csabay

Maybe I'm hopeless... Z_2^5 has got 32 elements. {a, b, c, d, e} are 5 elements. I lost the thread.

Fine. I was too cautious. It must be a "theorem" rather than a "conjecture". Let's name this theorem as "Uncool's theorem" Here I am confused a bit. Z_2^5 consists of 32 elements. I assign to them the names 0, 1, ..., 31. The Cayley-table of them is: (See appendix. z2^5.txt) I wonder if among them which ones could become "e" (perhaps 0?), "a", "b" &c. I don't understand the roles of the operations. This one above may have got the role of addition. Then the computational rules given by you -- they apply the multiplication, don't they? My perception is slow... Please try to give the both Cayley tables explicitly. Thanks -- Charles

I was perhaps unobservant as I formulated. I think -- as a conjecture -- that: The set of products in a SPRING always is less (i.e. real subset) than the spring itself. In other words: The multiplication contracts the spring onto a real subset of it. Uncool's theorem makes sure that this subset is a commutative subgroup under addition. No doubt that the set of products must be evidently a brutal ideal. I claim that the brutality of this ideal is an essential feature, i.e. the brutality is not trivial. (The brutality is trivial when I = R.) Charles P.S.: I'm curious about your 32-element-group.

Of course! But in this example one has to distinguish the above congruencies. Btw. in this example you cannot find a "normal" ideal except of the {0} ideal. In addition, the products build a "brutal" ideal. I call an ideal being "brutal" when from a, b in R follows that ab in I. For example let's define a product under the integers: a x b := 2ab where on the right you see the traditional product of integers. (Z, +, x) becomes a ring, in which the even integers form a brutal ideal. My conjecture is: In springs (my shorthand for strictly partial ring) products build always a brutal ideal. Charles

On the representation questions I have got no idea... As regards the modules, however,... When you intend to get an infinite example of pring, look at the following: Let be (F,+) a free group generated by elements x, x', y, y' and l. Rules are: x + x' = x' + x = y + y' = y' + y = l and x + l = l + x = x, x' + l = l + x' = x', y + l = l + y = y, y' + l = l + y' = y'. Elements of F — by contrast — will be written by letters p, q, r,… So it can be for example: p = xyyx, q = x'yy'x'y or r = l. Let's introduce the following integer functions on F: l(p), w(p); we assign them the names "length" and "weight" respectively. Recursive definitions are: l(l) = 0, l(x + p) = l(p) + 1, l(x' + p) = l(p) + 1, l(y + p) = l(p) + 1, l(y' + p) = l(p) + 1; w(l) = 0, w(x + p) = w(p) + 1, w(x' + p) = w(p) – 1, w(y + p) = w(p) + 1, w(y' + p) = w(p) – 1. In spite of the length which is not additive [l(xy) + l(y'x) = 4, l(xyy'x) = l(xx) = 2], the weight is. We also introduce a writing method. Let z be an integer, so we write: x0 := l xz+1 := x + xz xz–1 := x' + xz For example: x5 = xxxxx; x-3 = x'x'x'. It is easy to see that our "power" is also additive in the meaning: xu+v = xu + xv. So now we are ready to define our multiplication as: p • q := xw(p) w(q) Well, we have got the (F,+,•) pring. It is an interesting job to define congruency among F's elements. Let I be an ideal in the pring. So you get two several congruencies, namely p =L q if and only if by definition when –p + q in I p =R q if and only if by definition when p – q in I

You are right, your proof "3) -ac + (a+b)(c+d) + -bd = -ac + (a+b)c + (a+b)d + -bd= -ac + ac + bc + ad + bd + -bd = bc + ad = -ac + a(c+d) + b(c+d) + -bd = -ac + ac + ad + bc + bd + -bd = ad + bc" points out that the target of the commutativity is addition. I was fixed onto multiplication. Sorry. Charles

I found (3) being not true. Namely: Let (R, +, ×) be a ring and (G,+) be a group. I'll write the both addition as + and the both (additive) unity as a 0 without the risk of misunderstanding. The structure (R x G, +, ×) with the addition: (r, s) + (g, h) = (r + s, g + h) product: (r, s) × (g, h) = (r × s, 0) will be always a pring. (The proof is easy.) (Pring is my shorthand for partial ring.) Well, let now be our ring the Klein-ring (K), and our group the S3 group. When you perform the generating task written above you get a pring in which the set of products does not commute. (You need a non-commutative ring, the smallest one of which is K, and a non-commutative group, the smallest one of which is S3). Charles