$-vector space must be an endomorphism of $G$ as a group ($mathbb Z$-module)? First of all, we need to define the structure of a $mathbb Z$-module on the $mathbb F_p$-vector space $G$; how can we do it?
linear-algebra abstract-algebra group-theory vector-spaces modules
asked Aug 6 at 0:51
user437309
556212
 |Â
show 4 more comments
up vote
0
down vote
favorite
Let $p$ be a prime number. Consider a direct sum of $n$ copies of the cyclic group of order $p$, written $C_p$: $G=C_p^oplus n$. By a comment in this question, given an $R$-module $M$ and an ideal $Isubset R$, there is a structure of an $R/I$-module on $M$ given by $overline r x=rx$ ($rin R, xin M$) provided $IM=0$. In our case, $G$ is a $mathbb Z$-module; after setting $I=(p)$, we see that $IM=0$ is equivalent to $pG=0$, which holds. So there is a $mathbb Z/pmathbb Z$
-module structure on $G$. In other words, there is an $mathbb F_p$-vector space structure on $G$.
As I'm writing this, I came to realize that I don't really understand why is the dimension of this space $n$. (I believe this is so.) How to justify this?
Further, any group endomorphism of $G$ is by definition a $mathbb Z$-module endomorphism of $G$, and it is obvious that any such endomorphism is also an endomorphism as an endo of $mathbb F_p$-vector space. Why is the converse true? I.e., why is any endomorphism of $G$ as an $mathbb F_p$-vector space must be an endomorphism of $G$ as a group ($mathbb Z$-module)? First of all, we need to define the structure of a $mathbb Z$-module on the $mathbb F_p$-vector space $G$; how can we do it?
linear-algebra abstract-algebra group-theory vector-spaces modules
asked Aug 6 at 0:51
user437309
556212
Do you believe/understand why $mathbbR^n$ is $n$-diml over the reals?
– Randall
Aug 6 at 1:33
Yes I understand.
– user437309
Aug 6 at 1:35
It's the same idea. You have a basis consisting of vectors like $(0, ldots, 0, 1, 0, ldots, 0)$ where the single $1$ in the $i$th spot takes care of the $i$th factor of $C_p$ in $G$. This is your dimension question. The endo question is different.
– Randall
Aug 6 at 1:37
The endo question: a vector space map is also a group homomorphism (just forget the scaling for a moment).
– Randall
Aug 6 at 1:39
Oh, I didn't realize that the basis for the $mathbb Z_p$-module can be chosen the same as the basis for the $mathbb Z$-module. I did realize that the $e_i$'s form a basis for $G$ as a $mathbb Z$-module.
– user437309
Aug 6 at 1:39
 |Â
show 4 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $p$ be a prime number. Consider a direct sum of $n$ copies of the cyclic group of order $p$, written $C_p$: $G=C_p^oplus n$. By a comment in this question, given an $R$-module $M$ and an ideal $Isubset R$, there is a structure of an $R/I$-module on $M$ given by $overline r x=rx$ ($rin R, xin M$) provided $IM=0$. In our case, $G$ is a $mathbb Z$-module; after setting $I=(p)$, we see that $IM=0$ is equivalent to $pG=0$, which holds. So there is a $mathbb Z/pmathbb Z$
-module structure on $G$. In other words, there is an $mathbb F_p$-vector space structure on $G$.
As I'm writing this, I came to realize that I don't really understand why is the dimension of this space $n$. (I believe this is so.) How to justify this?
Further, any group endomorphism of $G$ is by definition a $mathbb Z$-module endomorphism of $G$, and it is obvious that any such endomorphism is also an endomorphism as an endo of $mathbb F_p$-vector space. Why is the converse true? I.e., why is any endomorphism of $G$ as an $mathbb F_p$-vector space must be an endomorphism of $G$ as a group ($mathbb Z$-module)? First of all, we need to define the structure of a $mathbb Z$-module on the $mathbb F_p$-vector space $G$; how can we do it?
linear-algebra abstract-algebra group-theory vector-spaces modules
asked Aug 6 at 0:51
user437309
556212
Let $p$ be a prime number. Consider a direct sum of $n$ copies of the cyclic group of order $p$, written $C_p$: $G=C_p^oplus n$. By a comment in this question, given an $R$-module $M$ and an ideal $Isubset R$, there is a structure of an $R/I$-module on $M$ given by $overline r x=rx$ ($rin R, xin M$) provided $IM=0$. In our case, $G$ is a $mathbb Z$-module; after setting $I=(p)$, we see that $IM=0$ is equivalent to $pG=0$, which holds. So there is a $mathbb Z/pmathbb Z$
-module structure on $G$. In other words, there is an $mathbb F_p$-vector space structure on $G$.
As I'm writing this, I came to realize that I don't really understand why is the dimension of this space $n$. (I believe this is so.) How to justify this?
