Partial rings

[uncool]

Recently, I have come up with an algebraic structure that I would like to call "Partial rings" with some interesting properties. They are similar to rings, as will be shown later. The rules for this structure are:

 

1) Group under addition

2) Another function, multiplication

3) mutiplication is distributive

4) multiplication is associative

 

5 (optional)) Noncommutative group under addition.

 

Some properties already found:

1) The identity under addition annihilates under multilplication (clearly)

2) The additive order of the product a*b divides the additive order of each a and b

3) The set of products is commutative

4) The set of products a*r, r is in R is a commutative group

5) If the order of the "partial ring" is squarefree, then the set of products is cyclic

 

Some conjectures:

1) The set of products is a group.

 

Are there any interesting applications of this "partial ring"?

[woelen]

What do you mean with "partial ring"? You mean the elements, which are of the form a*r, for a given a?

 

If I read your definition, then you are very close to the definition of a field. If the number of elements in your ring is a prime number, then you create the simple field Zn, with n a prime number. Such fields have applications in cryptography. E.g. the public key cryptosystem RSA is based on properties of fields Zp and of rings Zpq, with p and q distinct prime numbers.

[uncool]

by partial ring I mean the algebraic structure itself, and the order is the number of elements in it. This structure is not a ring until it is given an identity element (optional, depending on who you ask) and commutativity. Basically, I've taken the definition of a ring, and removed commutativity of addition.

=Uncool=

[woelen]

Could you give an example of such a ring-like structure, which is not a true ring, but is like a ring, except the commutivity of addition?

[uncool]

Alright, I'll take the smallest such struture, the nontrivial structure on D6, the dihedral group of order 6.

Let us call the elements e, a, 2a, b, b+a, and b+2a, where e is the identity, a means one rotation of the vertices, and b means a reflected. Note that addition is noncommutative, as b+a = 2a+b.

*     |e  a  2a  b  b+a  b+2a
------------------------
e     |e  e   e   e    e    e
a     |e  e   e   e    e    e
2a   |e  e   e    e    e    e
b     |e  e   e    b    b    b
b+a  |e  e   e   b     b    b
b+2a|e  e   e    b    b    b

This structure is associative, distributive, and has all the properties of a ring except commutativity of addition and identity of multiplication.

=Uncool=

[woelen]

Is this really the multiplicative table for D6? If a is a single rotation, and b is a reflection, then I get different results.

 

e*a is not equal to e, but equal to a. e is an identity element for D6. For any operation z, e*z = z*e = z. Also a*a*a = e and b*b = e. But a*a is not equal to e.

 

It is true that this group is non-abelian.

 

I do not understand the meaning of addition over here. What is e+a, what "physical" meaning should I assign to e+a?

 

Sorry, but I have the impression that I am something missing here . We're not yet "connected" .

[uncool]

Sorry, I forgot to specify. The "partial ring" is D6 under addition, and the table below defines multiplication.

=Uncool=

[uncool]

Perhaps a better way to state what I did:

 

+    | e    a    2a   b    b+a  b+2a
_______________________________________
e    | e    a    2a   b    b+a  b+2a
a    | a    2a   e    b+2a b    b+a
2a   | 2a   e    a    b+a  b+2a b
b    | b    b+a  b+2a e    a    2a
b+a  | b+a  b+2a b    2a   e    a
b+2a | b+2a b    b+a  a    2a   e

*    | e    a    2a   b    b+a  b+2a
______________________________________
e    | e    e    e    e    e    e
a    | e    e    e    e    e    e
2a   | e    e    e    e    e    e
b    | e    e    e    b    b    b
b+a  | e    e    e    b    b    b
b+2a | e    e    e    b    b    b

The "partial ring" is D6 under addition, and the multiplication table is shown below addition.

=Uncool-

[woelen]

Yes, now it is perfectly clear for this example. I am trying to match this with other algebraic structures for which I know applications.

 

In what direction are you thinking for applications of these algebraic structures? Mathematical/theoretical, or practical, such as crypto-systems, data-encoding, etc?

 

Can you tell something about the number of distinct products in general? Did you derive that number? The underlying group should have a fairly high additive order, otherwise the multiplication is rather boring, with the set of products being small. For practical applications the algebraic structure must be sufficiently "rich", both in addition and in multiplication.

[uncool]

I'm looking more for theoretical here, as I haven't really looked at practical uses of algebraic structures (besides cryptography a bit) recently.

