Mixing
Confusion about the quotient $mathbb Z^4/H$
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Let $f:mathbb Z^3tomathbb Z^4$ be the group homomorphism given by $$f(a,b,c)=(a+b+c,a+3b+c,a+b+5c,4a+8b).$$ Let $H$ be the image of $f$. Find an element of infinite order in $mathbb Z^4 /H$ and find the order of the torsion subgroup of that quotient.
This problem certainly has to do with presentation matrices, but I'm too confused. Let $A$ denote the matrix with rows $(1,1,1),(1,3,1),(1,1,5),(4,8,0)$. Then $Amathbb Z^3=H$. So $A$ must be a presentation matrix for $mathbb Z^4/H$ according to the definition from here (or from here, both of them agree with Artin's definition). So in the quotient module the relations $$v_1+v_2+v_3=0,\v_1+3v_2+v_3=0,\v_1+v_2+5v_3=0,\ 4v_1+8v_2=0$$ hold. But then according to Artin's text, the matrix should be $3times 4$, not $4times 3$ (see Example 14.5.2 in this question). So what is the size of a presentation matrix in this case? My size doesn't agree with Artin's notation, even though I use his definitions and conventions.
Example 14.5.2 The $mathbbZ-$module or an abelian group $V$ that is generated by three elements $v_1, v_2, v_3$ with the compete
set of relations
$$ 3v_1+2v_2+v_3=0\ 8v_1+4v_2+2v_3=0\ 7v_1+6v_2+2v_3=0\
9v_1+6v_2+v_3=0 $$
is presented by the matrix
$$
A=beginbmatrix
3 & 8 & 7 & 9 \
2 & 4 & 6 & 6 \
1 & 2 & 2 & 1 \
endbmatrix. $$
Its columns are the coefficients of the (above) relations: $(v_1, v_2,
v_3)A=(0, 0, 0, 0)$.
abstract-algebra group-theory modules quotient-group
edited Aug 3 at 21:20
k.stm
10.4k22149
asked Aug 3 at 21:12
user437309
534212
 |Â
show 5 more comments
up vote
2
down vote
favorite
Let $f:mathbb Z^3tomathbb Z^4$ be the group homomorphism given by $$f(a,b,c)=(a+b+c,a+3b+c,a+b+5c,4a+8b).$$ Let $H$ be the image of $f$. Find an element of infinite order in $mathbb Z^4 /H$ and find the order of the torsion subgroup of that quotient.
This problem certainly has to do with presentation matrices, but I'm too confused. Let $A$ denote the matrix with rows $(1,1,1),(1,3,1),(1,1,5),(4,8,0)$. Then $Amathbb Z^3=H$. So $A$ must be a presentation matrix for $mathbb Z^4/H$ according to the definition from here (or from here, both of them agree with Artin's definition). So in the quotient module the relations $$v_1+v_2+v_3=0,\v_1+3v_2+v_3=0,\v_1+v_2+5v_3=0,\ 4v_1+8v_2=0$$ hold. But then according to Artin's text, the matrix should be $3times 4$, not $4times 3$ (see Example 14.5.2 in this question). So what is the size of a presentation matrix in this case? My size doesn't agree with Artin's notation, even though I use his definitions and conventions.
Example 14.5.2 The $mathbbZ-$module or an abelian group $V$ that is generated by three elements $v_1, v_2, v_3$ with the compete
set of relations
$$ 3v_1+2v_2+v_3=0\ 8v_1+4v_2+2v_3=0\ 7v_1+6v_2+2v_3=0\
9v_1+6v_2+v_3=0 $$
is presented by the matrix
$$
A=beginbmatrix
3 & 8 & 7 & 9 \
2 & 4 & 6 & 6 \
1 & 2 & 2 & 1 \
endbmatrix. $$
Its columns are the coefficients of the (above) relations: $(v_1, v_2,
v_3)A=(0, 0, 0, 0)$.
abstract-algebra group-theory modules quotient-group
edited Aug 3 at 21:20
k.stm
10.4k22149
asked Aug 3 at 21:12
user437309
534212
What are $v_1, v_2, v_3$?
