Partial rings

[uncool]

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

x^z+1 := x + x^z

x^z–1 := x' + x^z

 

For example: x⁵ = xxxxx; x^-3 = x'x'x'.

 

It is easy to see that our "power" is also additive in the meaning: x^u+v = x^u + x^v.

 

So now we are ready to define our multiplication as:

 

p • q := x^{w(p) w(q)}

 

Well, we have got the (F,+,•) pring.

Interesting; seems like this could work for any homomorphism to the integers...

 

Let G be a group; let f be a nontrivial homomorphism from G to the integers. Then let g be an element which does not go to 0 (which would mean that the subgroup generated by g is infinite). Then define x • y as g^(f(x) + f(y)).

 

ETA: This also happens to work when it's going to any cyclic group.

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

 

The natural version of the ideal in a pring is the normal ideal - that is, a normal subgroup which is an ideal. For those, the two equivalences are the same.

=Uncool-

Edited December 27, 2010 by uncool

[csabay]

The natural version of the ideal in a pring is the normal ideal - that is, a normal subgroup which is an ideal. For those, the two equivalences are the same.

=Uncool-

 

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

[uncool]

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

Actually, I think there is a normal ideal here.

 

The normal closure of I is the set of all elements such that their abelianization is generated by x. This is a normal subgroup, and as it contains the brutal ideal, it is an ideal. Therefore, it is a normal ideal.

 

Also, the products are always a brutal ideal, if when you say "build" you mean generate. It is not true if you mean that they are a brutal ideal - there is a 32 element group in which the products do not form a subgroup (one is missing).

=Uncool-

[csabay]

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.

[uncool]

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.

That must be true. A spring is a partial ring which is not additively commutative, correct? If so, then the set of products cannot be the entire thing, as the set of products is commutative.

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.

See earlier:

 

Additively, it's (Z_2)^5 - it's a ring without unity. To be specific:

 

c, d, and e all have trivial multiplication. Then a*a = c, a*b = b*a = d, b*b = e.

 

Then, as it is commutative, and only 3 elements do not eliminate anything, we can see that there are only 6 products, not the 7 that would be necessary (c + d + e is not a product).

=Uncool-

[csabay]

That must be true. A spring is a partial ring which is not additively commutative, correct? If so, then the set of products cannot be the entire thing, as the set of products is commutative.

 

Fine. I was too cautious. It must be a "theorem" rather than a "conjecture". Let's name this theorem as "Uncool's theorem"

 

 

See earlier:

 

Additively, it's (Z_2)^5 - it's a ring without unity. To be specific:

 

c, d, and e all have trivial multiplication. Then a*a = c, a*b = b*a = d, b*b = e.

 

Then, as it is commutative, and only 3 elements do not eliminate anything, we can see that there are only 6 products, not the 7 that would be necessary (c + d + e is not a product).

=Uncool-

 

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

[uncool]

Giving the Cayley table would take a long time; writing the relations of the generators is enough.

 

Additively:

a + a = 0, b + b = 0, c + c = 0, d + d = 0, e + e = 0

 

a + b = b + a

a + c = c + a

a + d = d + a

a + e = e + a

 

b + c = c + b

b + d = d + b

b + e = e + b

 

c + d = d + c

c + e = e + c

 

d + e = e + d

 

fully describes the addition in the group.

 

Multiplicatively:

a*a = c

a*b = d

a*c = 0

a*d = 0

a*e = 0

 

b*a = d

b*b = e

b*c = 0

b*d = 0

b*e = 0

 

c*anything = 0

 

d*anything = 0

 

e*anything = 0

=Uncool-

[csabay]

Giving the Cayley table would take a long time; writing the relations of the generators is enough.

 

Additively:

a + a = 0, b + b = 0, c + c = 0, d + d = 0, e + e = 0

...

 

 

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

[uncool]

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

The 32 elements are:

 

0, a

b, a + b

 

c, a + c

b + c, a + b + c

 

 

d, a + d

b + d, a + b + d

 

c + d, a + c + d

b + c + d, a + b + c + d

 

 

 

 

e, a + e

b + e, a + b + e

 

c + e, a + c + e

b + c + e, a + b + c + e

 

 

d + e, a + d + e

b + d + e, a + b + d + e

 

c + d + e, a + c + d + e

b + c + d + e, a + b + c + d + e

=Uncool-