MATH 205B NOTES 2010 COMMUTATIVE ALGEBRA 7 1.3. Basic Algebraic Geometry. I will use exercises 15,17,21 from the book and other example from other books to review basic algebraic geometry. Definition 1.29. For any ring A let Spec(A) denote the set of prime ideals of A. 1.3.1. Zariski topology. If E is any subset of A let V (E) denote the set of all prime ideals p which contain E: V (E) = p Spec(A) E p { ∈ | ⊆ } This function has the property that it reverses inclusion: V (E) V (E) if E F and V (E) = V (a) = V (r(a)) if a = (E) the ideal generated⊃ by E⊂. In particular, V (x) = V ((x)) for any x A. ∈ Exercise 1.30. (1) V (0) = Spec(A) (2) V (1) = ∅ (3) V ( Ei) = V (Ei) (4) V (a) V (b) = V (a b) = V (ab) ! ∪ " ∩ This exercise shows that the collection of subsets V (E) of Spec(A) are the closed sets in a topology on Spec(A) which is{called}the Zariski topology. The closed points in Spec(A) are the maximal ideals. For homework, please do: Exercises 17 and 21. They are not hard but we don’t have time to do them in class. 1.3.2. examples. I will take the ring A = C[x, y] but I will draw R[x, y]. The ideal (x a, y b) where a, b C are maximal and we will see later − − ∈ that they are the only maximal ideals since C is algebraically closed. So, the closed point in Spec C[x, y] correspond to the points (a, b) in the xy-plane. The key observation is: Proposition 1.31. The point (a, b) is in the zero set of the polynomials fi C[x, y] iff ∈ (x a, y b) Spec( f ) − − ∈ { i} Example 1.32. The prime ideal (x) corresponds to the y axis. Or more precisely: V (x) = V ((x)) = (x), (x, y b) b C { − | ∈ } The closed set V (x) consists of the closed points on the y-axis and the “generic point” (x) Spec C[x, y]. ∈ 8 MATH 205B NOTES 2010 COMMUTATIVE ALGEBRA Example 1.33. The ideal a = (xy x3) is not prime. It is the inter- section (and product) of two prime −ideals: a = (xy x3) = (x)(y x2) = (x) (y x2) − − ∩ − This means that: V (xy x3) = V (x) V (y x2) − ∪ − This is the union of the y-axis and the parabola y = x2. The fact that the set of zeros of xy x3 is a union of two sets indicates that it is not prime. − A closed subset of any topological space is called irreducible if it is not the union of two proper closed subsets. Proposition 1.34. If p is prime then V (p) is irreducible. But the converse is not true since V (p2) = V (p) is irreducible but p2 is not in general prime. Example 1.35. a = (y2 x2y) = (y)(y x2) = (y) (y x2) − − ∩ − This means that: V (y2 x2y) = V (y) V (y x2) − ∪ − This leads to the following questions (which will lead us back to the text). (1) When does ab = a b? (2) Can we tell algebraically∩ that the two curves intersect trans- versely in the first example and tangentially in the second ex- ample? (3) What happens if the field is not algebraically closed? 1.3.3. coprime ideals. Definition 1.36. Two ideals a, b are called coprime if a + b = A. Equivalently, a + b = 1 for some a a, b b. ∈ ∈ Proposition 1.37. If a, b are coprime then ab = a b. ∩ Exercise 1.38. Give an example of two prime ideals a, b in K[x, y] which are coprime and draw the sets V (a), V (b). Proposition 1.39 (A-M 1.10.i). a and b are coprime iff V (a) V (b) = . ∩ ∅ MATH 205B NOTES 2010 COMMUTATIVE ALGEBRA 9 So, the prime ideals in the examples are not coprime. However, if we look at p1, p2 in the two examples we see that, in Example 1.33, (x) + (y x2) = (x, y) − is a maximal ideal corresponding to the point (0, 0). In Example 1.35 (y) + (y x2) = (x2, y) − is not maximal, reflecting the fact that the curves are tangent to each other. Proposition 1.40 (A-M 1.10.ii). Let a , a , , a be ideals and let 1 2 · · · n p : A A/a → i be the homomorphism whose i-coor#dinate is the quotient map A → A/ai. Then p is surjective iff the ideals ai are pairwise coprime. 1.4. extension and contraction. Example 1.41. Take the inclusion map f : R[x, y] C[x, y]. Then → (x2 + y2) is prime in R[x, y] but (x2 + y2) = (x + iy)(x iy) in C[x, y]. So − V (x2 + y2) = V (x + iy) V (x iy) ∪ − Definition 1.42. Let f : A B be a ring homomorphism. Then for any ideal a in A the ideal in→B generated by the image of a is called the extension of a: ae := Bf(a) = b f(a ) a a, b B i i | i ∈ i ∈ and for any ideal b in B its$in%verse image in A is an ideal& called the contraction of b: c 1 b = f − (b) c 1 If b is prime then the contraction b = f − (b) is prime. But the example above shows that extension of a prime ideal may not be prime. The example is the book is: f : Z Z[i]. The extension of the prime ideal (5) is → (5)e = (2 + i) (2 i) ∩ − If f : A B is surjective, then b B is prime iff bc is prime. → ⊂ Proposition 1.43 (A-M 1.17). (1) a aec, b bce (2) ae = aece ⊆ ⊇ (3) bc = bcec (4) Let C = bc = set of contracted ideals and E = ae be the set of extende{ }d ideals. Then there is a bijection C { }E given by a ae and be b ↔ ,→ ← 10 MATH 205B NOTES 2010 COMMUTATIVE ALGEBRA 1.5. unique factorization domains. This is a quick review of UFDs expanding on the comment I made in class about prime and irreducible elements. Definition 1.44. An element p A which is not a unit is called irreducible if it is not the product ∈of two nonunits. An element p A is called prime if the ideal (p) is prime. ∈ Any prime element is irreducible. However, the converse is not true. An example is given by A = Z[√ 5]. In this ring 2 is irreducible but not prime since 2 3 = (1 + √ 5)(1− √ 5) (2) but 1 √ 5 / (2). In class I used the· contrapositiv− e:−If an− elemen∈ t x is not± irreduc− ∈ ible then (x) is not a prime ideal. Exercise 1.45. Show: If an element of A can be written as a product of prime elements, then that expression is unique up to reordering and multiplication of each prime element by a unit. This leads to the definition: A is a UFD if every nonunit is a product of prime elements. Exercise 1.46. Show that a ring is a UFD iff every irreducible element is prime and any nonempty collection of principal ideals contains a maximal element (the ACC)..