– k.stm
Aug 3 at 21:22
They are generators of $mathbb Z^4/H$.
– user437309
Aug 3 at 21:23
Yeah, I'm not sure why Artin is using the transpose there (assuming everything has been transcribed correctly). A matrix describing a linear map $R^3to R^4$ for a commutative ring $R$ is a $4times 3$ matrix.
– jgon
Aug 3 at 21:25
That being said, the shape of the matrix is convention, and doesn't really affect the underlying mathematics.
– jgon
Aug 3 at 21:26
It doesn't affect the underlying mathematics, but the fact that following Artin's conventions doesn't give me an answer in his convention may mean that I misunderstand the underlying mathematics.
– user437309
Aug 3 at 21:33
 |Â
show 5 more comments
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Let $f:mathbb Z^3tomathbb Z^4$ be the group homomorphism given by $$f(a,b,c)=(a+b+c,a+3b+c,a+b+5c,4a+8b).$$ Let $H$ be the image of $f$. Find an element of infinite order in $mathbb Z^4 /H$ and find the order of the torsion subgroup of that quotient.
This problem certainly has to do with presentation matrices, but I'm too confused. Let $A$ denote the matrix with rows $(1,1,1),(1,3,1),(1,1,5),(4,8,0)$. Then $Amathbb Z^3=H$. So $A$ must be a presentation matrix for $mathbb Z^4/H$ according to the definition from here (or from here, both of them agree with Artin's definition). So in the quotient module the relations $$v_1+v_2+v_3=0,\v_1+3v_2+v_3=0,\v_1+v_2+5v_3=0,\ 4v_1+8v_2=0$$ hold. But then according to Artin's text, the matrix should be $3times 4$, not $4times 3$ (see Example 14.5.2 in this question). So what is the size of a presentation matrix in this case? My size doesn't agree with Artin's notation, even though I use his definitions and conventions.
Example 14.5.2 The $mathbbZ-$module or an abelian group $V$ that is generated by three elements $v_1, v_2, v_3$ with the compete
set of relations
$$ 3v_1+2v_2+v_3=0\ 8v_1+4v_2+2v_3=0\ 7v_1+6v_2+2v_3=0\
9v_1+6v_2+v_3=0 $$
is presented by the matrix
$$
A=beginbmatrix
3 & 8 & 7 & 9 \
2 & 4 & 6 & 6 \
1 & 2 & 2 & 1 \
endbmatrix. $$
Its columns are the coefficients of the (above) relations: $(v_1, v_2,
v_3)A=(0, 0, 0, 0)$.
abstract-algebra group-theory modules quotient-group
edited Aug 3 at 21:20
k.stm
10.4k22149
asked Aug 3 at 21:12
user437309
534212
Let $f:mathbb Z^3tomathbb Z^4$ be the group homomorphism given by $$f(a,b,c)=(a+b+c,a+3b+c,a+b+5c,4a+8b).$$ Let $H$ be the image of $f$. Find an element of infinite order in $mathbb Z^4 /H$ and find the order of the torsion subgroup of that quotient.
This problem certainly has to do with presentation matrices, but I'm too confused. Let $A$ denote the matrix with rows $(1,1,1),(1,3,1),(1,1,5),(4,8,0)$. Then $Amathbb Z^3=H$. So $A$ must be a presentation matrix for $mathbb Z^4/H$ according to the definition from here (or from here, both of them agree with Artin's definition). So in the quotient module the relations $$v_1+v_2+v_3=0,\v_1+3v_2+v_3=0,\v_1+v_2+5v_3=0,\ 4v_1+8v_2=0$$ hold. But then according to Artin's text, the matrix should be $3times 4$, not $4times 3$ (see Example 14.5.2 in this question). So what is the size of a presentation matrix in this case? My size doesn't agree with Artin's notation, even though I use his definitions and conventions.
