Let $ R $ be a p.i.d. and $ Ain M_n(R) $. If $ det(A)=1 $, prove or disprove that $ A $ can be expressed as...












3












$begingroup$



Let $ R $ be a p.i.d. and $ Ain M_n(R) $. If $ det(A)=1 $, prove or disprove that $ A $ can be expressed as products of elementary matrices.




I know that we can express $ A $ as products of elementary matrices when we require $ R $ to be Euclidean. Since we can get rid of matrices whose left uppermost elements are like $$ begin{pmatrix}x & s\ y & tend{pmatrix} .$$



(Note: Furthermore, I know that if we require $ R $ to be a field, and $ det(A)=1 $, we can even express $ A $ as products of transvection matrices. Transvection matrices generate SLn(R))




Now back to this question, it seems to me that it suffices to prove that invertible matrices of type: $$ begin{pmatrix}x & s\ y & tend{pmatrix} $$ where $ ax+by=d $, $ gcd(a, b)=d $ and $ s=bd^{-1}, t=-ad^{-1} $(Here the inverse is obtained by cancellation in $ R $) can be expressed into products of elementary matrices. Since $ det(A)=1 $, then $ d=1 $ and we have: $ gcd(a, b)=1, s=b, t=-a $. Well how to move on?




EDIT: I have changed my title and statement in a less misleading way since I find out that we can't express $ A $ in such a way under the given assumption generally.






abstract-algebra matrices principal-ideal-domains







share|cite|improve this question











$endgroup$












  • $begingroup$
    You're looking for Smith Normal Form
    $endgroup$
    – jgon
    Jan 23 at 15:35






  • 2




    $begingroup$
    @jgon Nope. We can't express a matrix over p.i.d. into products of elementary matrices in general. This might be helpful: math.stackexchange.com/q/99236/549397
    $endgroup$
    – user549397
    Jan 23 at 15:53










  • $begingroup$
    Oh I see, interesting, I'd forgotten about that. So you're trying to prove that the 2x2 matrices that arise are products of elementary matrices in this case?
    $endgroup$
    – jgon
    Jan 23 at 16:22






  • 1




    $begingroup$
    The counterexample in the P. M. Cohn article cited by @joriki in the linked thread is a $2times 2$-matrix of determinant $1$. So I don't think that condition will save you.
    $endgroup$
    – darij grinberg
    Jan 24 at 18:26






  • 1




    $begingroup$
    I'm getting $(3-θ)(5-2θ)-(2+θ)(-3-2θ)=-4θ+4θ^2+21=4underbrace{(θ^2-θ+5)}_{=0}+1=1$ instead.
    $endgroup$
    – darij grinberg
    Jan 25 at 2:28


















3












$begingroup$



Let $ R $ be a p.i.d. and $ Ain M_n(R) $. If $ det(A)=1 $, prove or disprove that $ A $ can be expressed as products of elementary matrices.




I know that we can express $ A $ as products of elementary matrices when we require $ R $ to be Euclidean. Since we can get rid of matrices whose left uppermost elements are like $$ begin{pmatrix}x & s\ y & tend{pmatrix} .$$



(Note: Furthermore, I know that if we require $ R $ to be a field, and $ det(A)=1 $, we can even express $ A $ as products of transvection matrices. Transvection matrices generate SLn(R))




Now back to this question, it seems to me that it suffices to prove that invertible matrices of type: $$ begin{pmatrix}x & s\ y & tend{pmatrix} $$ where $ ax+by=d $, $ gcd(a, b)=d $ and $ s=bd^{-1}, t=-ad^{-1} $(Here the inverse is obtained by cancellation in $ R $) can be expressed into products of elementary matrices. Since $ det(A)=1 $, then $ d=1 $ and we have: $ gcd(a, b)=1, s=b, t=-a $. Well how to move on?




