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A Course in Commutative Algebra

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A Course in Commutative Algebra Graduate Texts in Mathematics 256 A Course in Commutative Algebra Bearbeitet von Gregor Kemper 1. Auflage 2010. Buch. xii, 248 S. Hardcover ISBN 978 3 642 03544 9 Format (B x L): 0 x 0 cm Gewicht: 1200 g Weitere Fachgebiete > Mathematik > Algebra Zu Inhaltsverzeichnis schnell und portofrei erhältlich bei Die Online-Fachbuchhandlung beck-shop.de ist spezialisiert auf Fachbücher, insbesondere Recht, Steuern und Wirtschaft. Im Sortiment finden Sie alle Medien (Bücher, Zeitschriften, CDs, eBooks, etc.) aller Verlage. Ergänzt wird das Programm durch Services wie Neuerscheinungsdienst oder Zusammenstellungen von Büchern zu Sonderpreisen. Der Shop führt mehr als 8 Millionen Produkte. Chapter 1 Hilbert’s Nullstellensatz Hilbert’s Nullstellensatz may be seen as the starting point of algebraic geom- etry. It provides a bijective correspondence between affine varieties, which are geometric objects, and radical ideals in a polynomial ring, which are algebraic objects. In this chapter, we give proofs of two versions of the Nullstellensatz. We exhibit some further correspondences between geometric and algebraic objects. Most notably, the coordinate ring is an affine algebra assigned to an affine variety, and points of the variety correspond to maximal ideals in the coordinate ring. Before we get started, let us fix some conventions and notation that will be used throughout the book. By a ring we will always mean a commutative ring with an identity element 1. In particular, there is a ring R = {0},the zero ring,inwhich1=0.AringR is called an integral domain if R has no zero divisors (other than 0 itself) and R ={0}.Asubring of a ring R must contain the identity element of R, and a homomorphism R → S of rings must send the identity element of R to the identity element of S. If R is a ring, an R-algebra is defined to be a ring A together with a homomorphism α: R → A. In other words, by an algebra we will mean a commutative, associative algebra with an identity element. A subalgebra of an algebra A is a subring that contains the image α(R). If A and B are R-algebras with homomorphisms α and β,thenamapϕ: A → B is called a homomorphism of (R-)algebras if ϕ is a ring homomorphism, and ϕ ◦ α = β.IfA is a nonzero algebra over a field K, then the map α is injective, so we may view K asasubringofA. With this identification, a homomorphism of nonzero K-algebras is just a ring homomorphism fixing K elementwise. One of the most important examples of an R-algebra is the ring of polynomials in n indeterminates with coefficients in R, which is written as R[x1,...,xn]. If A is any R-algebra and a1,...,an ∈ A are elements, then there is a unique algebra homomorphism ϕ: R[x1,...,xn] → A with ϕ(xi)=ai, given by applying α to the coefficients of a polynomial and sub- stituting xi by ai. Clearly the image of ϕ is the smallest subalgebra of A G. Kemper, A Course in Commutative Algebra, Graduate Texts 7 in Mathematics 256, DOI 10.1007/978-3-642-03545-6 2, c Springer-Verlag Berlin Heidelberg 2011 8 1 Hilbert’s Nullstellensatz containing all ai, i.e., the subalgebra of A generated by the ai. We write this image as R[a1,...,an], which is consistent with the notation R[x1,...,xn] for a polynomial ring. We say that A is finitely generated if there exist a1,...,an with A = R[a1,...,an]. Thus an algebra is finitely generated if and only if it is isomorphic to the quotient ring R[x1,...,xn]/I of a polynomial ring by an ideal I ⊆ R[x1,...,xn]. By an affine (K-)algebra we mean a finitely generated algebra over a field K.Anaffine (K-)domain is an affine K-algebra that is an integral domain. Recall that the definition of a module over a ring is identical to the def- inition of a vector space over a field. In particular, an ideal in a ring R is the same as a submodule of R viewed as a module over itself. Recall that a module does not always have a basis (= a linearly independent generating set). If it does have a basis, it is called free.IfM is an R-module and S ⊆ M is a subset, we write (S)R =(S) for the submodule of M generated by S, i.e., the set of all R-linear