# Commutative Ring

• March 10th 2011, 05:28 AM
page929
Commutative Ring
Let I be any set and let R be the collection of all subsets of I. Define addition and multiplication of subsets A,B ⊆ I as follows:
A+B = (A∪B) ∩ [A∩B(with a line over this)] and AB = A∩B. Show that R is a commutative ring under this addition and multiplication.

I know that to be a commutative ring it has to have the following:
closure, associativity, commutativity, distribution, identity and inverses.

Closure - show well-defined
Assoc - (a+b)+c = a+(b+c) and a(bc) = (ab)c
Comm - a+b = b+a and ab = ba
Dist - a(b+c) = ab+ac and (a+b)c = ac+bc

I am not sure how to write them out using the defined subsets above.

• March 10th 2011, 05:55 AM
emakarov
So, $A + B = (A\cup B)\cap\overline{(A\cap B)}$. When talking about set operations, this is called the symmetric difference and is commonly denoted by $A\vartriangle B$.

Quote:

Closure - show well-defined
Do you have difficulties with this?

Quote:

Assoc - (a+b)+c = a+(b+c) and a(bc) = (ab)c
The associativity of intersection is pretty obvious. For addition, it's a little messy. Hint: prove that $x\in (a+b)+c$ iff x belongs to exactly one of a, b, c or to all three, and similarly for $x\in a+(b+c)$.

Quote:

Comm - a+b = b+a and ab = ba
Follows directly from the definition and commutativity of union and intersection.

Quote:

Dist - a(b+c) = ab+ac and (a+b)c = ac+bc
Draw a Venn diagram of a, b, c and then prove that $x\in a(b+c)$ iff $x\in ab+ac$.
• March 10th 2011, 07:33 AM
page929
Any help with identity or inverses?
• March 10th 2011, 07:49 AM
Plato
Quote:

Originally Posted by page929
Any help with identity or inverses?

What is $A\Delta A=~?$

What is $A\Delta \emptyset=~?$
• March 10th 2011, 08:19 AM
page929
Quote:

Originally Posted by Plato
What is $A\Delta A=~?$

What is $A\Delta \emptyset=~?$

$emptyset$
A complement
• March 10th 2011, 08:47 AM
Plato
The best way to view this operation is $A\Delta B=(A\cap\overline{B})\cup(B\cap\overline{A}).$
That is the same as $(A\cup B)\cap (\overline{A\cap B})$.

Now $A\Delta \emptyset = \left( {A \cap \overline \emptyset } \right) \cup \left( {\emptyset \cap \overline A } \right) = A$