Part I The Algebra Geometry Lexicon
2.2 Noetherian Rings and Modules
2.2 Noetherian Rings and Modules 39 (b) M is finitely generated.
In particular, every submodule of a finitely generatedR-module is also finitely generated.
Proof. We only need to show that (b) implies (a), since the converse impli- cation is a consequence of Theorem2.9. So let M = (m1, . . . , mk)R. We use induction onk. There is nothing to show fork= 0, so assumek >0. Consider the submodule
N := (m1, . . . , mk−1)R⊆M.
By induction,N is Noetherian. The homomorphism ϕ:R→M/N, a7→amk+N
is surjective, so M/N ∼= R/ker(ϕ). By hypothesis and by Proposition 2.4, R/ker(ϕ) is Noetherian, so M/N is Noetherian, too. Applying Proposi-
tion2.4again shows thatM is Noetherian. ut
The following theorem is arguably the most important result on Noethe- rian rings.
Theorem 2.11(Polynomial rings over Noetherian rings). Let R be a Noe- therian ring. Then the polynomial ring R[x]is Noetherian, too.
Proof. LetI⊆R[x] be an ideal. By Theorem2.9, we need to show thatI is finitely generated. For a non-zero integeri, set
Ji:=n ai∈R
there exista0, . . . , ai−1∈Rsuch that
i
X
j=0
ajxj∈Io . Clearly Ji ⊆ R is an ideal. Let ai ∈ Ji with f = Pi
j=0ajxj ∈ I. Then I3xf =Pi
j=0ajxj+1, soai ∈Ji+1. It follows that theJi form an ascending chain of ideals of R. By hypothesis, there exists annsuch that fori≥nwe have Ji=Jn. Again by hypothesis, everyJi is finitely generated, so
Ji= (ai,1, . . . , ai,mi)R for i≤n. (2.2) and
Ji=Jn= (an,1, . . . , an,mn)R for i > n. (2.3) By the definition of Ji, there exist polynomials fi,j ∈I of degree at mosti whosei-th coefficient isai,j. Set
I0:=
fi,j
i= 0, . . . , n, j= 1, . . . , mi
R[x]⊆I.
We claim that I = I0. To prove the claim, consider a polynomial f = Pd
i=0bixi ∈ I with deg(f) = d. We use induction on d. We first consider
the case d≤n. Sincebd ∈Jd, we can use (2.2) and write bd =Pmd j=1rjad,j withrj∈R. Then
fe:=f−
md
X
j=1
rjfd,j
lies inIand has degree less thand, so by inductionfe∈I0. This impliesf ∈I0. Now assumed > n. Then we can use (2.3) and write bd=Pmn
j=1rjan,j with rj∈R. So
fe:=f−
mn
X
j=1
rjxd−nfn,j
lies inIand has degree less thand, so by inductionfe∈I0. Again we conclude f ∈I0. So indeedI=I0 is a finitely generated ideal. ut The corresponding statement for formal power series rings is contained in Exercise2.4. By applying Theorem2.11repeatedly and using the second statement of Proposition2.4, we obtain
Corollary 2.12 (Finitely generated algebras). Every finitely generated al- gebra over a Noetherian ring is Noetherian. In particular, every affine algebra is Noetherian.
A special case is the celebrated Basis Theorem of Hilbert.
Corollary 2.13 (Hilbert’s Basis Theorem). Let K be a field. Then K[x1, . . . , xn] is Noetherian. In particular, every ideal in K[x1, . . . , xn] is finitely generated.
The name Basis Theorem comes from the fact that generating sets of ideals are sometimes calledbases. One consequence is that every affine variety X ⊆Kn is the solution set of afinite system of polynomial equations:X = V(f1, . . . , fm).
Exercises to Chapter 2
2.1 (A non-Noetherian ring providing many counter examples).
Consider the polynomial ringS=K[x, y] and the subalgebraR:=K+S·x given in Example 2.3. Show that R is not Noetherian. Conclude that R is not finitely generated as an algebra. Explain why this provides an example for the following caveats:
• Subrings of Noetherian rings need not be Noetherian.
• Subalgebras of finitely generated algebras need not be finitely generated.