Example 14.5.2 The $mathbbZ-$module or an abelian group $V$ that is generated by three elements $v_1, v_2, v_3$ with the compete
set of relations
$$ 3v_1+2v_2+v_3=0\ 8v_1+4v_2+2v_3=0\ 7v_1+6v_2+2v_3=0\
9v_1+6v_2+v_3=0 $$
is presented by the matrix
$$
A=beginbmatrix
3 & 8 & 7 & 9 \
2 & 4 & 6 & 6 \
1 & 2 & 2 & 1 \
endbmatrix. $$
Its columns are the coefficients of the (above) relations: $(v_1, v_2,
v_3)A=(0, 0, 0, 0)$.
abstract-algebra group-theory modules quotient-group
edited Aug 3 at 21:20
k.stm
10.4k22149
asked Aug 3 at 21:12
user437309
534212
edited Aug 3 at 21:20
k.stm
10.4k22149
edited Aug 3 at 21:20
k.stm
10.4k22149
edited Aug 3 at 21:20
k.stm
10.4k22149
10.4k22149
asked Aug 3 at 21:12
user437309
534212
asked Aug 3 at 21:12
user437309
534212
asked Aug 3 at 21:12
user437309
534212
534212
What are $v_1, v_2, v_3$?
– k.stm
Aug 3 at 21:22
They are generators of $mathbb Z^4/H$.
– user437309
Aug 3 at 21:23
Yeah, I'm not sure why Artin is using the transpose there (assuming everything has been transcribed correctly). A matrix describing a linear map $R^3to R^4$ for a commutative ring $R$ is a $4times 3$ matrix.
– jgon
Aug 3 at 21:25
That being said, the shape of the matrix is convention, and doesn't really affect the underlying mathematics.
– jgon
Aug 3 at 21:26
It doesn't affect the underlying mathematics, but the fact that following Artin's conventions doesn't give me an answer in his convention may mean that I misunderstand the underlying mathematics.
– user437309
Aug 3 at 21:33
 |Â
show 5 more comments
What are $v_1, v_2, v_3$?
– k.stm
Aug 3 at 21:22
They are generators of $mathbb Z^4/H$.
– user437309
Aug 3 at 21:23
Yeah, I'm not sure why Artin is using the transpose there (assuming everything has been transcribed correctly). A matrix describing a linear map $R^3to R^4$ for a commutative ring $R$ is a $4times 3$ matrix.
– jgon
Aug 3 at 21:25
That being said, the shape of the matrix is convention, and doesn't really affect the underlying mathematics.
– jgon
Aug 3 at 21:26
It doesn't affect the underlying mathematics, but the fact that following Artin's conventions doesn't give me an answer in his convention may mean that I misunderstand the underlying mathematics.
– user437309
Aug 3 at 21:33
What are $v_1, v_2, v_3$?
– k.stm
Aug 3 at 21:22
What are $v_1, v_2, v_3$?
– k.stm
Aug 3 at 21:22
They are generators of $mathbb Z^4/H$.
– user437309
Aug 3 at 21:23
They are generators of $mathbb Z^4/H$.
– user437309
Aug 3 at 21:23
Yeah, I'm not sure why Artin is using the transpose there (assuming everything has been transcribed correctly). A matrix describing a linear map $R^3to R^4$ for a commutative ring $R$ is a $4times 3$ matrix.
– jgon
Aug 3 at 21:25
Yeah, I'm not sure why Artin is using the transpose there (assuming everything has been transcribed correctly). A matrix describing a linear map $R^3to R^4$ for a commutative ring $R$ is a $4times 3$ matrix.
– jgon
Aug 3 at 21:25
That being said, the shape of the matrix is convention, and doesn't really affect the underlying mathematics.
– jgon
Aug 3 at 21:26
That being said, the shape of the matrix is convention, and doesn't really affect the underlying mathematics.
– jgon
Aug 3 at 21:26
It doesn't affect the underlying mathematics, but the fact that following Artin's conventions doesn't give me an answer in his convention may mean that I misunderstand the underlying mathematics.
– user437309
Aug 3 at 21:33
It doesn't affect the underlying mathematics, but the fact that following Artin's conventions doesn't give me an answer in his convention may mean that I misunderstand the underlying mathematics.