Further, any group endomorphism of $G$ is by definition a $mathbb Z$-module endomorphism of $G$, and it is obvious that any such endomorphism is also an endomorphism as an endo of $mathbb F_p$-vector space. Why is the converse true? I.e., why is any endomorphism of $G$ as an $mathbb F_p$-vector space must be an endomorphism of $G$ as a group ($mathbb Z$-module)? First of all, we need to define the structure of a $mathbb Z$-module on the $mathbb F_p$-vector space $G$; how can we do it?
linear-algebra abstract-algebra group-theory vector-spaces modules
asked Aug 6 at 0:51
user437309
556212
asked Aug 6 at 0:51
user437309
556212
asked Aug 6 at 0:51
user437309
556212
asked Aug 6 at 0:51
user437309
556212
556212
Do you believe/understand why $mathbbR^n$ is $n$-diml over the reals?
– Randall
Aug 6 at 1:33
Yes I understand.
– user437309
Aug 6 at 1:35
It's the same idea. You have a basis consisting of vectors like $(0, ldots, 0, 1, 0, ldots, 0)$ where the single $1$ in the $i$th spot takes care of the $i$th factor of $C_p$ in $G$. This is your dimension question. The endo question is different.
– Randall
Aug 6 at 1:37
The endo question: a vector space map is also a group homomorphism (just forget the scaling for a moment).
– Randall
Aug 6 at 1:39
Oh, I didn't realize that the basis for the $mathbb Z_p$-module can be chosen the same as the basis for the $mathbb Z$-module. I did realize that the $e_i$'s form a basis for $G$ as a $mathbb Z$-module.
– user437309
Aug 6 at 1:39
 |Â
show 4 more comments
Do you believe/understand why $mathbbR^n$ is $n$-diml over the reals?
– Randall
Aug 6 at 1:33
Yes I understand.
– user437309
Aug 6 at 1:35
It's the same idea. You have a basis consisting of vectors like $(0, ldots, 0, 1, 0, ldots, 0)$ where the single $1$ in the $i$th spot takes care of the $i$th factor of $C_p$ in $G$. This is your dimension question. The endo question is different.
– Randall
Aug 6 at 1:37
The endo question: a vector space map is also a group homomorphism (just forget the scaling for a moment).
– Randall
Aug 6 at 1:39
Oh, I didn't realize that the basis for the $mathbb Z_p$-module can be chosen the same as the basis for the $mathbb Z$-module. I did realize that the $e_i$'s form a basis for $G$ as a $mathbb Z$-module.
– user437309
Aug 6 at 1:39
Do you believe/understand why $mathbbR^n$ is $n$-diml over the reals?
– Randall
Aug 6 at 1:33
Do you believe/understand why $mathbbR^n$ is $n$-diml over the reals?
– Randall
Aug 6 at 1:33
Yes I understand.
– user437309
Aug 6 at 1:35
Yes I understand.
– user437309
Aug 6 at 1:35
It's the same idea. You have a basis consisting of vectors like $(0, ldots, 0, 1, 0, ldots, 0)$ where the single $1$ in the $i$th spot takes care of the $i$th factor of $C_p$ in $G$. This is your dimension question. The endo question is different.
– Randall
Aug 6 at 1:37
It's the same idea. You have a basis consisting of vectors like $(0, ldots, 0, 1, 0, ldots, 0)$ where the single $1$ in the $i$th spot takes care of the $i$th factor of $C_p$ in $G$. This is your dimension question. The endo question is different.
– Randall
Aug 6 at 1:37
The endo question: a vector space map is also a group homomorphism (just forget the scaling for a moment).
– Randall
Aug 6 at 1:39
The endo question: a vector space map is also a group homomorphism (just forget the scaling for a moment).
– Randall
Aug 6 at 1:39
Oh, I didn't realize that the basis for the $mathbb Z_p$-module can be chosen the same as the basis for the $mathbb Z$-module. I did realize that the $e_i$'s form a basis for $G$ as a $mathbb Z$-module.
– user437309
Aug 6 at 1:39
Oh, I didn't realize that the basis for the $mathbb Z_p$-module can be chosen the same as the basis for the $mathbb Z$-module. I did realize that the $e_i$'s form a basis for $G$ as a $mathbb Z$-module.
– user437309
Aug 6 at 1:39
 |Â
show 4 more comments
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Do you believe/understand why $mathbbR^n$ is $n$-diml over the reals?
– Randall
Aug 6 at 1:33
Yes I understand.
– user437309
Aug 6 at 1:35
It's the same idea. You have a basis consisting of vectors like $(0, ldots, 0, 1, 0, ldots, 0)$ where the single $1$ in the $i$th spot takes care of the $i$th factor of $C_p$ in $G$. This is your dimension question. The endo question is different.
– Randall
Aug 6 at 1:37
The endo question: a vector space map is also a group homomorphism (just forget the scaling for a moment).
– Randall
Aug 6 at 1:39
Oh, I didn't realize that the basis for the $mathbb Z_p$-module can be chosen the same as the basis for the $mathbb Z$-module. I did realize that the $e_i$'s form a basis for $G$ as a $mathbb Z$-module.
– user437309
Aug 6 at 1:39