Mixing
Is the general linear group generated by elementary matrices?
$begingroup$
(Cfr. Wikipedia for the definition of Elementary matrix).
Have a look at the following excerpt of Jacobson's Basic algebra vol.I, 2nd edition, pag.186.
There exist PID in which not every invertible matrix is a product of elementary ones. An example of this type is given in a paper by P.M.Cohn, On the structure of the $text{GL}_2$ of a ring, Institut des Hautes Etudes Scientifiques, #30 (1966), pp 5 - 54.
This leaves me puzzled. Take an invertible matrix $A$ over a PID. Then $A$ has a Smith normal form, that is, up to elementary row and columns operations it is equivalent to something like this
$$begin{bmatrix} d_1 & && \ & d_2 &&\ &&ddots&\ &&&d_nend{bmatrix}.$$
In particular $det A= d_1ldots d_n u$ for some unit element $u$. But $det A$ needs be unit, so all of $d_i$'s are units, which means that up to some other elementary row operation $A$ is equivalent to the identity matrix. It seems to me that we have just proven that $A$ is the product of elementary matrices, which is false as of Jacobson's claim.
There must be an error somewhere, but where?
Thank you.
abstract-algebra group-theory matrices
edited Jan 15 '12 at 11:24
joriki
171k10188349
asked Jan 15 '12 at 11:05
Giuseppe NegroGiuseppe Negro
17.4k332126
$endgroup$
|
show 2 more comments
$begingroup$
(Cfr. Wikipedia for the definition of Elementary matrix).
Have a look at the following excerpt of Jacobson's Basic algebra vol.I, 2nd edition, pag.186.
There exist PID in which not every invertible matrix is a product of elementary ones. An example of this type is given in a paper by P.M.Cohn, On the structure of the $text{GL}_2$ of a ring, Institut des Hautes Etudes Scientifiques, #30 (1966), pp 5 - 54.
This leaves me puzzled. Take an invertible matrix $A$ over a PID. Then $A$ has a Smith normal form, that is, up to elementary row and columns operations it is equivalent to something like this
$$begin{bmatrix} d_1 & && \ & d_2 &&\ &&ddots&\ &&&d_nend{bmatrix}.$$
In particular $det A= d_1ldots d_n u$ for some unit element $u$. But $det A$ needs be unit, so all of $d_i$'s are units, which means that up to some other elementary row operation $A$ is equivalent to the identity matrix. It seems to me that we have just proven that $A$ is the product of elementary matrices, which is false as of Jacobson's claim.
There must be an error somewhere, but where?
Thank you.
abstract-algebra group-theory matrices
edited Jan 15 '12 at 11:24
joriki
171k10188349
asked Jan 15 '12 at 11:05
Giuseppe NegroGiuseppe Negro
17.4k332126
$endgroup$
2
$begingroup$
I am not sure about this, but is it possible that the usual algorithm for reducing a matrix to Smith Normal Form requires a Euclidean domain? I know that the theory of finitely generated modules over a PID guarantees the existence of Smith Normal form, but that is an existence result, rather than a constructive one.
$endgroup$
– Geoff Robinson
Jan 15 '12 at 11:12
3
$begingroup$
It seems you mean this article, whose title is "On the structure of the $GL_2$ of a ring"?
$endgroup$
– joriki
Jan 15 '12 at 11:14
1
$begingroup$
The Wikipedia article you cite uses multiplication with a full $2times2$ matrix, so one explanation could be that this sometimes can't be written as the product of two elementary matrices.
$endgroup$
– joriki
Jan 15 '12 at 11:19
2
$begingroup$
Indeed Section 2 of the article makes just that distinction, between the set of invertible $2times2$ matrices and the set of $2times2$ matrices generated by elementary matrices, and states that the two coincide for Euclidean rings.
$endgroup$
– joriki
Jan 15 '12 at 11:23
1
$begingroup$
@GeoffRobinson, joriki: I think you're right, I had missed this point. In the non-Euclidean case reduction to SNF requires some other matrix besides elementary ones. This leads to failure of my argument above. Thank you both! If you want to publish your comment as an answer I will accept it.
$endgroup$
– Giuseppe Negro
Jan 15 '12 at 11:25
|
show 2 more comments
$begingroup$
(Cfr. Wikipedia for the definition of Elementary matrix).
