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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.

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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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  • $\begingroup$ Thanks! I don't have much of an idea about non-standard analysis, but I can easily believe that it's an example. :) $\endgroup$
    – user23211
    Jun 30, 2012 at 2:40
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    $\begingroup$ @ymar: I mentioned the example because it is used (though not for its ring-theoretic properties!) in hundreds of papers. So it is not an example just constructed to be an example. There are many such examples in function rings, the "divisibility" restriction is not a strong one. $\endgroup$
    – André Nicolas
    Jun 30, 2012 at 2:47
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    $\begingroup$ @ymar: Yes, it is an ideal, and I smiled while I was choosing the name (ideal, infinitesimal). $\endgroup$
    – André Nicolas
    Jun 30, 2012 at 2:54
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    $\begingroup$ @ymar: Well, the usual way is to start by constructing the ordered field $M$. In the usual construction, $M$ contains an isomorphic copy of the reals, plus, naturally, infinite elements. $\endgroup$
    – André Nicolas
    Jun 30, 2012 at 3:05
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    $\begingroup$ @ymar: Messed up, deleted that part. Was trying to use an example from an old paper of mine on ideals in rings of continuous functions, but it was kind of messy, and I falsely thought it simplified. What I think works is finite sums of the form $\sum a_i x^{e_i}$ where the $a_i$ are reals and the $e_i$ are positive reals. $\endgroup$
    – André Nicolas
    Jun 30, 2012 at 5:05
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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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    $\begingroup$ Do you mean $\{(x,y):x,y\in\Bbb R^+\}$ under $(\times,\times)$ as multiplication? If so, what is the addition - just componentwise? If so, wouldn't it need additive inverses...? $\endgroup$
    – anon
    Jun 30, 2012 at 2:27
  • $\begingroup$ Also, what does "the first" mean? $\endgroup$
    – user23211
    Jun 30, 2012 at 2:44
  • $\begingroup$ 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, ...) $\endgroup$
    – ZEE
    Jul 1, 2012 at 22:03

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