– user437309
Aug 3 at 21:33
 |Â
show 5 more comments
1 Answer
1
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oldest
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2
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As I mentioned in the comments, conventionally, the shape of the matrix ought to be $4times 3$ as you've worked out. So we have
$$A=newcommandbmatbeginpmatrixnewcommandematendpmatrixbmat 1 & 1 & 1 \ 1 & 3 & 1 \ 1 & 1 & 5 \ 4 & 8 & 0 emat.$$
Now row operations are isomorphisms of $BbbZ^3$, so we can freely row reduce this matrix and preserve the image of $ABbbZ^3$ as a subset of $newcommandZZBbbZZZ^4$. I.e. applying row operations to a matrix $A$ that presents $ZZ^4/AZZ^3$ to get a matrix $A'$ means that $A'$ also presents $ZZ^4/AZZ^3$.
Subtracting the first row from the others to make the other entries of the first column zero, we have
$$A'=bmat 1 & 1 & 1 \ 0 & 2 & 0 \ 0 & 0 & 4 \ 0 & 4 & -4 emat.$$
Then subtracting twice the second row and adding the third row to the fourth row, we get
$$A''=bmat 1 & 1 & 1 \ 0 & 2 & 0 \ 0 & 0 & 4 \ 0 & 0 & 0 emat.$$
Thus $$v_4=bmat 0 \ 0 \ 0 \ 1 emat$$ is never in the image of $A$, even over $BbbQ$, hence $v_4$ has infinite order in the cokernel.
Finally, if we don't care about the coordinates on $ZZ^4$, column operations on $A''$ correspond to applying automorphisms of $ZZ^4$, and hence preserve the isomorphism class of $ZZ^4/A''ZZ^3$. However, if we subtract the first column from the second and third, we get
$$A''' = bmat 1 & 0 & 0 \ 0 & 2 & 0 \ 0 & 0 & 4 \ 0 & 0 & 0 emat.$$
Then since $A'''$ is diagonal, $ZZ^4/A'''ZZ^3$ splits as a direct sum, $$ZZ/ZZ oplus ZZ/2ZZ oplus ZZ/4ZZoplus ZZ/(0),$$
so the torsion subgroup of $ZZ^4/AZZ^3$ has size 8.
answered Aug 3 at 21:49
jgon
8,53611335
How does the fact that $v_4$ is not in the image of $A$ imply that it has infinite order in the cokernel?
– user437309
yesterday
@user437309 the important thing is that it isn't in the $BbbQ$ image of $A$, since if it had finite order, then for some $n$, $nv_4$ would be in the $ZZ$ image of A, but then $v_4$ would be in the $BbbQ$ image.
– jgon
yesterday
What is the $mathbb Q$-image of $A$? Also I don't understand why if it had finite order, then $nv_4$ would be in the $mathbb Z$-image. (By $mathbb Z$ image I just understand the set of all vectors gotten by multiplication of $A$ by all vectors of an appropriate size.)
– user437309
yesterday
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
As I mentioned in the comments, conventionally, the shape of the matrix ought to be $4times 3$ as you've worked out. So we have
$$A=newcommandbmatbeginpmatrixnewcommandematendpmatrixbmat 1 & 1 & 1 \ 1 & 3 & 1 \ 1 & 1 & 5 \ 4 & 8 & 0 emat.$$
Now row operations are isomorphisms of $BbbZ^3$, so we can freely row reduce this matrix and preserve the image of $ABbbZ^3$ as a subset of $newcommandZZBbbZZZ^4$. I.e. applying row operations to a matrix $A$ that presents $ZZ^4/AZZ^3$ to get a matrix $A'$ means that $A'$ also presents $ZZ^4/AZZ^3$.
Subtracting the first row from the others to make the other entries of the first column zero, we have
$$A'=bmat 1 & 1 & 1 \ 0 & 2 & 0 \ 0 & 0 & 4 \ 0 & 4 & -4 emat.$$
Then subtracting twice the second row and adding the third row to the fourth row, we get
$$A''=bmat 1 & 1 & 1 \ 0 & 2 & 0 \ 0 & 0 & 4 \ 0 & 0 & 0 emat.$$
Thus $$v_4=bmat 0 \ 0 \ 0 \ 1 emat$$ is never in the image of $A$, even over $BbbQ$, hence $v_4$ has infinite order in the cokernel.
