Ideals and Quotient Rings
The ring counterpart of a normal subgroup is an ideal, and quotienting by it declares its elements to be zero. Whether an ideal is prime or maximal is read directly off the ring the quotient produces.
What a learner can do afterwards
- Check whether a subset of a ring is an ideal
- Describe the quotient of a polynomial ring by a principal ideal
- Distinguish prime from maximal in terms of the quotient ring
1 · Read
Throughout this lesson, R is a commutative ring with identity. A subset I is an ideal if it contains 0, closes under differences, and absorbs multiplication: ri belongs to I for every r in R and i in I. Equivalently, require nonemptiness and the two closures. The ambient ring matters because its elements are the allowed multipliers. If an ideal contains 1, absorption puts every r = r1 inside, so I = R.
A ring homomorphism f preserves addition and multiplication; here it also preserves 1. It preserves 0 and subtraction as consequences. Its kernel contains 0. If f(a) = f(b) = 0, then f(a - b) = 0. If f(a) = 0, then f(ra) = f(r)f(a) = 0 for every r. Thus the kernel is an ideal of the source ring, not generally a subfield.
In Z, (a, b) means all integer combinations as + bt and equals (gcd(a, b)). For example, (15, 25) = (5): all combinations are multiples of 5, and 5 = 2 times 15 - 25 is a combination. Intersections use a different rule: (a) intersect (b), for positive a and b, consists of their common multiples and equals (lcm(a, b)).
In R/I, a + I = b + I exactly when a - b is in I. Add or multiply representatives, then take classes. For nonconstant p in F[x], with F a field, divide leading terms, multiply p by that term and subtract. Repeat until the remainder is zero or has degree below d = degree(p). Its d coefficients give a unique representative: two such remainders cannot differ by a nonzero multiple of p. The relation p(x) = 0 also reduces higher powers.
In Q[x]/(x^2 + 2), let u be the class of x. Then u^2 = -2 and every class is a + bu. For a and b not both zero, its inverse is (a - bu)/(a^2 + 2b^2), because their product is 1 and the denominator is positive. Thus this quotient is a field. This direct calculation explains the field conclusion rather than relying only on the equation u^2 = -2.
For nonconstant p in F[x], with F a field, F[x]/(p) is a field iff p is irreducible: not a product of two positive-degree polynomials. Repeated division by remainders finds a monic gcd; back-substitution writes it as ug + vp. If irreducible p does not divide g, the gcd is 1, so u is an inverse modulo p. A quadratic is irreducible iff it has no root in F. A factorization p = ab with positive-degree factors gives nonzero lower-degree classes with zero product.
A prime ideal P is proper and satisfies ab in P implies a in P or b in P. Thus R/P is a nonzero integral domain. A maximal ideal M is proper with no ideal strictly between M and R; equivalently, R/M is a field. If a is outside M, maximality gives 1 = m + ra with m in M, so r is an inverse of its class. In Z, (0) is prime but not maximal: its quotient is Z, a domain but not a field.
For positive n, Z/(n) is Z_n. At n = 1, (n) = Z is not proper and the quotient is the zero ring. For prime n, the quotient is a field, so (n) is maximal. For composite n, its nonzero factor residues have zero product, so the quotient is not a field. Hence (n) is maximal exactly when n is prime. Every maximal ideal is prime because every field is a domain.
Test ideals using subtraction and absorption. Use the quotient ring to identify prime and maximal ideals.
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16 questions wait behind this lesson, each with its answer explained. Every answer feeds the sky: stars light as they are learned, and dim when it is time to come back.