Mixing
Linear Algebra over $mathbbZ/nmathbbZ$
Clash Royale CLAN TAG#URR8PPP
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We can see a direct product of $mathbbZ/nmathbbZ$ as a free $mathbbZ/nmathbbZ$ module. How does the results from linear algebra change if we change from $mathbbZ$ to $mathbbZ/nmathbbZ$.
Is for example the kernel of a matrix still well defined or do we need to be more careful?
Is there any reference for a book that covers this topic?
linear-algebra
asked Jul 24 at 6:54
Boki
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We can see a direct product of $mathbbZ/nmathbbZ$ as a free $mathbbZ/nmathbbZ$ module. How does the results from linear algebra change if we change from $mathbbZ$ to $mathbbZ/nmathbbZ$.
Is for example the kernel of a matrix still well defined or do we need to be more careful?
Is there any reference for a book that covers this topic?
linear-algebra
asked Jul 24 at 6:54
Boki
558
One difference is that you can have $AB=0$ with $det Ane0$ and $det Bne0$.
– Gerry Myerson
Jul 24 at 7:16
This question is too broad. It would be better to reformulate it as a reference request for a textbook that treats this subject (the linear algebra of $mathbb Z/nmathbb Z$), otherwise it should really be closed.
– Jack M
Jul 24 at 7:20
Many things behave the same way. If $n$ is prime, then you really end up with linear algebra. If $n$ is not prime, you have zero divisors, so the effect that Gerry Myerson is describing might occur. Still, a matrix is invertible if its determinant is invertible (i.e. is a unit in the ring $mathbbZ/mathbbZ$.
– Babelfish
Jul 24 at 7:24
I was reading recently that, over a finite field, the vector space is finite iff finite dimensional...
– Chris Custer
Jul 24 at 7:36
A finite direct product is free. This is true over any ring. I don't remember the argument, but I think that in general, an infinite product of modules is almost never free.
– tomasz
Jul 24 at 7:42
 |Â
show 5 more comments
up vote
1
down vote
favorite
up vote
1
down vote
favorite
We can see a direct product of $mathbbZ/nmathbbZ$ as a free $mathbbZ/nmathbbZ$ module. How does the results from linear algebra change if we change from $mathbbZ$ to $mathbbZ/nmathbbZ$.
Is for example the kernel of a matrix still well defined or do we need to be more careful?
Is there any reference for a book that covers this topic?
linear-algebra
asked Jul 24 at 6:54
Boki
558
We can see a direct product of $mathbbZ/nmathbbZ$ as a free $mathbbZ/nmathbbZ$ module. How does the results from linear algebra change if we change from $mathbbZ$ to $mathbbZ/nmathbbZ$.
Is for example the kernel of a matrix still well defined or do we need to be more careful?
Is there any reference for a book that covers this topic?
linear-algebra
asked Jul 24 at 6:54
Boki
558
edited Jul 24 at 7:23
asked Jul 24 at 6:54
Boki
558
asked Jul 24 at 6:54
Boki
558
asked Jul 24 at 6:54
Boki
558
558
One difference is that you can have $AB=0$ with $det Ane0$ and $det Bne0$.
– Gerry Myerson
Jul 24 at 7:16
This question is too broad. It would be better to reformulate it as a reference request for a textbook that treats this subject (the linear algebra of $mathbb Z/nmathbb Z$), otherwise it should really be closed.
– Jack M
Jul 24 at 7:20
Many things behave the same way. If $n$ is prime, then you really end up with linear algebra. If $n$ is not prime, you have zero divisors, so the effect that Gerry Myerson is describing might occur. Still, a matrix is invertible if its determinant is invertible (i.e. is a unit in the ring $mathbbZ/mathbbZ$.
– Babelfish
Jul 24 at 7:24
I was reading recently that, over a finite field, the vector space is finite iff finite dimensional...
– Chris Custer
Jul 24 at 7:36
A finite direct product is free. This is true over any ring. I don't remember the argument, but I think that in general, an infinite product of modules is almost never free.