Finally, if we don't care about the coordinates on $ZZ^4$, column operations on $A''$ correspond to applying automorphisms of $ZZ^4$, and hence preserve the isomorphism class of $ZZ^4/A''ZZ^3$. However, if we subtract the first column from the second and third, we get
$$A''' = bmat 1 & 0 & 0 \ 0 & 2 & 0 \ 0 & 0 & 4 \ 0 & 0 & 0 emat.$$
Then since $A'''$ is diagonal, $ZZ^4/A'''ZZ^3$ splits as a direct sum, $$ZZ/ZZ oplus ZZ/2ZZ oplus ZZ/4ZZoplus ZZ/(0),$$
so the torsion subgroup of $ZZ^4/AZZ^3$ has size 8.
answered Aug 3 at 21:49
jgon
8,53611335
How does the fact that $v_4$ is not in the image of $A$ imply that it has infinite order in the cokernel?
– user437309
yesterday
@user437309 the important thing is that it isn't in the $BbbQ$ image of $A$, since if it had finite order, then for some $n$, $nv_4$ would be in the $ZZ$ image of A, but then $v_4$ would be in the $BbbQ$ image.
– jgon
yesterday
What is the $mathbb Q$-image of $A$? Also I don't understand why if it had finite order, then $nv_4$ would be in the $mathbb Z$-image. (By $mathbb Z$ image I just understand the set of all vectors gotten by multiplication of $A$ by all vectors of an appropriate size.)
– user437309
yesterday
add a comment |Â
up vote
2
down vote
As I mentioned in the comments, conventionally, the shape of the matrix ought to be $4times 3$ as you've worked out. So we have
$$A=newcommandbmatbeginpmatrixnewcommandematendpmatrixbmat 1 & 1 & 1 \ 1 & 3 & 1 \ 1 & 1 & 5 \ 4 & 8 & 0 emat.$$
Now row operations are isomorphisms of $BbbZ^3$, so we can freely row reduce this matrix and preserve the image of $ABbbZ^3$ as a subset of $newcommandZZBbbZZZ^4$. I.e. applying row operations to a matrix $A$ that presents $ZZ^4/AZZ^3$ to get a matrix $A'$ means that $A'$ also presents $ZZ^4/AZZ^3$.
Subtracting the first row from the others to make the other entries of the first column zero, we have
$$A'=bmat 1 & 1 & 1 \ 0 & 2 & 0 \ 0 & 0 & 4 \ 0 & 4 & -4 emat.$$
Then subtracting twice the second row and adding the third row to the fourth row, we get
$$A''=bmat 1 & 1 & 1 \ 0 & 2 & 0 \ 0 & 0 & 4 \ 0 & 0 & 0 emat.$$
Thus $$v_4=bmat 0 \ 0 \ 0 \ 1 emat$$ is never in the image of $A$, even over $BbbQ$, hence $v_4$ has infinite order in the cokernel.
Finally, if we don't care about the coordinates on $ZZ^4$, column operations on $A''$ correspond to applying automorphisms of $ZZ^4$, and hence preserve the isomorphism class of $ZZ^4/A''ZZ^3$. However, if we subtract the first column from the second and third, we get
$$A''' = bmat 1 & 0 & 0 \ 0 & 2 & 0 \ 0 & 0 & 4 \ 0 & 0 & 0 emat.$$
Then since $A'''$ is diagonal, $ZZ^4/A'''ZZ^3$ splits as a direct sum, $$ZZ/ZZ oplus ZZ/2ZZ oplus ZZ/4ZZoplus ZZ/(0),$$
so the torsion subgroup of $ZZ^4/AZZ^3$ has size 8.
answered Aug 3 at 21:49
jgon
8,53611335
How does the fact that $v_4$ is not in the image of $A$ imply that it has infinite order in the cokernel?
– user437309
yesterday
@user437309 the important thing is that it isn't in the $BbbQ$ image of $A$, since if it had finite order, then for some $n$, $nv_4$ would be in the $ZZ$ image of A, but then $v_4$ would be in the $BbbQ$ image.
