Prove Ring with Identity on Set S with One Element x

[Stephen88]

On a set S with exactly one element x,
define x + x = x, x*x = x. Prove that S is a ring.
The way I think about this problem is be showing that it verifies certain axioms...like associativity,commutativity,identity,inverse for addition and commutativity for multiplication and a (b + c) = ab + ac .. (a + b) c = ac + bc.
For Addition the first two i think it is obvious since
1.x+x=x+x..
2.(x+x)+x=x+(x+x)
For Identity since x+x=x then 0_S=x.
For the inverse I don't see how since the set has only one element x which equal 0_S...I guess I don't have to check the last two axioms because S is not a ring.
Am I doing this right?

 

[Evgeny.Makarov]

We have x + (-x) = 0 where both -x and 0 are defined to be x, so there is no problem with an additive inverse.

 

[Stephen88]

uh sorry...yes that is true then for multiplication commutativity (x*x)*=x*(x*x) and also x(x+x)=x +x and (x+x)x=x+x again.Will this suffice or is there something else.?..because it seemed quite short.

 

[Evgeny.Makarov]

for multiplication commutativity (x*x)*=x*(x*x)

The fact (x * x) * x = x * (x * x) is called associativity.

x(x+x)=x +x and (x+x)x=x+x again.

Distributivity says x(x+x) = x * x + x * x and (x+x) * x = x * x + x * x.

Will this suffice or is there something else.?..because it seemed quite short.

Why don't you check the list of ring axioms, for example, in Wikipedia?

 

[blue_raver22]

Yes, you are on the right track. To prove that S is a ring, we need to show that it satisfies all the necessary axioms for a ring. Here is a step-by-step explanation of how we can prove that S is a ring:

1. Addition is associative: For any a, b, c in S, we have (a+b)+c = a+(b+c) = a+c = a. This follows from the definition of addition in S.

2. Addition is commutative: For any a and b in S, we have a+b = a = b+a = b. Again, this follows from the definition of addition in S.

3. Identity element for addition exists: We have already shown that 0_S is the identity element for addition in S.

4. Inverse element for addition exists: Since S only has one element x, we can see that x is its own additive inverse. This means that for any a in S, we have a+(-a) = a+a = a = 0_S.

5. Multiplication is associative: For any a, b, c in S, we have (a*b)*c = a*c = a and a*(b*c) = a*c = a. This follows from the definition of multiplication in S.

6. Multiplication is commutative: For any a and b in S, we have a*b = a = b*a = b. Again, this follows from the definition of multiplication in S.

7. Distributive property holds: For any a, b, c in S, we have a*(b+c) = a = a*b + a*c = a+a = a. Similarly, (a+b)*c = a = a*c + b*c = a+a = a. This follows from the definition of multiplication in S.

Since S satisfies all the necessary axioms for a ring, we can conclude that S is indeed a ring.