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MATH 614 LECTURE NOTES, FALL, 2015 by Mel Hochster MATH 614 LECTURE NOTES, FALL, 2015 by Mel Hochster Lecture of September 9 We assume familiarity with the notions of ring, ideal, module, and with the polynomial ring in one or finitely many variables over a commutative ring, as well as with homomor- phisms of rings and homomorphisms of R-modules over the ring R. As a matter of notation, N ⊆ Z ⊆ Q ⊆ R ⊆ C are the non-negative integers, the integers, the rational numbers, the real numbers, and the complex numbers, respectively, throughout this course. Unless otherwise specified, all rings are commutative, associative, and have a multiplica- tive identity 1 (when precision is needed we write 1R for the identity in the ring R). It is possible that 1 = 0, in which case the ring is f0g, since for every r 2 R, r = r ·1 = r ·0 = 0. We shall assume that a homomorphism h of rings R ! S preserves the identity, i.e., that h(1R) = 1S. We shall also assume that all given modules M over a ring R are unital, i.e., that 1R · m = m for all m 2 M. When R and S are rings we write S = R[θ1; : : : ; θn] to mean that S is generated as a ring over its subring R by the elements θ1; : : : ; θn. This means that S contains R and the elements θ1; : : : ; θn, and that no strictly smaller subring of S contains R and the θ1; : : : ; θn. It also means that every element of S can be written (not necessarily k1 kn uniquely) as an R-linear combination of the monomials θ1 ··· θn . When one writes S = R[x1; : : : ; xk] it often means that the xi are indeterminates, so that S is the polynomial ring in k variables over R. But one should say this. The main emphasis in this course will be on Noetherian rings, i.e., rings in which every ideal is finitely generated. Specifically, for all ideals I ⊆ R, there exist f1; : : : ; fk 2 R Pk such that I = (f1; : : : ; fk) = (f1; : : : ; fk)R = i=1 Rfi. We shall develop a very useful theory of dimension in such rings. This will be discussed further quite soon. We shall not be focused on esoteric examples of rings. In fact, almost all of the theory we develop is of great interest and usefulness in studying the properties of polynomial rings over a field or the integers, and homomorphic images of such rings. There is a strong connection between studying systems of equations, studying their solutions sets, which often have some kind of geometry associated with them, and studying commutative rings. Suppose the equations involve variables X1;:::;Xn with coefficients in K. The most important case for us will be when K is an algebraically closed field such as the complex numbers C. Suppose the equations have the form Fi = 0 where the Fi are polynomials in the Xj with coefficients in K. Let I be the ideal generated by the Fi in the polynomial ring K[X1;:::;Xn] and let R be the quotient ring K[X1;:::;Xn]=I. In R, the images xj of the variables Xj give a solution of the equations, a sort of \universal" 1 2 solution. The connection between commutative algebra and algebraic geometry is that algebraic properties of the ring R are reflected in geometric properties of the solution set, and conversely. Solutions of the equations in the field K give maximal ideals of R. This leads to the idea that maximal ideals of R should be thought of as points in a geometric object. Some rings have very few maximal ideals: in that case it is better to consider all of the prime ideals of R as points of a geometric object. We shall soon make this idea more formal. Before we begin the systematic development of our subject, we shall look at some very simple examples of problems, many unsolved, that are quite natural and easy to state. Suppose that we are given polynomials f and g in C[x], the polynomial ring in one variable over the complex numbers C. Is there an algorithm that enables us to tell whether f and g generate C[x] over C? This will be the case if and only if x 2 C[f; g], i.e., if and only if x can be expressed as a polynomial with complex coefficients in f and g. For example, suppose that f = x5 + x3 − x2 + 1 and g = x14 − x7 + x2 + 5. Here it is easy to see that f and g do not generate, because neither has a term involving x with nonzero coefficient. But if we change f to x5 + x3 − x2 + x + 1 the problem does not seem easy. The following theorem of Abhyankar and