– jgon
yesterday
What is the $mathbb Q$-image of $A$? Also I don't understand why if it had finite order, then $nv_4$ would be in the $mathbb Z$-image. (By $mathbb Z$ image I just understand the set of all vectors gotten by multiplication of $A$ by all vectors of an appropriate size.)
– user437309
yesterday
add a comment |Â
up vote
2
down vote
up vote
2
down vote
As I mentioned in the comments, conventionally, the shape of the matrix ought to be $4times 3$ as you've worked out. So we have
$$A=newcommandbmatbeginpmatrixnewcommandematendpmatrixbmat 1 & 1 & 1 \ 1 & 3 & 1 \ 1 & 1 & 5 \ 4 & 8 & 0 emat.$$
Now row operations are isomorphisms of $BbbZ^3$, so we can freely row reduce this matrix and preserve the image of $ABbbZ^3$ as a subset of $newcommandZZBbbZZZ^4$. I.e. applying row operations to a matrix $A$ that presents $ZZ^4/AZZ^3$ to get a matrix $A'$ means that $A'$ also presents $ZZ^4/AZZ^3$.
Subtracting the first row from the others to make the other entries of the first column zero, we have
$$A'=bmat 1 & 1 & 1 \ 0 & 2 & 0 \ 0 & 0 & 4 \ 0 & 4 & -4 emat.$$
Then subtracting twice the second row and adding the third row to the fourth row, we get
$$A''=bmat 1 & 1 & 1 \ 0 & 2 & 0 \ 0 & 0 & 4 \ 0 & 0 & 0 emat.$$
Thus $$v_4=bmat 0 \ 0 \ 0 \ 1 emat$$ is never in the image of $A$, even over $BbbQ$, hence $v_4$ has infinite order in the cokernel.
Finally, if we don't care about the coordinates on $ZZ^4$, column operations on $A''$ correspond to applying automorphisms of $ZZ^4$, and hence preserve the isomorphism class of $ZZ^4/A''ZZ^3$. However, if we subtract the first column from the second and third, we get
$$A''' = bmat 1 & 0 & 0 \ 0 & 2 & 0 \ 0 & 0 & 4 \ 0 & 0 & 0 emat.$$
Then since $A'''$ is diagonal, $ZZ^4/A'''ZZ^3$ splits as a direct sum, $$ZZ/ZZ oplus ZZ/2ZZ oplus ZZ/4ZZoplus ZZ/(0),$$
so the torsion subgroup of $ZZ^4/AZZ^3$ has size 8.
answered Aug 3 at 21:49
jgon
8,53611335
As I mentioned in the comments, conventionally, the shape of the matrix ought to be $4times 3$ as you've worked out. So we have
$$A=newcommandbmatbeginpmatrixnewcommandematendpmatrixbmat 1 & 1 & 1 \ 1 & 3 & 1 \ 1 & 1 & 5 \ 4 & 8 & 0 emat.$$
Now row operations are isomorphisms of $BbbZ^3$, so we can freely row reduce this matrix and preserve the image of $ABbbZ^3$ as a subset of $newcommandZZBbbZZZ^4$. I.e. applying row operations to a matrix $A$ that presents $ZZ^4/AZZ^3$ to get a matrix $A'$ means that $A'$ also presents $ZZ^4/AZZ^3$.
Subtracting the first row from the others to make the other entries of the first column zero, we have
$$A'=bmat 1 & 1 & 1 \ 0 & 2 & 0 \ 0 & 0 & 4 \ 0 & 4 & -4 emat.$$
Then subtracting twice the second row and adding the third row to the fourth row, we get
$$A''=bmat 1 & 1 & 1 \ 0 & 2 & 0 \ 0 & 0 & 4 \ 0 & 0 & 0 emat.$$
Thus $$v_4=bmat 0 \ 0 \ 0 \ 1 emat$$ is never in the image of $A$, even over $BbbQ$, hence $v_4$ has infinite order in the cokernel.