– tomasz
Jul 24 at 7:42
 |Â
show 5 more comments
One difference is that you can have $AB=0$ with $det Ane0$ and $det Bne0$.
– Gerry Myerson
Jul 24 at 7:16
This question is too broad. It would be better to reformulate it as a reference request for a textbook that treats this subject (the linear algebra of $mathbb Z/nmathbb Z$), otherwise it should really be closed.
– Jack M
Jul 24 at 7:20
Many things behave the same way. If $n$ is prime, then you really end up with linear algebra. If $n$ is not prime, you have zero divisors, so the effect that Gerry Myerson is describing might occur. Still, a matrix is invertible if its determinant is invertible (i.e. is a unit in the ring $mathbbZ/mathbbZ$.
– Babelfish
Jul 24 at 7:24
I was reading recently that, over a finite field, the vector space is finite iff finite dimensional...
– Chris Custer
Jul 24 at 7:36
A finite direct product is free. This is true over any ring. I don't remember the argument, but I think that in general, an infinite product of modules is almost never free.
– tomasz
Jul 24 at 7:42
One difference is that you can have $AB=0$ with $det Ane0$ and $det Bne0$.
– Gerry Myerson
Jul 24 at 7:16
One difference is that you can have $AB=0$ with $det Ane0$ and $det Bne0$.
– Gerry Myerson
Jul 24 at 7:16
This question is too broad. It would be better to reformulate it as a reference request for a textbook that treats this subject (the linear algebra of $mathbb Z/nmathbb Z$), otherwise it should really be closed.
– Jack M
Jul 24 at 7:20
This question is too broad. It would be better to reformulate it as a reference request for a textbook that treats this subject (the linear algebra of $mathbb Z/nmathbb Z$), otherwise it should really be closed.
– Jack M
Jul 24 at 7:20
Many things behave the same way. If $n$ is prime, then you really end up with linear algebra. If $n$ is not prime, you have zero divisors, so the effect that Gerry Myerson is describing might occur. Still, a matrix is invertible if its determinant is invertible (i.e. is a unit in the ring $mathbbZ/mathbbZ$.
– Babelfish
Jul 24 at 7:24
Many things behave the same way. If $n$ is prime, then you really end up with linear algebra. If $n$ is not prime, you have zero divisors, so the effect that Gerry Myerson is describing might occur. Still, a matrix is invertible if its determinant is invertible (i.e. is a unit in the ring $mathbbZ/mathbbZ$.
– Babelfish
Jul 24 at 7:24
I was reading recently that, over a finite field, the vector space is finite iff finite dimensional...
– Chris Custer
Jul 24 at 7:36
I was reading recently that, over a finite field, the vector space is finite iff finite dimensional...
– Chris Custer
Jul 24 at 7:36
A finite direct product is free. This is true over any ring. I don't remember the argument, but I think that in general, an infinite product of modules is almost never free.
– tomasz
Jul 24 at 7:42
A finite direct product is free. This is true over any ring. I don't remember the argument, but I think that in general, an infinite product of modules is almost never free.
– tomasz
Jul 24 at 7:42
 |Â
show 5 more comments
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One difference is that you can have $AB=0$ with $det Ane0$ and $det Bne0$.
– Gerry Myerson
Jul 24 at 7:16
This question is too broad. It would be better to reformulate it as a reference request for a textbook that treats this subject (the linear algebra of $mathbb Z/nmathbb Z$), otherwise it should really be closed.
– Jack M
Jul 24 at 7:20
Many things behave the same way. If $n$ is prime, then you really end up with linear algebra. If $n$ is not prime, you have zero divisors, so the effect that Gerry Myerson is describing might occur. Still, a matrix is invertible if its determinant is invertible (i.e. is a unit in the ring $mathbbZ/mathbbZ$.
– Babelfish
Jul 24 at 7:24
I was reading recently that, over a finite field, the vector space is finite iff finite dimensional...
– Chris Custer
Jul 24 at 7:36
A finite direct product is free. This is true over any ring. I don't remember the argument, but I think that in general, an infinite product of modules is almost never free.
– tomasz
Jul 24 at 7:42