Introduction to Ring Theory (1)

Recall that in groups, one has only a binary operation *. Rings are algebraic structures with addition and multiplication operations – and consistency is ensured by the distributive property.

Definition. A ring R is a set together with two binary operations: (R, +, ×), satisfying the following properties.

(R, +) is an abelian group, with identity 0 and inverse of r denoted by -r.

(R, ×) satisfy the following:

(identity) there exists 1 in R (called the unity) such that 1×r = r×1 = r;

(associative) for any r, s, t in R, we have (r×s)×t = r×(s×t).

(Distributive) We have r×(s+t) = (r×s)+(r×t) for any r, s, t in R.

In general, multiplication in a ring need not commute, i.e. xy ≠ yx in general. If xy=yx, then we say the ring is commutative.

The first order of the day is to prove the following very basic properties:

0×r = r×0 = 0 for any r;
(-r)×s = -(r×s) = r×(-s) for any r and s.

Quick proof

To show r×0 = 0, use the distributive property: r×0 + r×0 = r×(0+0) = r×0. So adding -(r×0) to both sides proves r×0 = 0.
To show -(r×s) = r×(-s), we use the distributive property again: r×(-s) + r×s = r×((-s)+s) = r×0 = 0. ♦

Following our intuition from elementary algebra, we use the following notations/conventions.

r×s is denoted as rs for short.
For a positive integer n, product of n terms of r is denoted by $r \times r\times \ldots \times r = r^n$ . Also, $r^0 = 1$ .
a+(-b) is denoted as a–b for short. Thus we have (a–b)c = (a+(-b))c = ac+(-b)c = ac+(-(bc)) = ac–bc by the above.
Product is always performed before addition.

Various conventions exist in different literature. For example, some books like Herstein’s Topics in Algebra or Hungerford’s Algebra (GTM 73) define a ring without the condition of the existence of unity (1). Thus by their definition, 2Z (the set of even numbers) forms a ring under the usual addition and multiplication. Whichever is preferred is a matter of taste, but we’ll stick to our version above, under which Z is a ring but 2Z is not. If we ever need to talk about rings without identity, we’ll call them rngs instead (rings without “i”dentity).

Anyone who’s done matrix arithmetic knows that rings come in all kinds of flavours. Let’s go through some technical definitions here below.

Definitions. Let R be a ring. First we look at the elements of R.

A zero-divisor is an element x, such that there exist non-zero y, z for which xy = zx = 0.

A unit is an element x, such that there exist y, z for which xy = zx = 1. Here, we must have y=z since y = (zx)y = z(xy) = z. We usually denote $y = x^{-1}$ .

Now let’s look at the whole ring R.

A trivial ring is where 1=0. If so, any r = r×1 = r×0 = 0, so there’s only 1 element: {0}.

If every non-zero element of the non-trivial ring R is a unit, we call the ring a division ring.

If R is a non-trivial commutative ring, and $xy = 0 \implies x=0 \text{ or } y=0$ , then we call the ring an integral domain (or just domain if we’re lazy).

A commutative division ring is called a field.

[ Another way of looking at fields: a field is a an algebraic structure where (R, +) is an abelian group, (R-{0}, ×) is an abelian group, and the distributive law holds. ] The reason we wish to exclude trivial rings in some of the above definitions, is akin to why we’d exclude 1 as a prime number. That being said, there’re people who’re interested in studying structures over the field with 1 element (for very deep reasons), but that’s another story for another day.

Examples

We hope that the reader is not put off by the large number of technical definitions above. If in doubt, feel free to ignore some of them, since we won’t be using them for now.

