Abstract

Let R be a commutative ring with identity. Then we prove <TEX>$M_n(R)=GL_n(R)$</TEX> <TEX>${\cup}$</TEX>{<TEX>$A{\in}M_n(R)\;{\mid}\;detA{\neq}0$</TEX> and det <TEX>$A{\neq}U(R)$</TEX>}<TEX>${\cup}Z(M-n(R))$</TEX> where U(R) denotes the set of all units of R. In particular, it will be proved that the full matrix ring <TEX>$M_n(F)$</TEX> over a field F is the disjoint union of the general linear group <TEX>$GL_n(F)$</TEX> of degree n over the field F and the set <TEX>$Z(M_n(F))$</TEX> of all zero-divisors of <TEX>$M_n(F)$</TEX>. Using the result and universal mapping property we prove that <TEX>$M_n(F)$</TEX> is its total ring of fractions.

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