Summary
In this thesis we study ideals in Dedekind domains, which factorize uniquely into a prod- uct of prime ideals. Since not every Dedekind domain is a unique factorization domain, a general element in these domains does not necessarily factorize uniquely, so it is interest- ing that ideals has this property.
Norsk sammendrag:
I denne teksten studerer vi idealer i Dedekind-omr˚ader, som faktoriseres til et unikt pro- dukt av primidealer. I et Dedekind-omr˚ade er det ikke nødvendigvis slik at et generelt element faktoriseres unikt, s˚a det er interessant at idealer har denne egenskapen.
Preface
Before you lies the work that concludes my time as a student at the Department of Math- ematical Science and at the Teacher Education program at NTNU. I have found working with this thesis to be very educative, and it has allowed me to focus on the parts of mathe- matics that I have come to enjoy the most. Of all the different parts of my study program that I have been through, this semester has been the most rewarding.
I would like to thank my supervisor Petter Andreas Bergh for all his guidance and help during the months of writing. Thank you for helping me making this time much less stressful than it could have been.
Also, a big thanks to my friends here in Trondheim, who have made these years very memorable, and the studies that much easier. Finally, thank you to my family for always supporting me and building me up.
Simen Einmo Engdal Trondheim, 01.12.16
Table of Contents
1 Introduction 1
2 Integral domains and ideals 3
2.1 Commutative rings . . . 3
2.2 Integral domains . . . 4
2.3 Ideals . . . 5
2.4 Prime ideals . . . 6
3 Noetherian domains 11 3.1 Modules . . . 11
3.2 Noetherian and Artinian rings . . . 12
4 Euclidean domains, PIDs and UFDs 15 4.1 Norms . . . 15
4.2 Euclidean domains . . . 17
4.3 Principal ideal domains . . . 18
4.4 Unique factorization domains . . . 19
5 Algebraic number fields 25 5.1 Integral elements . . . 25
5.2 Integral closure . . . 29
5.3 The ring of integers . . . 30
6 Ideal factorization in Dedekind domains 37 6.1 Dedekind domains . . . 37
6.2 Ideals in Dedekind domains . . . 39
6.3 Unique factorization into prime ideals . . . 42
Bibliography 45
Chapter 1
Introduction
It is well known that inZ, the ring of integers, elements factorize uniquely into a product of prime numbers. If we extend this ring with an element of the form√
nwheren∈Z, we obtain a ring which does not necessarily have the property of its elements factorizing uniquely. However, these rings will be Dedekind domains for certain values of n, and in Dedekind domains nonzero proper ideals factorize uniquely into a product of nonzero prime ideals. The main objective of this thesis is to prove this fact. This was first done by the German mathematician Julius Wilhelm Richard Dedekind, in connection with his work on algebraic number theory.
To that end, we start by looking at integral domains along with the aforementioned ideals and prime ideals in Chapter 2. In Chapter 3 we present the notion of Noetherian domains, which is of importance as it is part of both the definition of Dedekind domains, and the proof of our main theorem. We move on to discuss additional types of domains in Chapter 4, namely Euclidean domains, principal ideal domains and unique factorization domains, before we define an algebraic number field in Chapter 5, and its associated ring of integers. In Chapter 6 we prove that this ring is in fact a Dedekind domain, and we give the proof of our main theorem.
Chapter 2
Integral domains and ideals
2.1 Commutative rings
Assuming the reader to be familiar with basic algebraic concepts and group theory, we start by defining a ring. It needs to be pointed out that all rings in this thesis will be rings with unity, i.e. rings containing a multiplicative identity element1.
Definition. AringRis a nonempty set with binary operations+and·, such that for all elementsa,b,c∈Rwe have the following:
i) (R,+)is an abelian group.
ii) Multiplication is associative:(ab)c=a(bc).
iii) The distributive laws hold:a(b+c) =ab+acand(a+b)c=ac+bc.
iv) 1∈Ris the multiplicative identity such that1·a=a·1 =a From here we immediately define a commutative ring.
Definition. Acommutative ringRis a ring in which multiplication is commutative, that isab=bafor alla,b∈R.
Example 2.1.1. The set of integersZis clearly a ring. As multiplication by integers is known to be commutative,Zis also a commutative ring.
In this thesis we are only interested in commutative rings, and from this point on every ring stated is a commutative ring. In order to simplify the method of concluding that a given set is in fact a ring, we give the definition of a subring.
