T0Keh

There is multiple things wrong here.

1. 1 is not a prime number because it is a unit and hence by definition excluded from being a prime.

2. You probably don’t mean units but identity elements:

• A unit is an element that has a multiplicative inverse
• An identity element is an element 1 such that 1x =x1 = x for all x in your ring

There are more units in R than just 1, take for example -1(unless your ring has characteristic 2 in which case thi argument not always works; however for the case of real numbers this is not relevant). But there is always just one identity element, so there is at most one “1” in any ring. Indeed suppose you have two identities e,f. Then e = ef = f because e,f both are identities.

3. The property “their prime factorisaton only contains trivial prime factors” is a circular definition as this requires knowledge about “being prime”. A prime (in Z) is normally defined as an irreducible element, i.e. p is a prime number if p=ab implies that either a or b is a unit (which is exactly the property of only having the factors 1 and p itself (up to a unit)).

4. (R,×) is not a ring (at least not in a way I am aware of) and not even a group (unless you exclude 0).

5. What are those “general prime elements”? Do you mean prime elements in a ring (or irreducible elements?)? Or something completely different?

That’s why I wrote prime number instead of prime element to not add more confusion. I know that in general prime and irreducible are not equivalent.