Homomorphisms of a matrix ring.
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I have the following question, What are all possible right $R$-module $(R = mathbb{M}_2(mathbb{Z}))$ homomorphisms from $mathbb{M}_2(nmathbb{Z})$ to $mathbb{M}_2(mathbb{Z})$. I tried to extend maps from $nmathbb{Z}$ to $mathbb{Z}$ defined as $f_{n,a}(nx) = ax$ and found that these are right module homomorphisms. But are these the only ones like in case of $mathbb{Z}$ ??
abstract-algebra matrices ring-theory
asked Nov 15 at 17:41
Rick
406
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0
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favorite
I have the following question, What are all possible right $R$-module $(R = mathbb{M}_2(mathbb{Z}))$ homomorphisms from $mathbb{M}_2(nmathbb{Z})$ to $mathbb{M}_2(mathbb{Z})$. I tried to extend maps from $nmathbb{Z}$ to $mathbb{Z}$ defined as $f_{n,a}(nx) = ax$ and found that these are right module homomorphisms. But are these the only ones like in case of $mathbb{Z}$ ??
abstract-algebra matrices ring-theory
asked Nov 15 at 17:41
Rick
406
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I have the following question, What are all possible right $R$-module $(R = mathbb{M}_2(mathbb{Z}))$ homomorphisms from $mathbb{M}_2(nmathbb{Z})$ to $mathbb{M}_2(mathbb{Z})$. I tried to extend maps from $nmathbb{Z}$ to $mathbb{Z}$ defined as $f_{n,a}(nx) = ax$ and found that these are right module homomorphisms. But are these the only ones like in case of $mathbb{Z}$ ??
abstract-algebra matrices ring-theory
asked Nov 15 at 17:41
Rick
406
I have the following question, What are all possible right $R$-module $(R = mathbb{M}_2(mathbb{Z}))$ homomorphisms from $mathbb{M}_2(nmathbb{Z})$ to $mathbb{M}_2(mathbb{Z})$. I tried to extend maps from $nmathbb{Z}$ to $mathbb{Z}$ defined as $f_{n,a}(nx) = ax$ and found that these are right module homomorphisms. But are these the only ones like in case of $mathbb{Z}$ ??
abstract-algebra matrices ring-theory
abstract-algebra matrices ring-theory
asked Nov 15 at 17:41
Rick
406
asked Nov 15 at 17:41
Rick
406
asked Nov 15 at 17:41
Rick
406
asked Nov 15 at 17:41
Rick
406
asked Nov 15 at 17:41
Rick
406
406
add a comment |
add a comment |
1 Answer
1
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1
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Since $Bbb{Z}$ is an integral domain, the entries of the matrix $nx$ are uniquely determined by $x.$
Now, you may want to ask yourself:
Is there a homomorphism which assigns
$$begin{pmatrix}
1& 0\
0 &1
end{pmatrix}mapstobegin{pmatrix}
0 &1 \
1 &0
end{pmatrix} $$
that is defined by $f_{n,a}(nx)=ax?$
answered Nov 15 at 17:47
Chickenmancer
3,209723
Sry, I didn't get what you want to explain from the above answer. Do you want to say that $f_{n,a}$ is not a homo?
– Rick
Nov 15 at 17:56
No, the question is can you interchange the action of $n$ on $x$ with $a$ on $x$ to produce the map which I asked you about. You may want to try a few examples.
– Chickenmancer
Nov 15 at 17:58
Thanks, I got it now what you meant.
– Rick
Nov 17 at 21:24
I think the thing to try to appreciate is how scalar action of integers can't transpose the columns. Please consider clicking "accept" if you found this suitable. It is a check mark underneath the vote options. I'm glad it helped!
– Chickenmancer
Nov 19 at 0:41
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Since $Bbb{Z}$ is an integral domain, the entries of the matrix $nx$ are uniquely determined by $x.$
Now, you may want to ask yourself:
Is there a homomorphism which assigns
$$begin{pmatrix}
1& 0\
0 &1
end{pmatrix}mapstobegin{pmatrix}
0 &1 \
1 &0
end{pmatrix} $$
that is defined by $f_{n,a}(nx)=ax?$
answered Nov 15 at 17:47
Chickenmancer
3,209723
Sry, I didn't get what you want to explain from the above answer. Do you want to say that $f_{n,a}$ is not a homo?
– Rick
Nov 15 at 17:56
No, the question is can you interchange the action of $n$ on $x$ with $a$ on $x$ to produce the map which I asked you about. You may want to try a few examples.
– Chickenmancer
Nov 15 at 17:58
Thanks, I got it now what you meant.
– Rick
Nov 17 at 21:24
I think the thing to try to appreciate is how scalar action of integers can't transpose the columns. Please consider clicking "accept" if you found this suitable. It is a check mark underneath the vote options. I'm glad it helped!