Definition. Let (R,+,·) be a ring and S a nonempty subset of R. Then S is called a subringif (S,+,·) is itself a ring.
Example 2.1.2. Trivially,Zis a subring ofQ, the ring of rational numbers.
Theorem 2.1.3. A nonempty subsetSof a ringRis a subring if and only ifScontains the multiplicative identity1ofR, and for alla,b∈Swe havea−b∈Sandab∈S.
Example 2.1.4. LetR be a ring andR[x]the polynomial ring over R. LetS be a set consisting of polynomials of the form
a0+a2x2+· · ·+anxn
wheren= 0orn≥2andai∈R, i.e. polynomials overRfor which the coefficient ofx is zero. ThenSis a subset ofR[x], contains the multiplicative identity, and is closed under multiplication and subtraction, making it a subring ofR[x]by Theorem 2.1.3.
Next, we move on to a special class of rings which gives us a natural setting for study- ing divisibility.
2.2 Integral domains
As every ring will be a commutative ring, every integral domain will be a commutative integral domain, as stated in the following definition:
Definition. Anintegral domainDis a ring that has no divisors of zero, that is, ifab= 0 eithera= 0orb= 0.
Furthermore, if for everya∈Dwherea6= 0, there exitsb∈Dwithab = 1, every nonzero element inDhas an inverse. In this caseDis called afield. If a field is a subring of another field, it is called a subfield.
Theorem 2.2.1. Every subring of a field that contains the identity, is an integral domain.
Proof. LetR⊆Fbe a subring of a fieldF. Assume that forx, y∈Rwe havexy= 0.
Now, sincex, y∈F andRandFhas the same zero element, eitherx= 0ory= 0, and soRhas no divisors of zero. Since all our rings contain1,Ris an integral domain.
Note that an integral domain need not be a field, but from every integral domain we may construct a field by inverting all the nonzero elements, obtaining thequotient fieldof the integral domain.
Definition. LetDbe an integral domain. Thequotient fieldofD, denotedQuot(D)is defined as
Quot(D) ={a·b−1|a, b∈D, b6= 0}.
Example 2.2.2. In the ringQ, elements can be written as fractionsa/bwhereaandb6= 0 are integers. The additive inverse of such fractions are−a/b, and the multiplicative inverse isb/afora6= 0. It is clear thatQcontains no divisors of zero, and soQis a field.
In example 2.1.2 we stated thatZis subring ofQ, hence, by Theorem 2.2.1,Zis an integral domain. Also, we observe that
Quot(Z) ={a·b−1|a, b∈Z, b6= 0}=Q, soQis in fact the quotient field ofZ.
At this point we have already seen that Z is both a ring and an integral domain.
Throughout this thesis we will show several more properties ofZas we go along. Also, we will see the consequences of extendingZwith an element which is not part of the ring already. We will look at elements of the form√
n, wheren∈Z, and obtain the set Z[√
n] ={a+b√
n|a, b∈Z}.
Z[√
n]is an integral domain, as it is a subset ofC, which is a field. Then, sinceZ[√ n]is closed under subtraction and multiplication, and contains1, it is a subring ofC, thus an integral domain by Theorem 2.2.1.
In order for√
nnot to be inZalready, we considernto besquarefree, meaning that in the prime factorization ofn, no prime number occurs more than once. As an example ofnnotbeing squarefree, look atn= 4 = 2·2. Then
Z[√
4] =Z[2] =Z,
and we have not extendedZat all. Now, there are integersnthat arenotsquarefree for which√
nis not inZ, but in this caseZ[√
n]will always be a subring of a domainZ[√ m]
wheremis a squarefree integer. For example, in the case wheren = 24, which is not squarefree, we obtain
Z[√
24] =Z[2√
6]⊆Z[√ 6], andm= 6is squarefree. Also, note thatZ[√
n]will not be an extension ofZin the case wheren= 1, as
Z[√
1] =Z[1] =Z.
We now turn to a special class of subsets of rings, calledideals, which have the prop- erty of being closed under addition and multiplication by elements of the ring.
2.3 Ideals
Definition. A nonempty subsetIof a ringRis called anidealofRif i) a,b∈Iimpliesa−b∈I.
ii) a∈Iandr∈Rimplyra∈I.
Actually, the definition above gives only a left ideal, but recalling that all rings stated here are commutative, all left (and right) ideals are two-sided. We state that every ringR has at least two ideals, namely the trivial idealsRandh0i.
Definition. An idealIof a ringRis called aproper idealofRifI6=h1i.
