# Thread: Proof of the trivial subring

1. ## Proof of the trivial subring

I have to prove that the ring $(\{0\},+,\cdot)$ is a subring of any ring $(R,+,\cdot)$
Let S= $(\{0\},+,\cdot)$ and R= $(R,+,\cdot)$ then S is a subring of R iff $(R,+,\cdot)$ is a ring and $S \subseteq R$ and S is a ring with the same operations.
As we know S has an identity element of 0 --> 0+0=0
It has an additive inverse of 0 --> 0-0=0
Its commutative and associative
and its distributive over addition.
So my only problem is that I am having difficulties cleaning this up and putting it into a proof.
Maybe you can help me

2. ## Re: Proof of the trivial subring

since {0} is a subset of every ring R (since every ring R has an additive identity, called its 0), we need only show the following 2 things (since the set {x in R: x = 0} is non-empty):

1. for all x,y in S = {0}, x - y is in S
2. for all x,y in S, xy is in S

(we automatically get commutativity of +, associativity of +, associativity of *, and distributivity of * over + (both ways) because the operations + and * in S are just those of R, and S is a subset of R).

so:

0 - 0 = 0, which is in S = {0}
0*0 = 0, which is in S = {0}.

that is the entire proof.

### trival subring

Click on a term to search for related topics.