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I have recently come across two examples of an algebraic structure similar to a ring, but without commutative addition. It consists of a set of values and two operations $(\sigma, +, \cdot)$, with the following axioms ($a, b, c \in \sigma$):

  1. $(a+b)+c=a+(b+c)$ (associativity of addition)
  2. $a\cdot (b+c) = a\cdot b+a\cdot c$ (distributivity of multiplication)
  3. $\exists 0 \in \sigma : a+0=0+a=a$ (additive identity)
  4. $\exists 1 \in \sigma : a \cdot 1 = 1 \cdot a = a$ (multiplicative identity)
  5. $\exists -a \in \sigma : a+(-a) = 0$ (additive inverse)
  6. $\exists a^{-1} \in \sigma : a\cdot a^{-1} = 1, a \neq 0$ (multiplicative inverse)

Is there a name for this type of structure?

Here's an example:

We start with strings, defining $+$ as concatenation (so $AAB+BCA=AABBCA$). It is clear that the empty string $\emptyset$ is the additive identity. Then we add additive inverses by defining an anti-character for every character, written with an overbar (so the inverse of $A$ is $\overline{A}$ such that $A+\overline{A}=\emptyset$, inverse of $ABB$ is $\overline{B}\overline{B}\overline{A}$, etc.) To create multiplication, we re-imagine strings as paths through space, so $A$ points right, $B$ points down, $C$ points forwards, $D$ points in a 4-dimensional unit direction, etc. Then, to multiply two characters, you place both of their vectors in space, then rotate such that $A$ points towards the direction of the left argument, then just see where the right argument points after the transformation. Since $A$ is used as the direction to point to, that causes $A$ to be the multiplicative identity. So $A\cdot B=B$, $B\cdot B=\overline{A}$, $B\cdot C = B$. To multiply a character by a whole string, you just apply the multiplication to each character: $ABC\cdot B=B\overline{A}C$. Then you can extend this single-character multiplication by multiplying every character on the left by the whole string on the right, repeatedly, and summing them (or more succinctly, $a\cdot b=\sum_{n=1}^{|n|} a_n \cdot b$). You can then extend this to multiplicative inverses, by allowing divisions between any two strings (not just ones who happen to be multiples of other strings, of course excluding right-division of $\emptyset$). Then, left-dividing by $A$ produces the inverse of a string.

RobPratt
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  • You mean absolutely every element has a multiplicative inverse, including $0$? And multiplication doesn't have to be associative? – Mark Mar 12 '24 at 23:58
  • @Mark oops, forgot a≠0, that's fixed now. but multiplication not being commutative was intentional. – The Zip Creator Mar 13 '24 at 00:02
  • Possibly https://en.wikipedia.org/wiki/Near-ring – Randy Marsh Mar 13 '24 at 00:04
  • @RandyMarsh I don't have associative multiplication (only addition), so not a near-ring. – The Zip Creator Mar 13 '24 at 00:06
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    Don't think it has a special name. Can you give an example of such a structure though? Looks interesting. – Mark Mar 13 '24 at 00:10
  • How about non-associative near-ring with identity and inverses then? – Randy Marsh Mar 13 '24 at 00:13
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    Your axioms imply addition is commutative (if you have both left and right distributivity)- see here – Bill Dubuque Mar 13 '24 at 00:27
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    @BillDubuque You made me follow that link to check -- they indeed do not even use associativity of multiplication – Hagen von Eitzen Mar 13 '24 at 00:33
  • @Mark added an example – The Zip Creator Mar 13 '24 at 00:42
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    Isn't the inverse of $AB$ the string $\bar B \bar A$? – Ethan Bolker Mar 13 '24 at 00:48
  • @EthanBolker argh, typo, you're right, I fixed it – The Zip Creator Mar 13 '24 at 00:54
  • For $A \cdot B$, since $B$ points down, I rotate the plane of the first two co-ordinate axes by $\frac{\pi}{2}$ anti-clockwise so that the right argument $(B)$ points the direction $A$ usually points. Then, the left argument ($A$) points upwards, so we find $A \cdot B = \overline{B}$? I'm not sure I understand the multiplication you have defined. – legionwhale Mar 13 '24 at 01:00
  • From what you've said, I'd certainly agree that $A$ is a one-sided multiplicative identity on the right, but I don't know that it is on the left. – legionwhale Mar 13 '24 at 01:02
  • @legionwhale I'm not the greatest at explaining things; I just edited the string mulitplication explanation (and found a way to explain it less verbosely) and now it hopefully should be more clear? – The Zip Creator Mar 13 '24 at 01:15
  • It still seems to me that $A \cdot B= \overline{B}$. Edit: I get an extra overline for your first two computations, but not the third. Even if I'm making some kind of error here which switches orientation, then e.g. I would also get that $B \cdot A = \overline{B}$. Given that Bill Dubuque gave a proof that your axioms would seem to imply commutative addition, which is clearly not the case for your addition, there must be some place where the hypotheses there are not fulfilled here. I wonder if this might be it, but maybe I'm really missing something. – legionwhale Mar 13 '24 at 01:24

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