This means that a proper ideal of a ring R is an idealI such thatI ( R, as h1i generates the whole ring.
Proposition 2.3.1. IfIis an ideal in a ringR, thenI=Rif and only if1∈I.
Proof. SupposeI=R. Every ringRcontains the identity element, hence1∈I.
Now assume1∈I, and letr∈R. AsIis an ideal we have1·r=r∈I, henceI=R.
Example 2.3.2. LetI6=h0ibe an ideal of a fieldF, and leta∈I. Thena∈F. We know thatra∈Ffor anyr∈F. SinceFis a field,ahas an inversea−1, and soa−1a= 1∈I.
With the identity element ofFinIwe must have thatI=F, makingh0iandFthe only ideals ofF.
To move on, we look at some additional examples of ideals.
Example 2.3.3. LetRbe a ring andr∈R. ThenrR =(ra|a∈R) is the ideal ofR generated byr. We then denotehrito be theprincipal idealofRgenerated byr.
Definition. Aprincipal ideal ringis a ringRin which every ideal is principal.
Example 2.3.4. We want to show thatZis a principal ideal ring. LetIbe an ideal ofZ. IfI = {0}thenI =h0iis a principal ideal, so we assumeI 6= {0}. Then we have an elementa∈I, wherea6= 0. Since also−a∈Z, we may supposea >0, and henceIhas at least one positive integer. Now letndenote the least positive integer inI. By dividing abynwe can expressaas
a=nq+r
for someq, r ∈ Z, where0 ≤ r < n. Sinceaandnare elements of the idealI, so is r=a−nq. To avoid a contradiction with the fact thatnis the least positive integer inI, we must have thatr= 0, makinga=nq. HenceI=hni=nZ. This shows that every ideal inZis principal, makingZis a principal ideal ring.
In chapter4we will revisit the above example, and see thatZis an example of what we call a principal ideal domain. The next section of this chapter, we dedicate to prime ideals, which play a central role in this thesis.
2.4 Prime ideals
Definition. An idealP (Rof a ringRis called aprime idealifa,b∈Randab∈P impliesa∈P orb∈P.
Example 2.4.1. For each prime integerp, the idealhpiinZis a prime ideal.
Actually, the above holdsonlyfor prime integers, and for each n ∈ Znot a prime integer, we can show thathniis not a prime ideal.
Example 2.4.2. For8Zwe have2,4∈Zand2·4∈8Z, but2,4∈/8Z, so8Zis a proper ideal ofZ, but not a prime ideal.
Going back to integral domains, it is now clear thath0iis a prime ideal in any integral domainD. In fact, ifh0iis a prime ideal in a ringD, and we leta,b∈Dbe elements such thatab∈ h0i, then eithera∈ h0iorb∈ h0i. Nowab= 0which implies thata= 0 orb= 0, henceDis an integral domain. We end up with the equivalence:
A ringDis an integral domain ⇐⇒ h0iis a prime ideal inD.
Closely related to prime ideals we have maximal ideals.
Definition. A proper idealM of a ringRis called amaximal idealif for an idealIofR such thatM⊆I⊆R, eitherI=M orI=R.
Now that we have defined both a prime ideal and a maximal ideal, we will move on with some results regarding the two.
Theorem 2.4.3. LetIbe an ideal of the ringR. Then we have that R/Iis a field ⇐⇒ Iis maximal.
Proof. SupposeR/Iis a field and thatJ is an ideal ofRwith I(J⊆R.
Thus there existsb ∈J such thatb /∈I. Thenb+I is a nonzero element ofR/Iand therefore, asR/Iis a field, there exists an elementc+I∈R/Isuch that
(b+I)(c+I) =bc+I= 1 +I, and so
bc−1∈I(J. Sinceb∈J andc∈Rwe have
bc∈J.
Hence
1 =bc−(bc−1)∈J,
so thatJ=h1i=R. ThusIis a maximal ideal. Conversely, letIbe a maximal ideal, and assumeR/I is not a field. Then there exists an elementr+I ∈R/I not a unit, hence 1 +I /∈ hri+I, which implieshri(R. Sincer+I6= 0 +I,r /∈Iand then
I(hri+I(R,
which contradicts withIbeing maximal. HenceR/Iis a field.
For our next result we need to defineprime elements. We are familiar with the fact that a prime number can not be factorized into a product of two integers other than itself and1.