EDIT: I have changed my title and statement in a less misleading way since I find out that we can't express $ A $ in such a way under the given assumption generally.






abstract-algebra matrices principal-ideal-domains







share|cite|improve this question











$endgroup$












  • $begingroup$
    You're looking for Smith Normal Form
    $endgroup$
    – jgon
    Jan 23 at 15:35






  • 2




    $begingroup$
    @jgon Nope. We can't express a matrix over p.i.d. into products of elementary matrices in general. This might be helpful: math.stackexchange.com/q/99236/549397
    $endgroup$
    – user549397
    Jan 23 at 15:53










  • $begingroup$
    Oh I see, interesting, I'd forgotten about that. So you're trying to prove that the 2x2 matrices that arise are products of elementary matrices in this case?
    $endgroup$
    – jgon
    Jan 23 at 16:22






  • 1




    $begingroup$
    The counterexample in the P. M. Cohn article cited by @joriki in the linked thread is a $2times 2$-matrix of determinant $1$. So I don't think that condition will save you.
    $endgroup$
    – darij grinberg
    Jan 24 at 18:26






  • 1




    $begingroup$
    I'm getting $(3-θ)(5-2θ)-(2+θ)(-3-2θ)=-4θ+4θ^2+21=4underbrace{(θ^2-θ+5)}_{=0}+1=1$ instead.
    $endgroup$
    – darij grinberg
    Jan 25 at 2:28
















3












3








3


1



$begingroup$



Let $ R $ be a p.i.d. and $ Ain M_n(R) $. If $ det(A)=1 $, prove or disprove that $ A $ can be expressed as products of elementary matrices.




I know that we can express $ A $ as products of elementary matrices when we require $ R $ to be Euclidean. Since we can get rid of matrices whose left uppermost elements are like $$ begin{pmatrix}x & s\ y & tend{pmatrix} .$$



(Note: Furthermore, I know that if we require $ R $ to be a field, and $ det(A)=1 $, we can even express $ A $ as products of transvection matrices. Transvection matrices generate SLn(R))




Now back to this question, it seems to me that it suffices to prove that invertible matrices of type: $$ begin{pmatrix}x & s\ y & tend{pmatrix} $$ where $ ax+by=d $, $ gcd(a, b)=d $ and $ s=bd^{-1}, t=-ad^{-1} $(Here the inverse is obtained by cancellation in $ R $) can be expressed into products of elementary matrices. Since $ det(A)=1 $, then $ d=1 $ and we have: $ gcd(a, b)=1, s=b, t=-a $. Well how to move on?




EDIT: I have changed my title and statement in a less misleading way since I find out that we can't express $ A $ in such a way under the given assumption generally.






abstract-algebra matrices principal-ideal-domains







share|cite|improve this question











$endgroup$





Let $ R $ be a p.i.d. and $ Ain M_n(R) $. If $ det(A)=1 $, prove or disprove that $ A $ can be expressed as products of elementary matrices.




I know that we can express $ A $ as products of elementary matrices when we require $ R $ to be Euclidean. Since we can get rid of matrices whose left uppermost elements are like $$ begin{pmatrix}x & s\ y & tend{pmatrix} .$$



(Note: Furthermore, I know that if we require $ R $ to be a field, and $ det(A)=1 $, we can even express $ A $ as products of transvection matrices. Transvection matrices generate SLn(R))




Now back to this question, it seems to me that it suffices to prove that invertible matrices of type: $$ begin{pmatrix}x & s\ y & tend{pmatrix} $$ where $ ax+by=d $, $ gcd(a, b)=d $ and $ s=bd^{-1}, t=-ad^{-1} $(Here the inverse is obtained by cancellation in $ R $) can be expressed into products of elementary matrices. Since $ det(A)=1 $, then $ d=1 $ and we have: $ gcd(a, b)=1, s=b, t=-a $. Well how to move on?




EDIT: I have changed my title and statement in a less misleading way since I find out that we can't express $ A $ in such a way under the given assumption generally.






abstract-algebra matrices principal-ideal-domains




abstract-algebra matrices principal-ideal-domains






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 25 at 5:20







user549397

















asked Jan 23 at 13:12









user549397user549397

1,5061418




1,5061418












  • $begingroup$
    You're looking for Smith Normal Form
    $endgroup$
    – jgon
    Jan 23 at 15:35






  • 2




    $begingroup$
    @jgon Nope. We can't express a matrix over p.i.d. into products of elementary matrices in general. This might be helpful: math.stackexchange.com/q/99236/549397
    $endgroup$
    – user549397
    Jan 23 at 15:53