Have a look at the following excerpt of Jacobson's Basic algebra vol.I, 2nd edition, pag.186.
There exist PID in which not every invertible matrix is a product of elementary ones. An example of this type is given in a paper by P.M.Cohn, On the structure of the $text{GL}_2$ of a ring, Institut des Hautes Etudes Scientifiques, #30 (1966), pp 5 - 54.
This leaves me puzzled. Take an invertible matrix $A$ over a PID. Then $A$ has a Smith normal form, that is, up to elementary row and columns operations it is equivalent to something like this
$$begin{bmatrix} d_1 & && \ & d_2 &&\ &&ddots&\ &&&d_nend{bmatrix}.$$
In particular $det A= d_1ldots d_n u$ for some unit element $u$. But $det A$ needs be unit, so all of $d_i$'s are units, which means that up to some other elementary row operation $A$ is equivalent to the identity matrix. It seems to me that we have just proven that $A$ is the product of elementary matrices, which is false as of Jacobson's claim.
There must be an error somewhere, but where?
Thank you.
abstract-algebra group-theory matrices
edited Jan 15 '12 at 11:24
joriki
171k10188349
asked Jan 15 '12 at 11:05
Giuseppe NegroGiuseppe Negro
17.4k332126
$endgroup$
(Cfr. Wikipedia for the definition of Elementary matrix).
Have a look at the following excerpt of Jacobson's Basic algebra vol.I, 2nd edition, pag.186.
There exist PID in which not every invertible matrix is a product of elementary ones. An example of this type is given in a paper by P.M.Cohn, On the structure of the $text{GL}_2$ of a ring, Institut des Hautes Etudes Scientifiques, #30 (1966), pp 5 - 54.
This leaves me puzzled. Take an invertible matrix $A$ over a PID. Then $A$ has a Smith normal form, that is, up to elementary row and columns operations it is equivalent to something like this
$$begin{bmatrix} d_1 & && \ & d_2 &&\ &&ddots&\ &&&d_nend{bmatrix}.$$
In particular $det A= d_1ldots d_n u$ for some unit element $u$. But $det A$ needs be unit, so all of $d_i$'s are units, which means that up to some other elementary row operation $A$ is equivalent to the identity matrix. It seems to me that we have just proven that $A$ is the product of elementary matrices, which is false as of Jacobson's claim.
There must be an error somewhere, but where?
Thank you.
abstract-algebra group-theory matrices
abstract-algebra group-theory matrices
edited Jan 15 '12 at 11:24
joriki
171k10188349
asked Jan 15 '12 at 11:05
Giuseppe NegroGiuseppe Negro
17.4k332126
edited Jan 15 '12 at 11:24
joriki
171k10188349
asked Jan 15 '12 at 11:05
Giuseppe NegroGiuseppe Negro
17.4k332126
edited Jan 15 '12 at 11:24
joriki
171k10188349
edited Jan 15 '12 at 11:24
joriki
171k10188349
edited Jan 15 '12 at 11:24
joriki
171k10188349
171k10188349
asked Jan 15 '12 at 11:05
Giuseppe NegroGiuseppe Negro
17.4k332126
asked Jan 15 '12 at 11:05
Giuseppe NegroGiuseppe Negro
17.4k332126
asked Jan 15 '12 at 11:05
Giuseppe NegroGiuseppe Negro
17.4k332126
17.4k332126
2
$begingroup$
I am not sure about this, but is it possible that the usual algorithm for reducing a matrix to Smith Normal Form requires a Euclidean domain? I know that the theory of finitely generated modules over a PID guarantees the existence of Smith Normal form, but that is an existence result, rather than a constructive one.
$endgroup$
– Geoff Robinson
Jan 15 '12 at 11:12
3
$begingroup$
It seems you mean this article, whose title is "On the structure of the $GL_2$ of a ring"?
$endgroup$
– joriki
Jan 15 '12 at 11:14
1
$begingroup$
The Wikipedia article you cite uses multiplication with a full $2times2$ matrix, so one explanation could be that this sometimes can't be written as the product of two elementary matrices.
$endgroup$
– joriki
Jan 15 '12 at 11:19
2
$begingroup$
Indeed Section 2 of the article makes just that distinction, between the set of invertible $2times2$ matrices and the set of $2times2$ matrices generated by elementary matrices, and states that the two coincide for Euclidean rings.
