The concept of subrings follow naturally.

Let R be a ring. A

subringis a subset S of R such that:

- (S, +) is a subgroup of (R, +);
- S contains 1;
- for any a, b in S, ab is in S.

We can actually simplify the above conditions to just the following:

*S*contains 1;- for any
*a*,*b*in*S*, both*a-b*and*ab*lie in*S*.

Indeed, since *S* is non-empty and closed under subtraction, (*S*, +) must be a subgroup of (*R*, +) . The result then follows. By the way, this probably doesn’t need saying, but if *T* is a subring of *S* and *S* is a subring of *R*, then*T* is also a subring of *R*.

*Examples*

- We have a sequence of subrings and .
- In the case of polynomials we have: .
- For matrices, we have .
- Consider the ring
**R**[*x*] of polynomials in*x*with real coefficients. We have the subring**R**[*x*^{2}] of polynomials in*x*^{2}with coefficients as well as*S*, the set of all polynomials spanned by all monomials except*x*^{1}. - The set 2
**Z**of even integers is*not*a subring of**Z**since it doesn’t contain 1. - If
*R*is any ring with identity=1, let*R’*be the set of multiples of 1, i.e. … -(1+1+1+1), -(1+1+1), -(1+1), -1, 0, 1, 1+1, 1+1+1, 1+1+1+1, … . Clearly, multiplying*m*and*n*in*R’*corresponds to integer multiplication of*m*and*n*. Hence*R’*is a subring of*R*. Let*n*be the smallest positive integer for which*n*= 0 in*R*. If no such*n*exists, then*R’*is isomorphic to**Z**. Otherwise,*R’*is isomorphic to**Z**/*n*.

*Exercise* : Prove that in example 5, if *R* is an integral domain, then *n* is prime.

Note that *S *is a subgroup of (*R*, +) and since all additive groups are abelian here, we see that *S* is a normal subgroup of (*R*, +). Hence, the set *S* divides the ring *R* into cosets of the form *x*+*S*. Although true, this is actually not very useful in practice. One really prefers to write *x*+*S* when *S* is an “ideal” of *R* rather than a subring. This will be covered later.

*Also, one can define a “subrng” of a rng, in which case one simply drops the condition that 1 must be contained in S.*

–

The next topic is on *ring products*.

If

Ris a collection of rings, let_{i}R= ∏be the set-theoretic product. Then R is a ring under component-wise addition and component-wise multiplication. The identity element is the one where all components are 1._{I}R_{i}

The easiest non-trivial case is that of *R* × *S*, where:

- (1, 1) is the identity;
- (
*r*,*s*) + (*r’*,*s’*) := (*r*+*r’*,*s*+*s’*) for any*r*,*r’*in*R*and*s*,*s’*in*S*; - (
*r*,*s*) × (*r’*,*s’*) := (*r*×*r’*,*s*×*s’*).

One can also generalise this to the product of an arbitrary collection of rings, i.e.

, where is a collection of rings indexed by *λ*.

An element of *R* is a gigantic tuple , where .* *

Beware! The ring *R* × *S* contains a copy of *R* × {0} and {0} × *S*, which appear to be isomorphic to *R* and *S* respectively. However, these two subsets are *not* subrings because they fail to contain the identity element!

On the other hand, ignoring the identity element, it is clear that *R* × {0} and {0} × *S* are both *subrngs* of *R* × *S*. Furthermore, these subrngs are themselves rings since they contain the identity element. So we have the awkward situation where *a subrng which is a ring is not a subring in general. *This can trip you badly if you weren’t careful.

However, if *R* = *S*, then *R* × *R* does contain a copy of *R*, namely the set of “diagonal entries” (*r*, *r*) for *r* in *R*.