The integers Z form an integral domain under the usual addition/product, but not a field since 2 is not a unit.
The rationals Q, reals R, and complex numbers C each form a field since every non-zero element can be inverted.
The set of integers modulo prime p, given by Z/p, forms a field under addition and product mod p.
The set of integers modulo composite n, given by Z/n, forms a commutative ring but is not an integral domain or field. E.g. in Z/6, 2×3 = 0.
The set of 2×2 matrices with real entries forms a non-commutative ring. There are lots of examples of zero-divisors, e.g. $\begin{pmatrix}1 & -1\\ 1 & -1\end{pmatrix} \times \begin{pmatrix} 1 & 1 \\ 1 & 1\end{pmatrix}$ = $\begin{pmatrix}0 & 0\\{0}&0\end{pmatrix}$ = $\begin{pmatrix}1 & 1\\1 & 1\end{pmatrix}\begin{pmatrix}1 & 1\\-1 & -1\end{pmatrix}$ .
The set of quaternions is $\mathbf{H} = \{a+bi + cj + dk : a,b,c,d \in \mathbf{R}\}$ , where i, j, k satisfy the multiplication laws $i^2 = j^2 = k^2 = -1$ and ij = –ji = k, jk = –kj = i, ki = –ik = j. Then H is a non-commutative division ring – if you’re not familiar with quaternions, just take it as it is for now.
The set $\mathbf{R}\times\mathbf{R} = \{(x,y) : x, y\in\mathbf{R}\}$ of ordered pairs of real numbers is a ring under component-wise addition and multiplication. This is not an integral domain since (1, 0) × (0, 1) = (0, 0).
The set of polynomials in x with real coefficients, R[x], is an integral domain. It is not a field since x is not a unit. Likewise, the set Z[x] of polynomials in x with integer coefficients is an integral domain.

Exercises

Does the set of 3-vectors (x, y, z) form a ring under vector addition and scalar product? What about vector addition and cross product?
For each of the above examples, describe the set of units.
Take R, the set of real numbers, with the following operations: r # s = r + s + 1, r * s = rs + r + s. Does this form a ring under # and * ? What are the identity and additive inverse of r?
We can generalise example 7 as follows. Given rings R and S, prove that R × S is a ring under component-wise addition and multiplication.

Intuition

Here’re some instances where one has to be careful with intuition.

(1) “Cancellation” only works if the value cancelled is not a zero-divisor.

For example, if ac = bc, it would be too rash to say if c ≠ 0, then a=b. On the other hand, if c is not a zero-divisor, then $ac = bc \implies (a-b)c = 0$ , so since c is not a zero-divisor we have a=b. In particular, cancellation always works in an integral domain.

(2) In expanding an algebraic expression, the order of product must not change.

For example, (p+a)(q+b) = pq+pb+aq+ab. Writing pb as bp is a no-no unless your ring is commutative. In particular, most standard algebraic formulae like $x^2 - y^2 = (x+y)(x-y)$ only hold in commutative rings. That includes our favourite binomial formula as well, for if the ring is non-commutative, then we’re forced to write:

$(x+y)^2 = (x+y)(x+y) = x(x+y)+y(x+y) = x^2+xy+yx+y^2.$

The corresponding expansion for $(x+y)^3$ would have 8 terms.

(3) An integer n can be considered as an element of ring R.

Since R contains 1, we can consider any integer n as an element in R via n·1. The notation deserves some clarification. Since (R, +) is a group, for any element r of R, we have … -2r, –r, 0, r, 2r, … , which corresponds to successive applications of the group operation. So our n·1 refers to this, with r=1. To ensure consistency, one has to check that n·r (as successive addition) is the same as n×r (as ring product). The usual way to show this is by first considering the case n>0, n=0, and then take the negative to prove it for n<0 (left as an exercise).

(4) A non-zero integer can become zero inside the ring.

This is most obviously seen in the finite ring Z/n. For clarity, let’s consider a concrete case n=35. Then the unity is 1 mod 35, so each integer gets mapped to its representative mod 35. In particular, this means 35=0 in the ring. Now, the smallest positive integer n for which n becomes 0 in the ring is called the characteristic of the ring. It’s clear that if m also becomes 0 in the ring, then m is a multiple of n (the characteristic) since we can write m = qn+b, where 0 ≤ b < n. Then b also becomes 0, so the only possibility is that the integer b itself is zero.

Exercise. Prove that (1) a field is also an integral domain; (2) a finite integral domain is a field. [ Answer (highlight to read) : (1) If ab=0 and a≠0, then multiplying by a^-1 gives b=0. (2) Let a be a non-zero element. Since ab=ac implies b=c, we see that “multiplication-by-a” is an injective map. Since the ring is finite, it is in fact bijective. So there’s an element b such that ab=1. ]