$endgroup$
– joriki
Jan 15 '12 at 11:23
1
$begingroup$
@GeoffRobinson, joriki: I think you're right, I had missed this point. In the non-Euclidean case reduction to SNF requires some other matrix besides elementary ones. This leads to failure of my argument above. Thank you both! If you want to publish your comment as an answer I will accept it.
$endgroup$
– Giuseppe Negro
Jan 15 '12 at 11:25
|
show 2 more comments
2
$begingroup$
I am not sure about this, but is it possible that the usual algorithm for reducing a matrix to Smith Normal Form requires a Euclidean domain? I know that the theory of finitely generated modules over a PID guarantees the existence of Smith Normal form, but that is an existence result, rather than a constructive one.
$endgroup$
– Geoff Robinson
Jan 15 '12 at 11:12
3
$begingroup$
It seems you mean this article, whose title is "On the structure of the $GL_2$ of a ring"?
$endgroup$
– joriki
Jan 15 '12 at 11:14
1
$begingroup$
The Wikipedia article you cite uses multiplication with a full $2times2$ matrix, so one explanation could be that this sometimes can't be written as the product of two elementary matrices.
$endgroup$
– joriki
Jan 15 '12 at 11:19
2
$begingroup$
Indeed Section 2 of the article makes just that distinction, between the set of invertible $2times2$ matrices and the set of $2times2$ matrices generated by elementary matrices, and states that the two coincide for Euclidean rings.
$endgroup$
– joriki
Jan 15 '12 at 11:23
1
$begingroup$
@GeoffRobinson, joriki: I think you're right, I had missed this point. In the non-Euclidean case reduction to SNF requires some other matrix besides elementary ones. This leads to failure of my argument above. Thank you both! If you want to publish your comment as an answer I will accept it.
$endgroup$
– Giuseppe Negro
Jan 15 '12 at 11:25
2
2
$begingroup$
I am not sure about this, but is it possible that the usual algorithm for reducing a matrix to Smith Normal Form requires a Euclidean domain? I know that the theory of finitely generated modules over a PID guarantees the existence of Smith Normal form, but that is an existence result, rather than a constructive one.
$endgroup$
– Geoff Robinson
Jan 15 '12 at 11:12
$begingroup$
I am not sure about this, but is it possible that the usual algorithm for reducing a matrix to Smith Normal Form requires a Euclidean domain? I know that the theory of finitely generated modules over a PID guarantees the existence of Smith Normal form, but that is an existence result, rather than a constructive one.
$endgroup$
– Geoff Robinson
Jan 15 '12 at 11:12
3
3
$begingroup$
It seems you mean this article, whose title is "On the structure of the $GL_2$ of a ring"?
$endgroup$
– joriki
Jan 15 '12 at 11:14
$begingroup$
It seems you mean this article, whose title is "On the structure of the $GL_2$ of a ring"?
$endgroup$
– joriki
Jan 15 '12 at 11:14
1
1
$begingroup$
The Wikipedia article you cite uses multiplication with a full $2times2$ matrix, so one explanation could be that this sometimes can't be written as the product of two elementary matrices.
$endgroup$
– joriki
Jan 15 '12 at 11:19
$begingroup$
The Wikipedia article you cite uses multiplication with a full $2times2$ matrix, so one explanation could be that this sometimes can't be written as the product of two elementary matrices.
$endgroup$
– joriki
Jan 15 '12 at 11:19
2
2
$begingroup$
Indeed Section 2 of the article makes just that distinction, between the set of invertible $2times2$ matrices and the set of $2times2$ matrices generated by elementary matrices, and states that the two coincide for Euclidean rings.
$endgroup$
– joriki
Jan 15 '12 at 11:23
$begingroup$
Indeed Section 2 of the article makes just that distinction, between the set of invertible $2times2$ matrices and the set of $2times2$ matrices generated by elementary matrices, and states that the two coincide for Euclidean rings.
$endgroup$
– joriki
Jan 15 '12 at 11:23
1
1
$begingroup$
@GeoffRobinson, joriki: I think you're right, I had missed this point. In the non-Euclidean case reduction to SNF requires some other matrix besides elementary ones. This leads to failure of my argument above. Thank you both! If you want to publish your comment as an answer I will accept it.
