Algebraic geometry M 4 P 33

Exercise sheet 1

All rings R are commutative with 1.

1. Localization. A subset S ⊂ R is called multiplicative if it is closed under multiplication and contains 1. The localization S^{−1}R of R with respect to S is defined as the set of formal fractions \frac{a}{b}, with a ∈ R and b ∈ S, with usual rules for addition and multiplication, up to the equivalence relation: \frac{a}{b} = \frac{a_{1}}{b_{1}} iff (ab_{1} − a_{1}b)s = 0 for some s ∈ S.

(a) Show that S^{−1}R is not the zero ring iff 0 ∉ S. (Hint: prove that 1 ≠ 0 in S^{−1}R.)

(b) Prove that the natural map R → S^{−1}R given by a ↦ \frac{a}{1} is an injective homomorphism of rings iff S does not contain a zero divisor. (In that case we can think of R as a subset of S^{−1}R. Then S^{−1}R is simply the fractions with “restricted denominators”.)

(c) Prove that if I ⊂ R is an ideal, then S^{−1}I is an ideal in S^{−1}R.

(d) Check that the following are multiplicative systems: (1) all the elements of R that are not zero divisors, (2) all non-negative powers of an element r ∈ R, (3) R \ I, where I is an ideal, is a multiplicative system iff I is a prime ideal.

2. Integrally closed domains. An integral domain is called integrally closed if it coincides with its own integral closure in its fraction field.

(a) Check that Z is integrally closed.

(b) Prove that any PID (principal ideals domain – this means that every ideal is principal) is integrally closed. (Hint: Use the theorem that any PID is a UFD - unique factorization domain - which means that every non-zero element is uniquely a product of prime elements (up to order and multiplication by invertible elements). Imitate the proof for Z.)

(c) Let d be a square-free integer, d ≠ 1. Show that the integral closure of Z in the quadratic field Q(\sqrt{d}) is Z[\sqrt{d}] if d is not 1 mod 4, otherwise it is Z[(1 + \sqrt{d})/2].

(d) Deduce from (c) that Z[\sqrt{−3}] is not integrally closed. Now (b) says that this is not a PID (whereas Z[(1 + \sqrt{−3})/2] is, because this ring has Euclid’s algorithm, for details see some book on algebraic number theory).

(e) Z[x_{1}, …, x_{n}] and k[x_{1}, …, x_{n}], where k is a field, are important examples of integrally closed rings. (I don’t ask to prove this. If you are really interested, find a proof in a book.)

(f) Show that Z[x]/(x^{2} + 4) is not integrally closed. Therefore, quotients of integrally closed rings are not necessarily integrally closed.

(g) Prove that a localization of an integrally closed domain is integrally closed.

2. Noetherian rings. (a) Prove that any quotient of a Noetherian ring is Noetherian.

(b) Prove that any localization of a Noetherian ring is Noetherian.

(c) Check that a polynomial ring in infinitely many variables is not Noetherian.

3. Radical. (a) Show that the set of nilpotent elements of a ring R is an ideal. It is called the radical of R.

(b) Let R be a ring, I an ideal in R. Let \sqrt{I} be the set of x ∈ R such that x^{n} ∈ I for some n. Prove that \sqrt{I} is an ideal of R.

(c) Show that \sqrt{I} is the intersection of all prime ideals of R that contain I. [Hint: A nilpotent is contained in every prime ideal – just use the definition of a prime ideal. It is more tricky to prove that if x ∈ R is a non-nilpotent, then there is a prime ideal not containing x. You may argue as follows: show that the localization R[x^{−1}] is a non-zero ring, thus it contains a maximal ideal. Consider its preimage I with respect to the canonical homomorphism R → R[x^{−1}]. Now show that I is prime and does not contain x.]

4. Let k be a field. Let I and J be ideals in k[x_{1}, …, x_{n}]. Use the Nullstellensatz to prove that there is an inclusion of closed algebraic subsets Z(I) ⊂ Z(J) of the affine n-space if and only if \sqrt{J} ⊂ \sqrt{I}.

5. Let k be an algebraicaly closed field. Prove that a k-algebra A is isomorphic to the coordinate ring of an affine variety over k iff A is a k-algebra of finite type (i.e., A is generated by finitely many elements as an algebra, not as a module) and has no zero-divisors. (Hint: Use the Nullstellensatz.)