Moh, whose original proof was about 150 pages long, gives a method of attacking this sort of problem. Theorem (Abhyankar-Moh). Let f, g in C[x] have degrees d and e respectively. If C[f; g] = C[x], then either d j e or e j d, i.e., one of the two degrees must divide the other. Shorter proofs have since been given. Given this difficult result, it is clear that the specific f and g given above cannot generate C[x]: 5 does not divide 14. Now suppose instead that f = x5 + x3 − x2 + x + 1 and g = x15 − x7 + x2 + 5. With this choice, the Abhyankar-Moh result does not preclude the possibility that f and g generate C[x]. To pursue the issue further, note that in g − f 3 the degree 15 terms cancel, producing a polynomial of smaller degree. But when we consider f and g−f 3, which generate the same ring as f and g, the larger degree has decreased while the smaller has stayed the same. Thus, the sum of the degrees has decreased. In this sense, we have a smaller problem. We can now see whether the Abhyankar-Moh criterion is satisfied for this smaller pair. If it is, and the smaller degree divides the larger, we can subtract off a multiple of a power of the smaller degree polynomial and get a new pair in which the larger degree has decreased and the smaller has stayed the same. Eventually, either the criterion fails, or we get a constant and a single polynomial of degree ≥ 2, or one of the polynomials has degree 1. In the first two cases the original pair of polynomials does not generate. In the last case, they do generate. This is a perfectly general algorithm. To test whether f of degree d and g of degree n ≥ d are generators, check whether d divides n. If so and n = dk, one can choose a constant c such that g − cf k has degree smaller than n. If the leading coefficients of f and g are a 6= 0 and b 6= 0, take c = b=ak. The sum of the degrees for the pair f; g − cf k has decreased. Continue in the same way with the new pair, f, g −cf k. If one eventually reaches a pair in which one of the polynomials is linear, the original pair were generators. Otherwise, one reaches either a pair in which neither degree divides the other, or else a pair in which one 3 polynomial has degree ≥ 2 while the other is constant. In either of these cases, the two polynomials do not generate. The constant does not help, since we have all of C available anyway, and a polynomial g of degree d ≥ 2 cannot generate: when g is substituted into a polynomial of degree n, call it F , F (g) has a term of degree dn coming from gn, and no other term occurring can cancel it. Thus, one cannot have x = F (g). One can work backwards from a pair in which one of the polynomials is linear to get all pairs of generators. For example, one gets pairs of generators x; 0 ! x; 1 ! x + 5; 1 ! x + 5; (x + 5)7 + 1 ! 11 (x + 5)7 + 1 + x + 5; (x + 5)7 + 1. If one expands the last pair out, it is not very obvious from looking at the polynomials that they generate. Of course, applying the algorithm described above would enable one to see it. This gives a reasonably appealing method for telling whether two polynomials in one variable generate C[x]. The step of going from the problem of when two polynomials generate to C[x] over C to when three polynomials generate turns out to be a giant one, however! While algorithms are known based on the theory of Gr¨obnerbases, the process is much more complex. There are some elegant conjectures, but there is a lack of elegant theorems in higher dimension. One might hope that given three polynomials that generate C[x], say f, g, and h, with degrees d; e; n, respectively, that it might be true that one of the degrees has to be a sum of non-negative integer multiples of the other two, e.g., n = rd+se. Then one could reduce to a smaller problem (i.e., one where the sum of the degrees is smaller) by subtracting a constant times f rgs from h, while keeping the same f and g. But it is not true in the case of three polynomials that one of the degrees must be a sum of non-negative integer multiples of the other two. (See whether f = x5; g = x4 + x; and h = x3 generate C[x].) The problem of giving an elegant test for deciding when m polynomials generate the polynomial ring C[x1; : : : ; xn] in n variables over C seems formidable, but when m = n there is at least a tantalizing conjecture.
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