$endgroup$
– Giuseppe Negro
Jan 15 '12 at 11:25
$begingroup$
@GeoffRobinson, joriki: I think you're right, I had missed this point. In the non-Euclidean case reduction to SNF requires some other matrix besides elementary ones. This leads to failure of my argument above. Thank you both! If you want to publish your comment as an answer I will accept it.
$endgroup$
– Giuseppe Negro
Jan 15 '12 at 11:25
|
show 2 more comments
1 Answer
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$begingroup$
The argument fails because the reduction to Smith normal form may require a full $2times2$ matrix that can't be written as a product of elementary matrices. The cited paper gives an example of such a $2times2$ matrix over $mathbb Q(sqrt{-19})$ on page 23.
edited Jan 23 at 10:38
user549397
1,5081418
answered Jan 15 '12 at 11:30
jorikijoriki
171k10188349
$endgroup$
add a comment |
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$begingroup$
The argument fails because the reduction to Smith normal form may require a full $2times2$ matrix that can't be written as a product of elementary matrices. The cited paper gives an example of such a $2times2$ matrix over $mathbb Q(sqrt{-19})$ on page 23.
edited Jan 23 at 10:38
user549397
1,5081418
answered Jan 15 '12 at 11:30
jorikijoriki
171k10188349
$endgroup$
add a comment |
$begingroup$
The argument fails because the reduction to Smith normal form may require a full $2times2$ matrix that can't be written as a product of elementary matrices. The cited paper gives an example of such a $2times2$ matrix over $mathbb Q(sqrt{-19})$ on page 23.
edited Jan 23 at 10:38
user549397
1,5081418
answered Jan 15 '12 at 11:30
jorikijoriki
171k10188349
$endgroup$
add a comment |
$begingroup$
The argument fails because the reduction to Smith normal form may require a full $2times2$ matrix that can't be written as a product of elementary matrices. The cited paper gives an example of such a $2times2$ matrix over $mathbb Q(sqrt{-19})$ on page 23.
edited Jan 23 at 10:38
user549397
1,5081418
answered Jan 15 '12 at 11:30
jorikijoriki
171k10188349
$endgroup$
The argument fails because the reduction to Smith normal form may require a full $2times2$ matrix that can't be written as a product of elementary matrices. The cited paper gives an example of such a $2times2$ matrix over $mathbb Q(sqrt{-19})$ on page 23.
edited Jan 23 at 10:38
user549397
1,5081418
answered Jan 15 '12 at 11:30
jorikijoriki
171k10188349
edited Jan 23 at 10:38
user549397
1,5081418
edited Jan 23 at 10:38
user549397
1,5081418
edited Jan 23 at 10:38
user549397
1,5081418
1,5081418
answered Jan 15 '12 at 11:30
jorikijoriki
171k10188349
answered Jan 15 '12 at 11:30
jorikijoriki
171k10188349
answered Jan 15 '12 at 11:30
jorikijoriki
171k10188349
171k10188349
add a comment |
add a comment |
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$begingroup$
I am not sure about this, but is it possible that the usual algorithm for reducing a matrix to Smith Normal Form requires a Euclidean domain? I know that the theory of finitely generated modules over a PID guarantees the existence of Smith Normal form, but that is an existence result, rather than a constructive one.
$endgroup$
– Geoff Robinson
Jan 15 '12 at 11:12
3
$begingroup$
It seems you mean this article, whose title is "On the structure of the $GL_2$ of a ring"?
$endgroup$
– joriki
Jan 15 '12 at 11:14
1
$begingroup$
The Wikipedia article you cite uses multiplication with a full $2times2$ matrix, so one explanation could be that this sometimes can't be written as the product of two elementary matrices.
$endgroup$
– joriki
Jan 15 '12 at 11:19
2
$begingroup$
Indeed Section 2 of the article makes just that distinction, between the set of invertible $2times2$ matrices and the set of $2times2$ matrices generated by elementary matrices, and states that the two coincide for Euclidean rings.
$endgroup$
– joriki
Jan 15 '12 at 11:23
1
$begingroup$
@GeoffRobinson, joriki: I think you're right, I had missed this point. In the non-Euclidean case reduction to SNF requires some other matrix besides elementary ones. This leads to failure of my argument above. Thank you both! If you want to publish your comment as an answer I will accept it.
$endgroup$
– Giuseppe Negro
Jan 15 '12 at 11:25