In order to avoid giving basically the same definition twice, we define a Noetherian and an Artinian module simultaneously, presenting the conditions of an Artinian module in brackets, as our main concern will be the Noetherian case.
Definition. AnR-module is calledNoetherian(Artinian) if for every ascending (descend-ing) chain of submodules ofM,
M1⊆M2⊆M3⊆ · · · (M1⊇M2⊇M3⊇ · · ·) there exists a positive integerksuch thatMk =Mk+1=Mk+2=· · ·.
Example 3.2.1. We have shown that in the ringZevery ideal is principal, making any ascending chain of ideals ofZof the form
hn1i ⊆ hn2i ⊆ hn3i ⊆ · · ·
forn1, n2, n3, ...∈Z. Sincehnii ⊆ hni+1iimpliesni+1|ni, any ascending chain of ideals inZstarting withn1will have a finite number of distinct terms, makingZas aZ-module Noetherian. On the other hand, the descending chain
hni ⊇ hn2i ⊇ hn3i ⊇ · · · is infinite, showing thatZas aZ-module isnotArtinian.
Definition. A ringR is a Noetherian (Artinian) ring ifR regarded as anR-module is Noetherian (Artinian).
Now that we have defined what it means for both a module and a ring to be Noetherian (Artinian), we will present one of the main results for Noetherian (Artinian) modules with a proof for the Noetherian case, and then rewrite the result for Noetherian (Artinian) rings.
Theorem 3.2.2. ForManR-module, the following are equivalent:
i) M is Noetherian (Artinian).
ii) Every submodule (quotient module) ofMis finitely generated (cogenerated).
iii) Every nonempty setSof submodules ofM has a maximal (minimal) element.
Proof. (i) =⇒ (ii): LetM be a Noetherian module andN a submodule ofM, whereN is assumed not to be finitely generated. For a positive integerkletn1, ..., nk ∈N. Then (n1, ..., nk)6=N, and we choosenk+1 ∈N such thatnk+1∈/(n1, ..., nk). This gives an infinite ascending chain of submodules ofM
(n1)((n1, n2)(· · ·((n1, ..., nk)((n1, ..., nk+1)(· · ·, contradicting withM being Noetherian. HenceNis finitely generated.
(ii) =⇒ (iii): LetM be anR-module where every submodule is finitely generated, and letS be a nonempty set of submodules ofM. Then, if an elementN1 ∈ S is not maximal, it is contained in another submoduleN2∈S. IfShas no maximal elements, we obtain an infinite ascending chain of submodules
N1(N2(· · ·
ofM. LetN =N1∪N2∪ · · ·, and letx, y∈Nandr∈R. Thenx∈Niandy∈Njfor i, j∈ {1,2, ...}andi6=j. Now, since eitherNi⊆NjorNj⊆Ni, bothxandylie inNi orNj, hencex−yandrxlie in the same submodule. This again impliesx−y∈Nand rx∈N, makingN a submodule ofM. From(ii)N is finitely generated, i.e. there exist elementsa1, a2, ..., an∈Nsuch thatN = (a1, a2, ..., an). There exists a submoduleNk such that allal∈Nk forl= 1,2, ..., n. SinceNk ⊆NandNis the smallest submodule containing allal, we must have thatNk =N. ThenNk = Nk+1 = · · ·, contradicting withSnot having a maximal element, henceShas a maximal element.
(iii) =⇒ (i): Assume we have an ascending chain of submodules ofM M1⊆M2⊆M3⊆ · · ·.
By(iii)this chain has a maximal elementMk, implyingMk =Mk+1=· · ·. HenceM is Noetherian.
Theorem 3.2.3. LetRbe a ring. Then the following are equivalent:
i) Ris Noetherian (Artinian).
ii) ForIan ideal ofR, we have thatI(R/I)is finitely generated (cogenerated).
iii) Every nonempty setSof ideals ofRhas a maximal (minimal) element.
We see that Noetherian (Artinian) rings provides us with the useful property that every set of ideals of the ring has a maximal (minimal) element. The same holds for Noetherian (Artinian) modules and the set of their submodules, as we just proved. Obviously it is desirable to be able to determine whenever a ring or a module is Noetherian (Artinian).
From [1, section 19.2] we make the following remarks.
Remark. 1. IfRis an Artinian ring it is also Noetherian.
2. Every principal ideal ring is a Noetherian ring.
Theorem 3.2.4. Every submodule of a Noetherian (Artinian) module is Noetherian (Ar-tinian).
Proof. The result follows immediately from Theorem 3.2.2.
Example 3.2.5. ConsiderQ. In example 2.2.2 we saw thatQis a field, and so by example 2.3.2 the only ideals ofQareh0iandQitself. Clearly the chainQ⊇ h0ihas a minimal element, and soQis Artinian. Then we know thatQis also Noetherian.
Now look atZwhich is a subring ofQ, as we have seen earlier. In example 3.2.1 we saw thatZas aZ-module is Noetherian, but not Artinian. Then, from the definition of an Artinian ring we can conclude thatZisnotan Artinian ring, even though it is a subring of an Artinian ring.
We finish this chapter by defining Noetherian domains.
Definition. ANoetherian domainDis an integral domain which is Noetherian.
Example 3.2.6. We have shown earlier thatZis an integral domain, and that it is Noethe-rian. HenceZis a Noetherian domain.
In the following chapter we will define three more domains, namely Euclidean do-mains, principal ideal dodo-mains, and unique factorization domains.
Chapter 4
Euclidean domains, PIDs and UFDs
4.1 Norms
Leading up to defining Euclidean domains, principal ideal domains and unique factoriza-tion domains, we start by defining the norm of elements inZ[√
n].
Definition. The normof an elementx ∈ Z[√
n], where nis a squarefree integer and x= (a+b√
n)for somea, b∈Z, is the integer defined by N(x) =|x·x|=|(a+b√
n)(a−b√
n)|=|a2−nb2|.
Notice thatxrepresents theconjugateofx, and not the complex conjugate, asxis not a complex number whenn >0. Also, in the case whereb = 0, we havex∈Z, and the norm simply becomesN(x) =a2. If botha= 0andb = 0, that isx= 0, we obviously obtainN(x) = 0, but sincenis a squarefree integer the implication goes the other way as well. ForN(x) = 0we have
|a2−nb2|= 0,
implying a2 = nb2. Since n is squarefree it factorizes intok distinct primes, that is n=p1· · ·pk, wherepi6=pjfori6=jandi, j∈ {1, ..., k}. We write
a2=p1· · ·pk·b2.
Now, ifb = 0we geta= 0and sox= 0. Therefore, assume thatb 6= 0. Then we may write
a2
b2 =p1· · ·pk, which after taking the square root of each side yields
a b =√
p1· · ·pk.
This however, is a contradiction asa/b∈Q, but√ but first we present two results regarding the norm.
Theorem 4.1.1. Forx, y∈Z[√
n], wherenis a squarefree integer, we have that N(xy) =N(x)· N(y).
Now we look at the norm ofxandy, and multiply them together.
N(x)· N(y) =|a2−nb2| · |c2−nd2|
=|(a2−nb2)(c2−nd2)|
=|(ac)2−n(ad)2−n(bc)2+ (nbd)2|.
We see thatN(xy) =N(x)· N(y).
Theorem 4.1.2. Foru∈Z[√
n], andna squarefree integer, we have that uis a unit ⇐⇒ N(u) = 1. integers. For the converse we assume thatN(u) = 1. Then
N(u) =|(a+b√
Now we turn to our different types of domains, starting with Euclidean domains.