 

I cannot really say much about the number of distinct products, except that it is limited by the size of the largest commutative subgroup of the original group.

 

Some other stuff I have found out about it: We may be able to form a version of a quotient partial ring by using what I'd call a "normal partial ideal" - that is, a normal subgroup which is also ideal (any element of the original group multiplied by any element in the subgroup is in the subgroup). This follows all the normal isomorphism theorems - If you have a "partial ring" homomorphism from R onto R' with kernel N, then R/N is isomorphic to R' (I'm pretty sure of this; haven't entirely checked though), etc.

 

A conjecture I have come up with, and checked on a few of these "partial rings," is the following:

Let N be the set of all elements which always multiply to 0, that is:

N = {r in R: r*b = b*r = e for all b in R}. Then N is a normal partial ideal - easy to prove - and R/N is commutative - not proven yet. I'm trying to figure out how to show this.

 

A note: I have found the smallest partial ring with noncommutative addition and noncommutative multiplication. It is as follows:

D8 under addition, with b as reflection and a as 90 degree rotation.
*    | e    a    2a   3a   b    b+a  b+2a b+3a
______________________________________________
e    | e    e    e    e    e    e    e    e    
a    | e    2a   e    2a   e    2a   e    2a
2a   | e    e    e    e    e    e    e    e    
3a   | e    2a   e    2a   e    2a   e    2a
b    | e    2a   e    2a   e    2a   e    2a   
b+a  | e    e    e    e    e    e    e    e    
b+2a | e    2a   e    2a   e    2a   e    2a   
b+3a | e    e    e    e    e    e    e    e    

I am still looking to see if I can find a partial ring such that the set of products is not a group. As shown before, a square must divide it. I know how to create it, as long as I can make sure it is associative; there is where I run into trouble.

=Uncool-

[uncool]

Proof for the properties:

Some properties already found:

1) The identity under addition annihilates under multilplication (clearly)

2) The additive order of the product a*b divides the additive order of each a and b

3) The set of products is commutative

4) The set of products a*r, r is in R is a commutative group

5) If the order of the "partial ring" is squarefree, then the set of products is cyclic

1) 0*a = 0*a + 0*a + -(0*a) = (0+0)*a + -(0*a) = 0*a + -(0*a) = 0

2) ab+ab+ab+...+ab (o(a) times) = (a+a+a+a+...+a)b = eb = e, so o(ab)|o(a). Similar argument for b.

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.

4) Commutative has already been shown. All we need show is closure and inverse to show it is a group. a*r1 + a*r2 = a*(r1+r2), so it is closed. a*r1 + a*-r1 = a*(r1 + -r1) = a*e = e, so we have inverse and therefore identity, so a*r is a group.

5) This one is long. It shall be proven later.

=Uncool-

[uncool]

For the 5th:

Let us assume we have a "partial ring" of order [math]p_{1} p_{2}...p_{n}[/math]. Then, by Cauchy's theorem, there is an element of order [math]p_{i}[/math] for each p. Let them be known as [math]a_{1}, a_{2},...,a_{n}[/math]. Let us look at the set of squares of the as, that is [math]a_{1}^{2} = a_{1}*a_{1}, a_{2}^{2},...,a_{n}^{2}[/math]. By 2) above, we know that each either has order 1 or [math]p_{i}[/math]. Let us take the sum of these squares. We know the additive orders of each are relatively prime. We also know, by 3), that those elements are commutative. Therefore, the order of the sum is the lcm of each of the orders - the product of each of the [math]p_{i}[/math] such that [math]a_{i}^2 \neq e[/math]. However, this is also the necessary additive order to include all possible combinations of the [math]a_{i}^2[/math] as a group. Clearly, every product can be shown to be a sum of the [math]a_{i}^2[/math], as every product can be written as [math](m_{1}a_{1}+m_{2}a_{2}+...+m_{n}a_{n})*(o_{1}a_{1}+o_{2}a_{2}+...+o_{n}a_{n}) = m_{1}o_{1}a_{1}^2+m_{2}o_{2}a_{2}^2+...+m_{n}o_{n}a_{n}^2[/math] as the products of the different orders are all e, where the m and o are all integers.

