## Basic Algebra I.3 (Subrings and Ring Products)

The concept of subrings follow naturally.

Let R be a ring. A subring is 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, thenT is also a subring of R.

Examples

1. We have a sequence of subrings $\mathbf{Z} \subset \mathbf{Z}[i] \subset \mathbf{C}$ and $\mathbf{Z} \subset \mathbf{Q} \subset \mathbf{R} \subset \mathbf{C}$.
2. In the case of polynomials we have: $\mathbf{R} \subset \mathbf{R}[x] \subset \mathbf{R}[x, y] \subset \dots$.
3. For matrices, we have $D(n, \mathbf{R}) \subset B(n, \mathbf{R}) \subset M(n, \mathbf{R})$.
4. Consider the ring R[x] of polynomials in x with real coefficients. We have the subring R[x2] of polynomials in x2 with coefficients as well as S, the set of all polynomials spanned by all monomials except x1.
5. The set 2Z of even integers is not a subring of Z since it doesn’t contain 1.
6. 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 Ri is a collection of rings, let R = ∏I Ri 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.

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;
• (rs) × (r’s’) := (r×r’s×s’).

One can also generalise this to the product of an arbitrary collection of rings, i.e. $R = \prod_\lambda R_\lambda$, where $\{R_\lambda\}$ is a collection of rings indexed by λ.

An element of R is a gigantic tuple $(r_\lambda)$, where $r_\lambda \in R_\lambda$. 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.

This entry was posted in Basic Algebra Notes and tagged , , , , , . Bookmark the permalink.