In this series, we will cover some common algebraic structures – other than groups, which had been amply covered in the Group Theory series. However, due to the considerable depth of many of the topics, we can only provide a brief overview of each of them. Thus, if the Group Theory series were a 2-hour movie, each of the following topics is more like a 5-minute preview. In some cases, it’s closer to a 30-second teaser; in fact, the topic of commutative rings is so deep that we can barely even begin to do it justice here.
As prerequisites, the reader is assumed to be familiar with Chapters I – VI of group theory, and preferably some exposure to Chapter XII as well. Abstract diagrams and universal properties are not so critical here, but they’re indispensable by the time we reach commutative algebra.
Ok, let’s begin. Start with the definition of rings, which are algebraic structures with addition and multiplication.
Definition. A rng (not a misspelling!) is a set R together with operations + and × (for brevity, r × s is often denoted by rs) such that the following hold.
- (R, +) is a group, whose identity is denoted by 0 and inverse is denoted by –r;
- (distributive) for any r, s, t in R, we have: (r + s)t = rt + st;
- (distributive) for any r, s, t in R, we have: t(r + s) = tr + ts;
- (associative) for any r, s, t in R, we have: (rs)t = r(st).
A ring is a rng if it has an element 1 (called the identity or unity) such that 1 × r = r = r× 1. Thus, one sees that a rng is a ring without the identity. [ No, I did not just make that up. ] If rs = sr for all r and s, then we say the rng/ring is commutative.
Our focus will be on rings rather than rngs; hence if we forget to mention, assume R is a ring and not merely a rng.
Take a minute out to prove the following basic properties for a rng R:
- If 1 exists, then it is unique.
- 0 × r = 0 = r × 0 for any r.
- For any r and s, (-r) × s = -(r × s) = r × (-s), and (-r) × (-s) = r × s.
Notation : multiplying an element r by itself n>0 times, we get r × … × r, denoted rn. Furthermore, r0 is defined to be 1 when r≠0. From this, we have:
rm × rn = rm+n and (rm)n = rmn
for any non-negative integers m and n.
Note : for rings, one can multiply and expand expressions as per normal, as long as one is careful with the order of multiplication. E.g. for any elements a, b of the ring,
(a+b)3 = a3 + a2b + aba + ab2 + ba2 + bab + b2a + b3,
and if our ring were commutative, the order of multiplication doesn’t matter and the above simplifies to a3 + 3a2b + 3ab2 + b3.
Note that the symbols 3, 5 and -4 etc refer to 1+1+1, 1+1+1+1+1 and -(1+1+1+1) respectively, where 1 is the unity in R, i.e. these are not integers but specific elements of the ring R. Seemingly weird things can then happen, e.g. in the ring Z/10 of integers modulo 10 (see next section), we have 20 = 0, 25 = 15 = 5 etc. Certainly 20 ≠ 0 as integers, but when we use these symbols to denote elements of R, 20 = 0 can happen.
Anyway, expansion works in commutative rings! Similarly, factoring generally works, e.g.
a3 + b3 + c3 – 3abc = (a + b + c)(a2 + b2 + c2 – ab – bc – ca)
holds in commutative rings since we can expand the RHS by merely using the above laws.
Example of Errors
A student wants to solve 2X2 – 14XY + 24Y2 = 0 in the ring R. He reasons as follows: “The expression is equivalent to 2(X – 4Y)(X – 3Y) = 0. Hence, one of the terms 2, X-4, Y-3 is equal to zero. Since 2 ≠ 0, we must have X = 4Y or X = 3Y.” …
… to which the teacher points out 3 mistakes.
- In equating 2X2 – 14XY + 24Y2 and 2(X – 4Y)(X – 3Y), he assumed XY = YX.
- Also 2=0 can occur in a ring.
- Finally, it is not true that ab = 0 implies a = 0 or b = 0.
We will go through these properties in greater detail in the next section.