=Uncool-

[woelen]

You are looking at the sum of the squares of the [math]a_i[/math]. The order of [math]a_i^2[/math] either is 1, or [math]p_i[/math]. In your examples, I, however, only see either squares, which are equal to e, or I see squares, which are equal to an element with order 2. If this always is the case, then your "partial ring" will only have either e-elements or order 2 elements in the set of products, which is not very exciting. Did you find an example, with squares, which have an higher order than 2, i.e. an order [math]p_i[/math], with [math]p_i[/math] being the order of [math]a_i[/math]?

 

I also see squares of elements of order of a prime squared (4), with the square having the order of the prime itself (2). This is the case in your last example, for [math]a^2[/math] being equal to [math]2a[/math]. If this could be true for a 9*2 order group, with [math]a^2[/math] having order 3, while [math]a[/math] has order 9, then that could also be interesting.

 

Summarizing, in order to be really interesting, the set of products must also have higher order elements.

[uncool]

Alright...I can create a group with any number for the order of a square element, and still have a cyclic product set. I shall provide in the case for a prime.

 

Let us take a prime p. Then let us find a prime q such that p | q-1. This necessarily does exist. Then we can create a nonabelian group of order pq. Then let a^2 = a, b^2 = e, where a is of order p and b is of order q, and everything else follows.

I believe that the same occurs when you have any squarefree m. Just take a prime q such that m|q-1, and you can create a nonabelian group of order mq where there is a cyclic group of order m, and let a^2 = a where a is one element of order m, and b^2 = e where b is an element of order q.

D6 is an example where m = 2, q = 3. We can also create it for D2n, for any integer n.

Any other example is too large to be written here easily, though.

The elements are of order 2 because those are all that can be written for such small partial rings.

=Uncool-

[uncool]

I also see squares of elements of order of a prime squared (4), with the square having the order of the prime itself (2). This is the case in your last example, for [math]a^2[/math] being equal to [math]2a[/math]. If this could be true for a 9*2 order group, with [math]a^2[/math] having order 3, while [math]a[/math] has order 9, then that could also be interesting.

Well, the reason it is of order 2 is because 2a must annihilate everything...with something of order 27 you could probably find such an a.

=Uncool-

[uncool]

Recently found: 32 element ring (with no identity) such that the set of products is not a group.

Under addition, it is Z2^5. Let the basis be named a, b, c, d, and e.

Under multiplication: c, d, and e annihilate the ring (their product with anything is 0). a*a = c, a*b = b*a = d, b*b = e. This is distributive, associative (as all triple products are 0), etc., and c + d + e is not generated.

=Uncool-

[uncool]

The 0th and 1st Isomorphism theorem works in these partial rings, that is:

0th: The kernel of any "partial ring" homomorphism is a partial normal ideal, that is, it is a normal subgroup under addition and is an ideal under multiplication. Any partial normal ideal N of R also has a homomorphism to R/N with kernel N.

The definition of a "partial ring" homomorphism is that it takes sums to sums and products to products.

1st: If you have an onto ring homomorphism from R to R' with kernel N, then R/N is isomorphic to R'.

I believe that this also implies the second and third isomorphism theorems, namely, H+N/N is isomorphic to H/H intersect N for any subring H and normal partial ideal N of R, and also that G/N1 = (G/N2)/(N1/N2) where N2 and N1 are normal partial ideals of G, and N2 is in N1.

=Uncool-

[csabay]

[...]

 

Some properties already found:

 

[...]

 

3) The set of products is commutative

 

[...]

 

 

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

[zheng sheng ming]

I think that the group is too ............to understand?how to understand it?

[uncool]

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

They commute under addition (since every pair of elements has a product which is effectively in the ring), which was my point - even though the elements of the pring may not commute under addition, the elements of the set of products will.

=Uncool-

Edited December 22, 2010 by uncool

[zheng sheng ming]

I feel that the group theory is too diffculd to understand.

[csabay]

They commute under addition (since every pair of elements has a product which is effectively in the ring), which was my point - even though the elements of the pring may not commute under addition, the elements of the set of products will.

=Uncool-

 

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

[zheng sheng ming]

I think group theory so hard!

[uncool]

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

No problem; I like feedback on this.

 

I'm trying to figure out whether these partial rings are the symmetries for any objects at all. Also wondering if it's possible to create pring-modules...

=Uncool-

[csabay]

No problem; I like feedback on this.

 

I'm trying to figure out whether these partial rings are the symmetries for any objects at all. Also wondering if it's possible to create pring-modules...

=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.

 

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