combinations of S. (The index R may be omitted if it is clear which ring we have in mind.) If S = {m1,...,mk} is finite, we write (S)R =(m1,...,mk)R =(m1,...,mk). In particular, if a1,...,ak ∈ R are ring elements, then (a1,...,ak)R =(a1,...,ak) denotes the ideal generated by them. 1.1 Maximal Ideals Let a ∈ A be an element of a nonzero algebra A over a field K.Asinfield theory, a is said to be algebraic (over K) if there exists a nonzero polynomial f ∈ K[x]withf(a)=0.WesaythatA is algebraic (over K) if every element from A is algebraic. Almost everything that will be said about affine algebras in this book has its starting point in the following lemma. Lemma 1.1 (Fields and algebraic algebras). Let A be an algebra over a field K. (a) If A is an integral domain and algebraic over K,thenA is a field. (b) If A is a field and is contained in an affine K-domain, then A is algebraic. Proof. (a) We need to show that every a ∈ A \{0} is invertible in A.For this, it suffices to show that K[a] is a field. We may therefore assume that A = K[a]. With x an indeterminate, let I ⊆ K[x] be the kernel of the map ∼ K[x] → A, f → f(a). Then A = K[x]/I.SinceA is an integral domain, I is a prime ideal, and since a is algebraic over K, I is nonzero. Since K[x] is a principal ideal domain, it follows that I =(f)withf ∈ K[x] ∼ irreducible, so I is a maximal ideal. It follows that A = K[x]/I is a field. (b) By way of contradiction, assume that A has an element a1 that is not algebraic. By hypothesis, A iscontainedinanaffineK-domain B = K[a1,...,an] (we may include a1 in the set of generators). We 1.1 Maximal Ideals 9 can reorder a2,...,an in such a way that {a1,...,ar} forms a maxi- mal K-algebraically independent subset of {a1,...,an}. Then the field of fractions Quot(B)ofB is a finite field extension of the subfield L := K(a1,...,ar). For b ∈ Quot(B), multiplication by b gives an L-linear endomorphism of Quot(B). Choosing an L-basis of Quot(B), we obtain amapϕ:Quot(B) → Lm×m assigning to each b ∈ Quot(B) the repre- sentation matrix of this endomorphism. Let g ∈ K[a1,...,ar] \{0} be a common denominator of all the matrix entries of all ϕ(ai), i =1,...,n. −1 m×m So ϕ(ai) ∈ K[a1,...,ar,g ] for all i.Sinceϕ preserves addition and multiplication, we obtain −1 m×m ϕ(B) ⊆ K[a1,...,ar,g ] . K[a1,...,ar] is isomorphic to a polynomial ring and therefore factorial (see, for example, Lang [33, Chapter V, Corollary 6.3]). Take a factoriza- tion of g,andletp1,...,pk be those irreducible factors of g that happen to lie in K[a1]. Let p ∈ K[a1] be an arbitrary irreducible element. Then −1 −1 p ∈ A ⊆ B since K[a1] ⊆ A and A is a field. Applying ϕ to p yields a diagonal matrix with all entries equal to p−1, so there exists a nonneg- −1 −s s ative integer s and an f ∈ K[a1,...,ar]withp = g · f,sog = p · f. By the irreducibility of p, it follows that p is a K-multiple of one of the pi. Since this holds for all irreducible elements p ∈ K[a1], every element from K[a1] \ K is divisible by at least one of the pi. But none of the k pi divides i=1 pi + 1. This is a contradiction, so all elements of A are algebraic. The following proposition is an important application of Lemma 1.1.A particularly interesting special case of the proposition is that A ⊆ B is a subalgebra and ϕ is the inclusion. Proposition 1.2 (Preimages of maximal ideals). Let ϕ: A → B be a homo- morphism of algebras over a field K,andletm ⊂ B be a maximal ideal. If B is finitely generated, then the preimage ϕ−1(m) ⊆ A is also a maximal ideal. Proof. The map A → B/m,a→ ϕ(a)+m,haskernelϕ−1(m)=:n.SoA/n is isomorphic to a subalgebra of B/m. By Lemma 1.1(b), B/m is algebraic over K. Hence the same is true for the subalgebra A/n,andA/n is also an integral domain. By Lemma 1.1(a), A/n is a field and therefore n is maximal. Example 1.3. We give a simple example that shows that intersecting a max- imal ideal with a subring does not always produce a maximal ideal. Let A = K[x] be a polynomial ring over a field and let B = K(x) be the rational function field. Then m := {0}⊂B is a maximal ideal, but A ∩ m = {0} is not maximal in A. Before drawing a “serious” conclusion from Proposition 1.2 in Proposi- tion 1.5, we need a lemma.
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