Finally, if we don't care about the coordinates on $ZZ^4$, column operations on $A''$ correspond to applying automorphisms of $ZZ^4$, and hence preserve the isomorphism class of $ZZ^4/A''ZZ^3$. However, if we subtract the first column from the second and third, we get
$$A''' = bmat 1 & 0 & 0 \ 0 & 2 & 0 \ 0 & 0 & 4 \ 0 & 0 & 0 emat.$$
Then since $A'''$ is diagonal, $ZZ^4/A'''ZZ^3$ splits as a direct sum, $$ZZ/ZZ oplus ZZ/2ZZ oplus ZZ/4ZZoplus ZZ/(0),$$
so the torsion subgroup of $ZZ^4/AZZ^3$ has size 8.
answered Aug 3 at 21:49
jgon
8,53611335
answered Aug 3 at 21:49
jgon
8,53611335
answered Aug 3 at 21:49
jgon
8,53611335
answered Aug 3 at 21:49
jgon
8,53611335
8,53611335
How does the fact that $v_4$ is not in the image of $A$ imply that it has infinite order in the cokernel?
– user437309
yesterday
@user437309 the important thing is that it isn't in the $BbbQ$ image of $A$, since if it had finite order, then for some $n$, $nv_4$ would be in the $ZZ$ image of A, but then $v_4$ would be in the $BbbQ$ image.
– jgon
yesterday
What is the $mathbb Q$-image of $A$? Also I don't understand why if it had finite order, then $nv_4$ would be in the $mathbb Z$-image. (By $mathbb Z$ image I just understand the set of all vectors gotten by multiplication of $A$ by all vectors of an appropriate size.)
– user437309
yesterday
add a comment |Â
How does the fact that $v_4$ is not in the image of $A$ imply that it has infinite order in the cokernel?
– user437309
yesterday
@user437309 the important thing is that it isn't in the $BbbQ$ image of $A$, since if it had finite order, then for some $n$, $nv_4$ would be in the $ZZ$ image of A, but then $v_4$ would be in the $BbbQ$ image.
– jgon
yesterday
What is the $mathbb Q$-image of $A$? Also I don't understand why if it had finite order, then $nv_4$ would be in the $mathbb Z$-image. (By $mathbb Z$ image I just understand the set of all vectors gotten by multiplication of $A$ by all vectors of an appropriate size.)
– user437309
yesterday
How does the fact that $v_4$ is not in the image of $A$ imply that it has infinite order in the cokernel?
– user437309
yesterday
How does the fact that $v_4$ is not in the image of $A$ imply that it has infinite order in the cokernel?
– user437309
yesterday
@user437309 the important thing is that it isn't in the $BbbQ$ image of $A$, since if it had finite order, then for some $n$, $nv_4$ would be in the $ZZ$ image of A, but then $v_4$ would be in the $BbbQ$ image.
– jgon
yesterday
@user437309 the important thing is that it isn't in the $BbbQ$ image of $A$, since if it had finite order, then for some $n$, $nv_4$ would be in the $ZZ$ image of A, but then $v_4$ would be in the $BbbQ$ image.
– jgon
yesterday
What is the $mathbb Q$-image of $A$? Also I don't understand why if it had finite order, then $nv_4$ would be in the $mathbb Z$-image. (By $mathbb Z$ image I just understand the set of all vectors gotten by multiplication of $A$ by all vectors of an appropriate size.)
– user437309
yesterday
What is the $mathbb Q$-image of $A$? Also I don't understand why if it had finite order, then $nv_4$ would be in the $mathbb Z$-image. (By $mathbb Z$ image I just understand the set of all vectors gotten by multiplication of $A$ by all vectors of an appropriate size.)
– user437309
yesterday
add a comment |Â
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What are $v_1, v_2, v_3$?
– k.stm
Aug 3 at 21:22
They are generators of $mathbb Z^4/H$.
– user437309
Aug 3 at 21:23
Yeah, I'm not sure why Artin is using the transpose there (assuming everything has been transcribed correctly). A matrix describing a linear map $R^3to R^4$ for a commutative ring $R$ is a $4times 3$ matrix.
– jgon
Aug 3 at 21:25
That being said, the shape of the matrix is convention, and doesn't really affect the underlying mathematics.
– jgon
Aug 3 at 21:26
It doesn't affect the underlying mathematics, but the fact that following Artin's conventions doesn't give me an answer in his convention may mean that I misunderstand the underlying mathematics.
– user437309
Aug 3 at 21:33