  • $begingroup$
    Oh I see, interesting, I'd forgotten about that. So you're trying to prove that the 2x2 matrices that arise are products of elementary matrices in this case?
    $endgroup$
    – jgon
    Jan 23 at 16:22






  • 1




    $begingroup$
    The counterexample in the P. M. Cohn article cited by @joriki in the linked thread is a $2times 2$-matrix of determinant $1$. So I don't think that condition will save you.
    $endgroup$
    – darij grinberg
    Jan 24 at 18:26






  • 1




    $begingroup$
    I'm getting $(3-θ)(5-2θ)-(2+θ)(-3-2θ)=-4θ+4θ^2+21=4underbrace{(θ^2-θ+5)}_{=0}+1=1$ instead.
    $endgroup$
    – darij grinberg
    Jan 25 at 2:28




















  • $begingroup$
    You're looking for Smith Normal Form
    $endgroup$
    – jgon
    Jan 23 at 15:35






  • 2




    $begingroup$
    @jgon Nope. We can't express a matrix over p.i.d. into products of elementary matrices in general. This might be helpful: math.stackexchange.com/q/99236/549397
    $endgroup$
    – user549397
    Jan 23 at 15:53










  • $begingroup$
    Oh I see, interesting, I'd forgotten about that. So you're trying to prove that the 2x2 matrices that arise are products of elementary matrices in this case?
    $endgroup$
    – jgon
    Jan 23 at 16:22






  • 1




    $begingroup$
    The counterexample in the P. M. Cohn article cited by @joriki in the linked thread is a $2times 2$-matrix of determinant $1$. So I don't think that condition will save you.
    $endgroup$
    – darij grinberg
    Jan 24 at 18:26






  • 1




    $begingroup$
    I'm getting $(3-θ)(5-2θ)-(2+θ)(-3-2θ)=-4θ+4θ^2+21=4underbrace{(θ^2-θ+5)}_{=0}+1=1$ instead.
    $endgroup$
    – darij grinberg
    Jan 25 at 2:28


















$begingroup$
You're looking for Smith Normal Form
$endgroup$
– jgon
Jan 23 at 15:35




$begingroup$
You're looking for Smith Normal Form
$endgroup$
– jgon
Jan 23 at 15:35




2




2




$begingroup$
@jgon Nope. We can't express a matrix over p.i.d. into products of elementary matrices in general. This might be helpful: math.stackexchange.com/q/99236/549397
$endgroup$
– user549397
Jan 23 at 15:53




$begingroup$
@jgon Nope. We can't express a matrix over p.i.d. into products of elementary matrices in general. This might be helpful: math.stackexchange.com/q/99236/549397
$endgroup$
– user549397
Jan 23 at 15:53












$begingroup$
Oh I see, interesting, I'd forgotten about that. So you're trying to prove that the 2x2 matrices that arise are products of elementary matrices in this case?
$endgroup$
– jgon
Jan 23 at 16:22




$begingroup$
Oh I see, interesting, I'd forgotten about that. So you're trying to prove that the 2x2 matrices that arise are products of elementary matrices in this case?
$endgroup$
– jgon
Jan 23 at 16:22




1




1




$begingroup$
The counterexample in the P. M. Cohn article cited by @joriki in the linked thread is a $2times 2$-matrix of determinant $1$. So I don't think that condition will save you.
$endgroup$
– darij grinberg
Jan 24 at 18:26




$begingroup$
The counterexample in the P. M. Cohn article cited by @joriki in the linked thread is a $2times 2$-matrix of determinant $1$. So I don't think that condition will save you.
$endgroup$
– darij grinberg
Jan 24 at 18:26




1




1




$begingroup$
I'm getting $(3-θ)(5-2θ)-(2+θ)(-3-2θ)=-4θ+4θ^2+21=4underbrace{(θ^2-θ+5)}_{=0}+1=1$ instead.
$endgroup$
– darij grinberg
Jan 25 at 2:28






$begingroup$
I'm getting $(3-θ)(5-2θ)-(2+θ)(-3-2θ)=-4θ+4θ^2+21=4underbrace{(θ^2-θ+5)}_{=0}+1=1$ instead.
$endgroup$
– darij grinberg
Jan 25 at 2:28












1 Answer
1






active

oldest

votes


















1












$begingroup$

As pointed out by @darijgrinberg in the comment, we have already had a counterexample in the cited article. (p23):



Consider the ring of integers in $ mathbb Q(sqrt{-19}) $: $ I=mathbb Zleft[frac{1+sqrt{-19}}{2}right] $ which is also a p.i.d. (Ref: An example of a PID which is not a Euclidean), then according to Theorem (6.1) in the article, we know that $ I $ is not a $ GE_2 $-ring.