In Exercise7.4we will also see that Krull’s Principal Ideal Theorem7.4fails forR. In Exercise2.6we explore whether Example2.3is, in some sense, the smallest of its kind. (Solution on page231)
Exercises 41 2.2 (An Artinian module that is not Noetherian). Let p ∈ N be a prime number and consider theZ-modules
Zp:=n
a/pn∈Q|a, n∈Z
o⊂Q and M :=Zp/Z. Show thatM is Artinian but not Noetherian. (Solution on page232) 2.3 (Modules over an Artinian ring). Show that a finitely generated moduleM over an Artinian ringRis Artinian. (Solution on page232) 2.4 (The Noether property for formal power series rings). LetRbe a Noetherian ring and
R[[x]] :=nX∞
i=0
aixi|ai∈Ro
the formal power series ring overR. Show thatR[[x]] is Noetherian.(Solution on page232)
2.5 (Separating subsets). LetKbe a field andA⊆K[x1, . . . , xn] a sub- algebra of a polynomials algebra (which, as we have seen in Example 2.3, need not be finitely generated). Every polynomial from A defines a func- tion Kn → K. A subset S ⊆ A is called (A-)separating if for all points P1, P2∈Kn we have:
If there existsf∈Awithf(P1)6=f(P2), then there existsf ∈Swithf(P1)6=
f(P2).
(Loosely speaking, this means thatS has the same capabilities of separating points asA.)
(a) Show that if S⊆AgeneratesA as an algebra, thenS is separating. (In other words, “separating” is a weaker condition than “generating”. It is seen in (b) and (c) that it is in fact substantially weaker.)
*(b) Show thatA has a finite separating subset.
(c) Exhibit a finite R-separating subset of the algebra R ⊂ K[x, y] from Example2.3.
(Solution on page234)
*2.6 (Subalgebras ofK[x]). LetKbe a field and K[x] a polynomial ring in one indeterminate. Is every subalgebra ofK[x] finitely generated? Give a proof or a counter example. (Solution on page234)
2.7 (Graded rings). A ringR is called graded if it has a direct sum de- composition
R=R0⊕R1⊕R2⊕ · · ·= M
d∈N0
Rd
(as an Abelian group) such that for all a ∈ Ri and b ∈ Rj one has ab ∈ Ri+j. Then an element from Rd is called homogeneous of degree d. A standard example isR=K[x1, . . . , xn] withRdthe space of all homogeneous polynomials of degreed(including the zero-polynomial). LetRbe graded and set
I= M
d∈N>0
Rd,
which obviously is an ideal.I is sometimes called theirrelevant ideal. Show the equivalence of the following statements.
(a) Ris Noetherian.
(b) R0 is Noetherian andI is finitely generated.
(c) R0 is Noetherian andRis finitely generated as anR0-algebra.
Remark: By Corollary 2.12, a finitely generated algebra over a Noetherian rings is Noetherian. However, Noetherian algebras are not always finitely generated. So graded rings constitute a special case where this converse holds.
(Solution on page235)
2.8 (The Noether property and subrings). In Exercise2.1we have seen that in general the Noether property does not go down to subrings. In this exercise we look at a situation where it does.
(a) LetS be a Noetherian ring andR⊆S a subring such that there exists a homomorphismϕ:S→RofR-modules withϕ|R = idR. Show thatR is Noetherian, too.
(b) Show that for a ringR, the following three statements are equivalent: (i) Ris Noetherian; (ii)R[x] is Noetherian; (iii)R[[x]] is Noetherian.
(Solution on page235)
2.9 (Right or wrong?). Decide if the following statements are true or false.
Give reasons for your answers.
(a) Every finitely generated module over an Artinian ring is Artinian.
(b) Every Artinian module is finitely generated.
(c) Every ring has a module that is both Noetherian and Artinian.
(d) The set of all ideals of a ring, together with the ideal sum and ideal product, forms a commutative semiring (i.e., we have an additive and a multiplicative commutative monoid, and a distributive law).
(Solution on page236)
2.10(The ring of analytic functions). LetR be the ring of all analytic functions R → R, i.e., all functions which are given by power series that converge on all ofR. Show thatR is not Noetherian.
Can your argument be used for showing that other classes of functions R→Rform non-Noetherian rings, too? (Solution on page236)
Chapter 3
The Zariski Topology
In this chapter we will put a topology on Kn and on affine varieties. This topology is quite weak, but surprisingly useful. We will define an analogous topology on Spec(R). In both cases, there are correspondences between closed sets and radical ideals. As a consequence of some general topological con- siderations, affine varieties can be decomposed into irreducible components.
Another consequence is that a Noetherian ring only contains finitely many minimal prime ideals.
Readers who are unfamiliar with the language of topology can find all that is needed for this book in any textbook on topology (for example Bourbaki [6]), usually on the first few pages.