More generally, for a primepwherep=ab,aorbmust be a unit. InZthe only units are
±1, and the prime elements are exactly the prime numbers. There is a second property of prime elements though, stating that ifp|ab, then eitherp|aorp|b. These two properties of a prime element are equivalent inZ, but not in general. However, the second one can be shown to always imply the first one, and so we define prime elements as follows.
Definition. A nonzero elementp∈Ris aprimein the ringRifpis not a unit, and ifp|ab for somea,b∈R, then eitherp|aorp|b.
Theorem 2.4.4. LetRbe a ring. Letp∈Rbe such thatp6= 0and not a unit. Then hpiis a prime ideal ofR ⇐⇒ pis a prime inR.
Proof. Lethpibe a prime ideal ofR, and leta, b∈Rbe such thatp|ab, i.e.ab∈ hpi. As hpiis a prime ideal,a∈ hpiorb∈ hpi, leading top|aorp|b. Hencepis prime inR.
Conversely, assume thatpis a prime inR, and leta∈Randb∈Rsuch thatab∈ hpi.
Then there existsc∈Rsuch thatab=pc, and thenp|ac. Sincepis a prime we have that p|aorp|b. Suppose nowp|a. Then there existsd∈Rsuch thata=pdand soa∈ hpi, hencehpiis a prime ideal. In the same way, ifp|bthere exists an elemente∈Rsuch that b=pe, and sob∈ hpi, makinghpia prime ideal in this case as well.
Theorem 2.4.5. For a ringRand an idealIofRwe have that R/Iis an integral domain ⇐⇒ Iis a prime ideal.
Proof. AssumeR/Ito be an integral domain. Leta, b∈Rsuch thatab∈I. Then (a+I)(b+I) =ab+I= 0 +I
is the zero element ofR/I. Since an integral domain has no zero divisors, we must have a+I= 0 +Iorb+I= 0 +I. This means thata∈Iorb∈I, soIis a prime ideal.
Conversely, suppose thatIis a prime ideal ofR. AsIthen is a proper ideal ofR,R/I is a ring with identity1 +I. Now leta+I∈R/Iandb+I∈R/Isuch that
(a+I)(b+I) = 0 +I.
Thenab+I=Iso thatab∈I. SinceIis prime eithera∈Iorb∈I, that is,a+I= 0+I orb+I= 0 +I, henceR/Ihas no zero divisors, and soR/Iis an integral domain.
Theorem 2.4.6. For a ringRa maximal idealIis also a prime ideal.
Proof. ForIa maximal ideal ofR, we have by Theorem 2.4.3 thatR/Iis a field, which is always an integral domain. By Theorem 2.4.5Iis then also a prime ideal ofR.
Before giving two additional results, we define multiplication of ideals.
Definition. LetIandJbe ideals in a ringR. Then theproduct ofIandJ, denotedIJis defined by
IJ={x∈R|x=i1j1+· · ·+irjrfor somer∈N, somei1, ..., ir∈I, and somej1, ..., jr∈J}.
IfI =hiiandJ =hjiare principal ideals, thenIJ =hiji. In addition, we state the following properties for idealsI,J andKof a ringR:
1. IJis itself an ideal ofR.
2. IJ=J I. 3. (IJ)K=I(J K).
4. Ih0i=h0i.
5. Ih1i=I.
We end this chapter with our two last results concerning prime ideals.
Theorem 2.4.7. LetPbe a proper ideal of a ringR. Then we have that
P is prime ⇐⇒ for idealsA,BofRsatisfyingAB⊆P, eitherA⊆P orB⊆P.
Proof. AssumePto be a proper ideal ofRmeeting the requirement of idealsA,BofR AB⊆P⇒A⊆P orB⊆P.
Leta,b∈Rsuch thatab∈P. Now letA=haiandB=hbi, makingAB=habi ⊆P.
From here we obtain thathai ⊆Porhbi ⊆P, and soa∈P orb∈P, makingP a prime ideal. For the converse, letP not meet the requirement above. Then there exist idealsA andBofR, such thatA*P,B*P andAB⊆P. Next, leta∈A,a /∈Pandb∈B, b /∈ P. Thenab ∈ AB ⊆ P, buta /∈ P,b /∈ P, so P is not a prime ideal, and this completes the proof.
Theorem 2.4.8. LetDandEbe rings whereD⊆E. LetP be a prime ideal ofE. Then P∩Dis a prime ideal ofD.
