Part I The Algebra Geometry Lexicon
2.1 The Noether and Artin Property for Rings and Modules
Definition 2.1. LetR be a ring and M anR-module.
(a) M is called Noetherian if the submodules of M satisfy the ascend- ing chain condition, i.e., for submodules M1, M2, M3, . . . ⊆ M with Mi ⊆ Mi+1 for all positive integers i, there exists an integer n such that Mi = Mn for all i ≥ n. In other words, every strictly ascending chain of submodules is finite.
(b) R is called Noetherian if R is Noetherian as a module over itself. In other words,Ris Noetherian if the ideals ofRsatisfy the ascending chain condition.
(c) M is calledArtinianif the submodules ofM satisfy thedescending chain condition, i.e., for submodulesM1, M2, M3, . . .⊆M withMi+1⊆Mi for
1“This is theology and not mathematics”
33
all positive integers i, there exists an integer n such that Mi =Mn for alli≥n.
(d) Ris calledArtinian ifR is Artinian as a module over itself, i.e., if the ideals ofR satisfy the descending chain condition.
Example 2.2. (1) The ringZof integers is Noetherian, since ascending chains of ideals correspond to chains of integersa1, a2, . . .withai+1a divisor of ai. So the well-ordering of the natural numbers yields the result.
(2) By the same argument, a polynomial ringK[x] over a field is Noetherian.
More trivially, every field is Noetherian.
(3) Let X be an infinite set and K a field (in fact, any non-zero ring will do). The setR :=KX of all functions from X to K forms a ring with point-wise operations. For every subsetY ⊆X, the set
IY :={f ∈R|f vanishes onY}
is an ideal of R. Since there are infinite strictly descending chains of subsets ofX, there are also infinite strictly ascending chains of ideals in R. SoRis not Noetherian.
(4) The ringsZandK[x] considered above are not Artinian.
(5) Every field and every finite ring or module is Artinian.
(6) The ringKX, as defined in (3), is Artinian if and only ifX is a finite set.
(7) Let R :=K[x] be a polynomial ring over a field. Then S := R/(x2) is
Artinian.S is also Artinian as anR-module. /
The ring from Example2.2(3) is a rather pathological example of a non- Noetherian ring. In particular, it is not an integral domain. The following provides a less pathological counter example.
Example 2.3. LetS:=K[x, y] be the polynomial ring in two indeterminates over a fieldK. Consider the subalgebra
R:=K+S·x=K[x, xy, xy2, xy3, . . .].
It is shown in Exercise2.1thatR is not Noetherian. / The following proposition shows that the Noether property and the Artin property behave well with submodules and quotient modules.
Proposition 2.4 (Submodules and quotient modules). Let M be a module over a ringR, and letN ⊆M be a submodule. Then the following statements are equivalent.
(a) M is Noetherian.
(b) Both N and the quotient moduleM/N are Noetherian.
In particular, every quotient ring of a Noetherian ring is Noetherian.
All statements of this proposition hold with “Noetherian” replaced by “Ar- tinian”.
2.1 The Noether and Artin Property for Rings and Modules 35 Proof. First assume that M is Noetherian. It follows directly from Defini- tion 2.1 that N is Noetherian, too. To show that M/N is Noetherian, let U1, U2, . . .⊆M/Nbe an ascending chain of submodules. Withϕ:M →M/N the canonical epimorphism, set Mi := ϕ−1(Ui). This yields an ascending chain of submodules ofM. By hypothesis, there exists ann withMi =Mn for i≥n. Since ϕ(Mi) =Ui, it follows thatUi =Un for i≥n. So we have shown that (a) implies (b).
Now assume that (b) is satisfied. To show (a), let M1, M2, . . . ⊆ M be an ascending chain of submodules. We obtain an ascending chain ϕ(M1), ϕ(M2), . . .⊆M/N of submodules of M/N. Moreover, the intersections N∩ Mi⊆N yield an ascending chain of submodules ofN. By hypothesis, there exists annsuch that fori≥nwe haveϕ(Mi) =ϕ(Mn) andN∩Mi=N∩Mn. We claim that also Mi =Mn for alli≥n. Indeed, let m∈Mi. Then there exists anm0 ∈Mn withϕ(m) =ϕ(m0), so
m−m0 ∈N∩Mi =N∩Mn⊆Mn.
We concludem=m0+ (m−m0)∈Mn. So the equivalence of (a) and (b) is proved.
To show the statement on quotient rings, observe that the ideals of a quotient ring R/I are precisely the submodules of R/I viewed as an R- module.
To get the proof for the case of Artinian modules, replace every occur- rence of the word “ascending” in the above argument by “descending”, and exchange “Mi” and “Mn” in the proof ofMi=Mn. ut
We need the following definition to push the theory further.
Definition 2.5 (Ideal product). Let R be a ring, I ⊆R and ideal, and M an R-module.
(a) The product of I and M is defined to be the abelian group generated by all productsa·mof elements fromI and elements fromM. So
IM =nXn
i=1
aimi
n∈N, ai∈I, andmi∈Mo . ClearlyIM ⊆M is a submodule.
