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Let $ R $ be a p.i.d. and $ A\in 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 $$ \begin{pmatrix}x & s\\ y & t\end{pmatrix} .$$

(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: $$ \begin{pmatrix}x & s\\ y & t\end{pmatrix} $$ 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.

Bach
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  • You're looking for Smith Normal Form – jgon Jan 23 '19 at 15:35
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    @jgon Nope. We can't express a matrix over p.i.d. into products of elementary matrices in general. This might be helpful: https://math.stackexchange.com/q/99236/549397 – Bach Jan 23 '19 at 15:53
  • 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? – jgon Jan 23 '19 at 16:22
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    The counterexample in the P. M. Cohn article cited by @joriki in the linked thread is a $2\times 2$-matrix of determinant $1$. So I don't think that condition will save you. – darij grinberg Jan 24 '19 at 18:26
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    I'm getting $(3-θ)(5-2θ)-(2+θ)(-3-2θ)=-4θ+4θ^2+21=4\underbrace{(θ^2-θ+5)}_{=0}+1=1$ instead. – darij grinberg Jan 25 '19 at 02:28

1 Answers1

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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 Z\left[\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 $ n\times n $ matrix is a product of elementary $ n\times n $ matrices.

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

Bach
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  • 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). – Bach Jan 25 '19 at 05:24