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Let $M(R)$ be the set of 2x2 matrices with elements of ring $R$. $M(R)$ is obviously a ring also. But if $R$ is a Euclidean domain, is $M(R)$ also a eucliean domain? If it is, how do you perform the division algorithm, and what is the notion of "degree"?

My understanding of a Euclidean domain:

  • a ring $R$, with $+$ and $\cdot$ operations, $\mathbb{0}$ and $\mathbb{1}$ values, an additive inverse operation, and distributive properties
  • a division operator suitable for use with an extended $\gcd$ algorithm. (for any $a$ and nonzero $b$ in $R$, it produces $q$ and $r$ such that $a = qb + r$ and the "degree" of r is smaller than the "degree" of b so that the extended gcd algorithm is guaranteed to terminate).

The extended gcd algorithm, given any $a$ and $b$ in $R$, uses the division operator of a Euclidean domain to produce $(g, x, y, a', b')$ such that:

  • $g = \gcd(a,b)$
  • $a = ga'$
  • $b = gb'$
  • $a'x + b'y = \mathbb{1}$ and $ax + by = g$.

I'm trying to create a division algorithm for $M(R)$.

Any matrix $\mathbf{B} \in M(R)$ can be factored into Smith Normal Form where $\mathbf{B} = \mathbf{X}\mathbf{B'}\mathbf{Y}$, $\mathbf{X}$ and $\mathbf{Y}$ are easily invertible and have determinant of $\mathbb{1}$, and $\mathbf{B'}$ is diagonal. This can be used to reduce the complexity of the division problem to only the case where the denominator is a diagonal matrix.

I'm having trouble getting past this point. Given arbitrary $\mathbf{A}$ and nonzero diagonal matrix $\mathbf{B}$, how do you find $\mathbf{Q}$ and $\mathbf{R}$ such that $\mathbf{A} = \mathbf{Q}\mathbf{B} + \mathbf{R}$ and what is the notion of degree such that $\mathrm{degree}(\mathbf{R}) < \mathrm{degree}(\mathbf{B})$?

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    $\begingroup$ $M(R)$ is never a domain for $n\ge2$. $\endgroup$ – user228113 Mar 7 '18 at 21:21
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    $\begingroup$ A Euclidean domain is an integral domain, which is commutative. en.wikipedia.org/wiki/Euclidean_domain $\endgroup$ – Thomas Andrews Mar 7 '18 at 21:21
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$M_2(\mathbb Z)$ is not commutative and has zero divisors. Very far from a domain.

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  • $\begingroup$ Thanks. I spent a lot of time on this, thus demonstrating the hazard of learning mathematical topics from Wikipedia. $\endgroup$ – NovaDenizen Mar 7 '18 at 21:39

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