Definition of $ mathbf{GE_n} $: Every invertible $ ntimes n $ matrix is a product of elementary $
ntimes n $ matrices.




Furthermore, on the top of page23, the author gives an explicit matrix in $ GL_2(I) $ but not in $ GE_2(I) $:
$$ begin{pmatrix} 3-theta& 2+theta\ -3-2theta& 5-2theta end{pmatrix} $$ where $ theta^2-theta+5=0 $. By direct computation, we know that $$ detbegin{pmatrix} 3-theta& 2+theta\ -3-2theta& 5-2theta end{pmatrix}=1 .$$ So generally we cannot express a matrix with determinant $ 1 $ in $ M_n(R) $ as products of elementary matrices.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    The reference(cf. [9], p213) given by Cohn on page 23 for the case $ d=19 $ doesn't prove that $ I $ is a p.i.d.. So I have found a new one and attached it to my post(An example of a PID which is not a Euclidean).
    $endgroup$
    – user549397
    Jan 25 at 5:24











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1 Answer
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1 Answer
1






active

oldest

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active

oldest

votes






active

oldest

votes









1












$begingroup$

As pointed out by @darijgrinberg in the comment, we have already had a counterexample in the cited article. (p23):



Consider the ring of integers in $ mathbb Q(sqrt{-19}) $: $ I=mathbb Zleft[frac{1+sqrt{-19}}{2}right] $ which is also a p.i.d. (Ref: An example of a PID which is not a Euclidean), then according to Theorem (6.1) in the article, we know that $ I $ is not a $ GE_2 $-ring.




Definition of $ mathbf{GE_n} $: Every invertible $ ntimes n $ matrix is a product of elementary $
ntimes n $ matrices.




Furthermore, on the top of page23, the author gives an explicit matrix in $ GL_2(I) $ but not in $ GE_2(I) $:
$$ begin{pmatrix} 3-theta& 2+theta\ -3-2theta& 5-2theta end{pmatrix} $$ where $ theta^2-theta+5=0 $. By direct computation, we know that $$ detbegin{pmatrix} 3-theta& 2+theta\ -3-2theta& 5-2theta end{pmatrix}=1 .$$ So generally we cannot express a matrix with determinant $ 1 $ in $ M_n(R) $ as products of elementary matrices.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    The reference(cf. [9], p213) given by Cohn on page 23 for the case $ d=19 $ doesn't prove that $ I $ is a p.i.d.. So I have found a new one and attached it to my post(An example of a PID which is not a Euclidean).
    $endgroup$
    – user549397
    Jan 25 at 5:24
















1












$begingroup$

As pointed out by @darijgrinberg in the comment, we have already had a counterexample in the cited article. (p23):



Consider the ring of integers in $ mathbb Q(sqrt{-19}) $: $ I=mathbb Zleft[frac{1+sqrt{-19}}{2}right] $ which is also a p.i.d. (Ref: An example of a PID which is not a Euclidean), then according to Theorem (6.1) in the article, we know that $ I $ is not a $ GE_2 $-ring.




Definition of $ mathbf{GE_n} $: Every invertible $ ntimes n $ matrix is a product of elementary $
ntimes n $ matrices.