Proof. First we must verify thatP∩Dis in fact an ideal ofD. For elementsa, b∈P∩D we havea, b ∈P anda, b ∈ D. Now, sinceP is an ideal,a+b ∈ P anda+b ∈ D.
Hencea+b ∈P∩D. Now leta∈P∩Dandd∈D. Then, sinceP is an ideal ofE, a∈Pandd∈D,da∈P, and asa∈Dandd∈D,da∈Das well, sinceDis closed under multiplication. Henceda∈P∩D, andP∩Dis an ideal ofD.
Now it only remains to show thatP∩Dis a prime ideal. Leta, b∈Dandab∈P∩D.
Thena, b∈Eandab∈P. SincePis a prime ideal ofE,a∈Porb∈P, showing that P∩Dis a prime ideal, and thus completing the proof.
Chapter 3
Noetherian domains
In this chapter we will defineNoetherian domains, which will reappear in our definition of a Dedekind domain in the final chapter. We start by defining modules, and see how they give rise to Noetherian rings.
3.1 Modules
Definition. LetRbe a ring,M an additive abelian group and(r, m)7→rma mapping of R×M 7→M such that
i) r(m1+m2) =rm1+rm2, ii) (r1+r2)m=r1m+r2m, iii) (r1r2)m=r1(r2m), iv) 1m=m,
for allr, r1, r2∈Randm, m1, m2∈M. ThenM is called a leftR-module.
Again, considering only commutative rings saves us the trouble of distinguishing be- tween left and right, and we consider a left and right R-module to be the same thing, simply denoting themR-modules.
Example 3.1.1. A ringRbecomes itself anR-module by definingamfora, m∈Rto be the product ofaandmas elements of the ringR.
Definition. For a ringRand anR-moduleM, a subgroupN ofM is called asubmodule ofM ifrn∈Nfor allr∈Randn∈N.
Example 3.1.2. If, like in the previous example, the ringRis considered anR-module, the submodules ofRare the ideals ofR.
Before moving on to Noetherian and Artinian modules, we define afinitely generated module.
Definition. AnR-moduleM isfinitely generatedifM is generated by some finite set of elements ofM.
This means that in order to be a finitely generatedR-module,M need to have finitely many elementsx1, ..., xn ∈ M such that eachx∈ M can be expressed as
n
P
i=1
rixiwith coefficientsri∈R.
3.2 Noetherian and Artinian rings
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 domains, 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√
p1· · ·pk ∈/Q. Hence the only conclu- sion is thata=b= 0, makingx= 0. This shows that
N(x) = 0 ⇐⇒ x= 0.
We will make use of the norm later, when we look at elements ofZ[√
n]being irreducible, 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).
Proof. Letx= (a+b√
n)andy= (c+d√
n), wherea, b, c, d∈Z. Then we obtain xy= (a+b√
n)(c+d√
n) = (ac+bdn) + (ad+bc)√ n, so that
xy= (ac+bdn)−(ad+bc)√ n.
The norm then becomes
N(xy) =|xy·xy|=|(ac+bdn)2−n(ad+bc)2|
=|(ac)2+n(2abcd) + (nbd)2−n(ad)2−n(2abcd)−n(bc)2|
=|(ac)2−n(ad)2−n(bc)2+ (nbd)2|.
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.
Proof. Letu = (a+b√
n) ∈ Z[√
n]and assume uis a unit. Then uhas an inverse u−1 ∈Z[√
n]such thatuu−1 = 1. This means thatN(1) = N(uu−1), and we obtain N(uu−1) = N(u)· N(u−1)from Theorem 4.1.1. Now, sinceN(1) = 12 = 1, also N(u)·N(u−1) = 1, which impliesN(u) =N(u−1) = 1, since they are both nonnegative integers. For the converse we assume thatN(u) = 1. Then
N(u) =|(a+b√
n)(a−b√
n)|= 1.
This implies that (a+b√
n)(a−b√
n) = ±1, and so±(a−b√
n) is the inverse of u= (a+b√
n). Henceuis a unit.
Example 4.1.3. Look atZ[√
10]and consider the elementx= (19 + 6√
10). Then the norm becomesN(x) =|192−10·62|=|361−360|= 1, and we conclude thatxis a unit inZ[√
10].
Now we turn to our different types of domains, starting with Euclidean domains.
4.2 Euclidean domains
Definition. An integral domainEis called aEuclidean domainif there exists a function φ:E→Zsuch that:
i) for alla, b∈E∗ =E\ {0}, we haveφ(ab)≥φ(a), and
ii) for each pair of elementsa, b∈E,b6= 0, there existq, r∈Esuch thata=bq+r andφ(r)< φ(q).
