Rings, Fields and Their First Properties
Two operations instead of one. Rings cover the integers and polynomials, fields are the rings where division works, and the difference explains why some equations are solvable and others are not.
What a learner can do afterwards
- Decide whether a given structure is a ring, an integral domain or a field
- Give a ring with zero divisors and say what breaks
- Explain why the integers modulo p form a field exactly when p is prime
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Here a ring has a multiplicative identity 1. Addition forms an abelian group: sums commute and associate, 0 is an identity, and additive inverses exist. Multiplication is closed and associative, with 1a = a = a1. Both distributive laws hold: a(b + c) = ab + ac and (a + b)c = ac + bc. Multiplication need not commute. Subtraction is always available as addition of an additive inverse.
An integral domain is a commutative ring with 1 not equal to 0 and no nonzero factors whose product is zero. A field is a commutative ring with 1 not equal to 0 in which every nonzero element has a multiplicative inverse. Rationals form a field. Integers form a domain, but division by an integer need not stay integral. Even integers lack a multiplicative identity and are not a ring under our convention.
A nonzero element a is a zero divisor when ab = 0 for some nonzero b. In a commutative ring, cancellation of a nonzero factor works exactly when there are no zero divisors: ab = ac gives a(b - c) = 0. In a field, multiply ab = 0 by a inverse when a is nonzero, obtaining b = 0. Thus every field is an integral domain. Zero itself is excluded from our zero-divisor definition.
In Z8, 2 times 4 = 0 although both residues are nonzero. Hence 2 times 4 equals 2 times 0, but 4 does not equal 0. Cancellation of 2 fails. A zero divisor cannot be invertible: multiplying 2 times 4 = 0 by an inverse of 2 would force 4 = 0.
For a positive integer n, Z_n has residues 0 through n - 1, with sums and products reduced modulo n. These operations form a commutative ring. A residue a is invertible exactly when gcd(a, n) = 1. Bezout's identity then gives integers s and t with sa + tn = 1; reducing modulo n gives sa = 1, so s is an inverse. Conversely, an inverse equation shows every common divisor of a and n divides 1.
Modulo 9, the identity 7 times 4 - 3 times 9 = 1 gives inverse 7 for residue 4. You can also check products directly until one is 1. Z1 is the zero ring: its only residue is 0, so 1 = 0. It is not a field.
For prime p, each residue 1 through p - 1 has gcd 1 with p, so Bezout gives every nonzero residue an inverse. Thus Z_p is a field. If n is composite, write n = ab with 1 < a, b < n. The two nonzero residues a and b have product 0, so Z_n is not a field. Together with the Z1 case, Z_n is a field exactly when the positive modulus n is prime.
Polynomials Q[x] form a commutative ring with identity. The product of two nonzero polynomials has degree equal to the sum of their degrees, so it is nonzero. Therefore Q[x] is a domain. It is not a field: multiplying the polynomial x by any nonzero polynomial gives positive degree, never the constant 1. Constants such as nonzero rational numbers still have inverses.
Rings have addition and multiplication. Domains allow cancellation of nonzero factors, and fields allow division by nonzero elements. Z_n is a field exactly when the positive modulus n is prime.
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