Furthermore, on the top of page23, the author gives an explicit matrix in $ GL_2(I) $ but not in $ GE_2(I) $:
$$ begin{pmatrix} 3-theta& 2+theta\ -3-2theta& 5-2theta end{pmatrix} $$ where $ theta^2-theta+5=0 $. By direct computation, we know that $$ detbegin{pmatrix} 3-theta& 2+theta\ -3-2theta& 5-2theta end{pmatrix}=1 .$$ So generally we cannot express a matrix with determinant $ 1 $ in $ M_n(R) $ as products of elementary matrices.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    The reference(cf. [9], p213) given by Cohn on page 23 for the case $ d=19 $ doesn't prove that $ I $ is a p.i.d.. So I have found a new one and attached it to my post(An example of a PID which is not a Euclidean).
    $endgroup$
    – user549397
    Jan 25 at 5:24














1












1








1





$begingroup$

As pointed out by @darijgrinberg in the comment, we have already had a counterexample in the cited article. (p23):



Consider the ring of integers in $ mathbb Q(sqrt{-19}) $: $ I=mathbb Zleft[frac{1+sqrt{-19}}{2}right] $ which is also a p.i.d. (Ref: An example of a PID which is not a Euclidean), then according to Theorem (6.1) in the article, we know that $ I $ is not a $ GE_2 $-ring.




Definition of $ mathbf{GE_n} $: Every invertible $ ntimes n $ matrix is a product of elementary $
ntimes n $ matrices.




Furthermore, on the top of page23, the author gives an explicit matrix in $ GL_2(I) $ but not in $ GE_2(I) $:
$$ begin{pmatrix} 3-theta& 2+theta\ -3-2theta& 5-2theta end{pmatrix} $$ where $ theta^2-theta+5=0 $. By direct computation, we know that $$ detbegin{pmatrix} 3-theta& 2+theta\ -3-2theta& 5-2theta end{pmatrix}=1 .$$ So generally we cannot express a matrix with determinant $ 1 $ in $ M_n(R) $ as products of elementary matrices.






share|cite|improve this answer









$endgroup$



As pointed out by @darijgrinberg in the comment, we have already had a counterexample in the cited article. (p23):



Consider the ring of integers in $ mathbb Q(sqrt{-19}) $: $ I=mathbb Zleft[frac{1+sqrt{-19}}{2}right] $ which is also a p.i.d. (Ref: An example of a PID which is not a Euclidean), then according to Theorem (6.1) in the article, we know that $ I $ is not a $ GE_2 $-ring.




Definition of $ mathbf{GE_n} $: Every invertible $ ntimes n $ matrix is a product of elementary $
ntimes n $ matrices.




Furthermore, on the top of page23, the author gives an explicit matrix in $ GL_2(I) $ but not in $ GE_2(I) $:
$$ begin{pmatrix} 3-theta& 2+theta\ -3-2theta& 5-2theta end{pmatrix} $$ where $ theta^2-theta+5=0 $. By direct computation, we know that $$ detbegin{pmatrix} 3-theta& 2+theta\ -3-2theta& 5-2theta end{pmatrix}=1 .$$ So generally we cannot express a matrix with determinant $ 1 $ in $ M_n(R) $ as products of elementary matrices.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 25 at 5:16









user549397user549397

1,5061418




1,5061418












  • $begingroup$
    The reference(cf. [9], p213) given by Cohn on page 23 for the case $ d=19 $ doesn't prove that $ I $ is a p.i.d.. So I have found a new one and attached it to my post(An example of a PID which is not a Euclidean).
    $endgroup$
    – user549397
    Jan 25 at 5:24


















  • $begingroup$
    The reference(cf. [9], p213) given by Cohn on page 23 for the case $ d=19 $ doesn't prove that $ I $ is a p.i.d.. So I have found a new one and attached it to my post(An example of a PID which is not a Euclidean).
    $endgroup$
    – user549397
    Jan 25 at 5:24
















$begingroup$
The reference(cf. [9], p213) given by Cohn on page 23 for the case $ d=19 $ doesn't prove that $ I $ is a p.i.d.. So I have found a new one and attached it to my post(An example of a PID which is not a Euclidean).
$endgroup$
– user549397
Jan 25 at 5:24




$begingroup$
The reference(cf. [9], p213) given by Cohn on page 23 for the case $ d=19 $ doesn't prove that $ I $ is a p.i.d.. So I have found a new one and attached it to my post(An example of a PID which is not a Euclidean).
$endgroup$
– user549397
Jan 25 at 5:24


















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