– Chickenmancer
Nov 19 at 0:41
add a comment |
up vote
1
down vote
accepted
Since $Bbb{Z}$ is an integral domain, the entries of the matrix $nx$ are uniquely determined by $x.$
Now, you may want to ask yourself:
Is there a homomorphism which assigns
$$begin{pmatrix}
1& 0\
0 &1
end{pmatrix}mapstobegin{pmatrix}
0 &1 \
1 &0
end{pmatrix} $$
that is defined by $f_{n,a}(nx)=ax?$
answered Nov 15 at 17:47
Chickenmancer
3,209723
Sry, I didn't get what you want to explain from the above answer. Do you want to say that $f_{n,a}$ is not a homo?
– Rick
Nov 15 at 17:56
No, the question is can you interchange the action of $n$ on $x$ with $a$ on $x$ to produce the map which I asked you about. You may want to try a few examples.
– Chickenmancer
Nov 15 at 17:58
Thanks, I got it now what you meant.
– Rick
Nov 17 at 21:24
I think the thing to try to appreciate is how scalar action of integers can't transpose the columns. Please consider clicking "accept" if you found this suitable. It is a check mark underneath the vote options. I'm glad it helped!
– Chickenmancer
Nov 19 at 0:41
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Since $Bbb{Z}$ is an integral domain, the entries of the matrix $nx$ are uniquely determined by $x.$
Now, you may want to ask yourself:
Is there a homomorphism which assigns
$$begin{pmatrix}
1& 0\
0 &1
end{pmatrix}mapstobegin{pmatrix}
0 &1 \
1 &0
end{pmatrix} $$
that is defined by $f_{n,a}(nx)=ax?$
answered Nov 15 at 17:47
Chickenmancer
3,209723
Since $Bbb{Z}$ is an integral domain, the entries of the matrix $nx$ are uniquely determined by $x.$
Now, you may want to ask yourself:
Is there a homomorphism which assigns
$$begin{pmatrix}
1& 0\
0 &1
end{pmatrix}mapstobegin{pmatrix}
0 &1 \
1 &0
end{pmatrix} $$
that is defined by $f_{n,a}(nx)=ax?$
answered Nov 15 at 17:47
Chickenmancer
3,209723
answered Nov 15 at 17:47
Chickenmancer
3,209723
answered Nov 15 at 17:47
Chickenmancer
3,209723
answered Nov 15 at 17:47
Chickenmancer
3,209723
3,209723
Sry, I didn't get what you want to explain from the above answer. Do you want to say that $f_{n,a}$ is not a homo?
– Rick
Nov 15 at 17:56
No, the question is can you interchange the action of $n$ on $x$ with $a$ on $x$ to produce the map which I asked you about. You may want to try a few examples.
– Chickenmancer
Nov 15 at 17:58
Thanks, I got it now what you meant.
– Rick
Nov 17 at 21:24
I think the thing to try to appreciate is how scalar action of integers can't transpose the columns. Please consider clicking "accept" if you found this suitable. It is a check mark underneath the vote options. I'm glad it helped!
– Chickenmancer
Nov 19 at 0:41
add a comment |
Sry, I didn't get what you want to explain from the above answer. Do you want to say that $f_{n,a}$ is not a homo?
– Rick
Nov 15 at 17:56
No, the question is can you interchange the action of $n$ on $x$ with $a$ on $x$ to produce the map which I asked you about. You may want to try a few examples.
– Chickenmancer
Nov 15 at 17:58
Thanks, I got it now what you meant.
– Rick
Nov 17 at 21:24
I think the thing to try to appreciate is how scalar action of integers can't transpose the columns. Please consider clicking "accept" if you found this suitable. It is a check mark underneath the vote options. I'm glad it helped!
– Chickenmancer
Nov 19 at 0:41
Sry, I didn't get what you want to explain from the above answer. Do you want to say that $f_{n,a}$ is not a homo?
– Rick
Nov 15 at 17:56
Sry, I didn't get what you want to explain from the above answer. Do you want to say that $f_{n,a}$ is not a homo?
– Rick
Nov 15 at 17:56
No, the question is can you interchange the action of $n$ on $x$ with $a$ on $x$ to produce the map which I asked you about. You may want to try a few examples.
– Chickenmancer
Nov 15 at 17:58
No, the question is can you interchange the action of $n$ on $x$ with $a$ on $x$ to produce the map which I asked you about. You may want to try a few examples.
– Chickenmancer
Nov 15 at 17:58
Thanks, I got it now what you meant.
– Rick
Nov 17 at 21:24
Thanks, I got it now what you meant.
– Rick
Nov 17 at 21:24
I think the thing to try to appreciate is how scalar action of integers can't transpose the columns. Please consider clicking "accept" if you found this suitable. It is a check mark underneath the vote options. I'm glad it helped!
– Chickenmancer
Nov 19 at 0:41
I think the thing to try to appreciate is how scalar action of integers can't transpose the columns. Please consider clicking "accept" if you found this suitable. It is a check mark underneath the vote options. I'm glad it helped!
– Chickenmancer
Nov 19 at 0:41
add a comment |
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