(b) An interesting special case is the case whereM =J is another ideal ofR.
Then the productIJ is called the ideal product. Clearly the formation of the ideal product is commutative and associative, and we have
IJ⊆I∩J and
√ IJ=
√ I∩J (check this!).
(c) Forn∈N0,In denotes the product ofn copies ofI, with I0:=R.
The following lemma gives a connection between ideal powers and radical ideals.
Lemma 2.6 (Ideal powers and radical ideals). LetRbe a ring andI, J⊆R ideals. If I is finitely generated, we have the equivalence
I⊆√
J ⇐⇒ there exists k∈N0 such that Ik⊆J.
Proof. We have I = (a1, . . . , an). Suppose that I ⊆ √
J. Then there exists m >0 withami ∈J fori= 1, . . . , n. Setk:=n·(m−1) + 1. We need to show that the product ofkarbitrary elements fromIlies inJ. So letx1, . . . , xk ∈I and write
xi=
n
X
j=1
ri,jaj with ri,j∈R.
When we multiply out the productx1· · ·xk, we find that every summand has some amj as a subproduct. Thereforex1· · ·xk ∈J. This shows thatIk⊆J.
The converse statement is clear (and does not require finite generation of
I). ut
Theorem2.8, which we start proving now, gives a comparison between the Noether property and the Artin property for rings. Readers who are mainly interested in the Noether property can continue with reading Section 2.2.
Theorem2.8will not be used before Chapter 7.
Lemma 2.7. Let R be a ring andm1, . . . ,mn∈Specmax(R)maximal ideals (which are not assumed to be distinct) such that for the ideal product we have
m1· · ·mn={0}.
ThenR is Artinian if and only if it is Noetherian. Moreover, Spec(R) ={m1, . . . ,mn}.
Proof. Setting
Ii:=m1· · ·mi, we get a chain
{0}=In⊆In−1⊆ · · · ⊆I2⊆I1⊆I0:=R
of ideals. Applying Proposition 2.4repeatedly, we see that R is Noetherian (Artinian) if and only if every quotient module Ii−1/Ii is Noetherian (Ar- tinian). Butmi·(Ii−1/Ii) = {0}, so Ii−1/Ii is a vector space over the field Ki := R/mi, and a subset of Ii−1/Ii is an R-submodule if and only if it is a Ki-subspace. So both the Noether and the Artin property forIi−1/Ii are equivalent to dimKi(Ii−1/Ii)<∞. This yields the claimed equivalence.
To prove the second claim, takeP ∈Spec(R). By hypothesis,m1· · ·mn⊆ P. From the primality ofP and the definition of the ideal product, we con- clude that there existsiwithmi⊆P, soP =mi. ut
2.1 The Noether and Artin Property for Rings and Modules 37 Theorem 2.8 (Artinian and Noetherian rings). Let R be a ring. Then the following statements are equivalent.
(a) Ris Artinian.
(b) Ris Noetherian and every prime ideal ofR is maximal.
Using the concept of dimension as defined in Definition 5.1, the condi- tion (b) in Theorem 2.8 can be rephrased as: “R is Noetherian and has dimension 0 or -1” (where -1 occurs if and only if R is the zero ring). We only prove the implication “(a) ⇒ (b)” here and postpone the proof of the converse to the end of Chapter3(see page52).
Proof of “(a)⇒ (b)”. Suppose thatRis Artinian. The first claim is thatR has only finitely many maximal ideals. Assume the contrary. Then there exist infinitely many pairwise distinct maximal idealsm1,m2,m3, . . .∈Specmax(R).
Setting Ii :=Ti
j=1mj yields a descending chain of ideals, so by hypothesis there exists nsuch that In+1 =In. This implies Tn
j=1mj ⊆mn+1, so there exists j≤n withmj =mn+1, a contradiction. We conclude that there exist finitely many maximal idealsm1, . . . ,mk. Setting
I:=m1· · ·mk,
we obtain a descending chain of idealsIi,i∈N0, so there existsn∈N0 with
Ii =In=:J for i≥n. (2.1)
By way of contradiction, assumeJ 6={0}. Then the set M:={J0⊆R|J0 is an ideal andJ0J 6={0}}
is non-empty. There exists a minimal element Jbin M, since otherwise M would contain an infinite, strictly descending chain of ideals. Pick an x∈Jb withxJ 6={0}. ThenJb= (x) by the minimality. Moreover, (2.1) implies that J2=J, so
(x)J·J = (x)J2= (x)J 6={0},
so (x)J = (x) again by the minimality of Jb. Therefore there exists y ∈ J withxy=x. By the definition ofJ,y lies in every maximal ideal ofR, and so y−1 lies in no maximal ideal. This means that y−1 is invertible, and (y−1)x= 0 implies x= 0. This contradicts xJ 6={0}. We conclude that J ={0}. So we can apply Lemma2.7and get thatRis Noetherian and that
every prime ideal is maximal. ut
Theorem 2.8 raises the question whether it is also true that every Ar- tinian module over a ring is Noetherian. This is answered in the negative by Exercise2.2.