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In this answer, Fortuon Paendrag provides an example of a ring without unity such that every element is a product of some two elements. The example has zero divisors. Can a ring without a unity and without non-zero zero divisors satisfy this condition if it's

(a) commutative,

(b) non-commutative?

Added: A related question.

2 Answers 2

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Here is a useful commutative example that one actually meets in the wild. Let $M$ be a non-standard model of analysis, and let our ring $I$ be the collection of infinitesimals in $M$, together with $0$.

One can make the example sound more explicit by constructing $M$ via the ultrapower.

One can also construct many function ring examples. One type of example is the unitless ring of all finite sums $\sum a_i x^{e_i}$, where the $a_i$ range over the reals, and the $e_i$ range over the positive reals. Or else we can restrict the $e_i$ o positive rationals, or positive dyadic rationals.

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    @ymar: You are presumably right about that.$I$have minimal intuition about non-commutative rings.2012-06-30
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For commutative (Abeliano) you can define a domain such as |R+(pairs, *) (positive pairs with multiplication) and it will be a a Domain without unity, is commutative and every element can be achieved as a product of other (except the first... or consider $2*2^0$... this is arguable of course)

For non-commutative the elements would have to gain some positional properties so commutativity do not work... for example : $ ( A \ op \ B ) <> ( B \ op \ A ) $ only if the position alter the meaning of the symbols (either a value or any other piece of information the simbols A and B represent)

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    the first element of ($R+$ pairs eg: $2, 4, 6, 8, ...$) is the element $2$... you can only consider it as a product of $2*2^0$, but 0 do not pertain to the domain... so we must exclude the first element; like the number 1 exclusion of the Number Theory for primes as numbers building blocks. I guess $R+ pairs$ can cause some confusion... consider instead $N+ pairs$ (eg: 2, 4, 6, 8, ...)2012-07-01