The functionφin the the definition of a Euclidean domain is central. Asφsatisfies the axioms in the definition, it is called theEuclidean function, but note that it is not part of the Euclidean domain itself. Actually, a single integral domain may be a Euclidean domain given several different Euclidean functions, but as we see, we only demand that there exists at least one. Next we present two examples.
Example 4.2.1. Zis a Euclidean domain forφ(a) = |a|, wherea ∈ Z. It is clear that φ(ab) ≥φ(a)when bothaandbare nonzero integers. Also, for any two integersaand b whereb is different from zero, it is known that the ordinary division algorithm yields integersqandrsatisfying (ii) in the definition of a Euclidean domain.
Example 4.2.2. We want to show thatZ[√
2]is a Euclidean domain given the norm defined in the previous section,
N(a+b√
2) =|a2−2b2|.
Letx, y∈Z[√
2], wherex=a1+b1
√
2andy =a2+b2
√
2 6= 0. Then,y 6= 0implies N(y)6= 0, and so1≤ N(y). Also
N(x)≤ N(x)N(y) =N(xy), and so (i) is verified. Next we want to verify (ii). Note that inQ[√
2]we have x
y =c1+c2
√ 2 where
c1=a1b1−2a2b2
b21−2b22 , c2=a2b1−a1b2
b21−2b22 .
Now letq1be the integer closest toc1, andq2the integer closest toc1, i.e.|c1−q1| ≤1/2 and|c2−q2| ≤1/2. Next, letr=q1+q2
√2. Certainlyr∈Z[√ 2]. Let s= (c1−q1) + (c2−q2)√
2.
Then we have
s= x y −r
so thatsy =x−r. Denotesy =u, and we obtainx=ry+u, as required of (ii) in the definition. We now need to show thatN(u)<N(y). Note that by the triangle inequality we have
N(s) =|(c1−q1)2−2(c2−q2)2| ≤ |(c1−q1)2|+| −2(c2−q2)2|.
Thus we have that
N(s)≤(c1−q1)2+ 2(c2−q2)2≤(1/2)2+ 2(1/2)2= 3/4,
and in particular N(u) ≤ 34N(y), making (ii) satisfied. HenceZ[√
2]is a Euclidean domain with respect to the norm.
Next we look at principal ideal domains, or PIDs. We will also see the relation between Euclidean domains and PIDs, and later also unique factorization domains.
4.3 Principal ideal domains
In section2.4we looked at principal ideal rings. Now, similarly we will define a principal ideal domain, denoted PID, and present several results that the property of PIDs provides us.
Definition. An integral domainD is called aprincipal ideal domain, denoted PID, if every ideal inDis principal.
Now, recalling our definition of a Noetherian domain and how finitely generated ideals relates to Noetherian rings provides us with the following result.
Theorem 4.3.1. Every PID is a Noetherian domain.
Proof. LetD be a PID, making every ideal ofD principal. Then every ideal is finitely generated, and so, by Theorem 3.2.3,Dis Noetherian.
Note that the converse of Theorem 4.3.1 is not true, as Noetherian domains may con- tain ideals generated by more than one element. Next, we look at an example of a PID.
Example 4.3.2. Recall that we in example 2.3.4 showed that inZ, all ideals are of the formnZ, which are all principal, makingZa principal ideal ring. In example 2.2.2 we concluded thatZis an integral domain. HenceZis a PID.
We now revisit maximal and prime ideals, and state the following result.
Theorem 4.3.3. ForDa PID, and a proper idealIofD, the following holds:
Iis a maximal ideal ⇐⇒ Iis a prime ideal.
Proof. The right implication is Theorem 2.4.6. Now for the left implication assumeIis a prime ideal inDthat is not maximal, meaning there exists an idealJofDsuch that
I(J (D.
SinceIandJ are both ideals of a PID, we have thatI =haiandJ =hbi, fora, b∈D both nonzero. Sincehai(hbi, there must exist an elementr∈Dsuch thatrb=a∈I.
Now, fromI being a prime ideal we must have that eitherr ∈Iorb∈I. Forb∈ Iwe obtain the chain
hbi=J ⊆I(J
which does not hold, and we must have r ∈ I. Thenr = safor somes ∈ D, hence a= (sa)b=a(sb), and sincea6= 0, we getsb= 1. This makesba unit, i.e.
hbi=J =D,
contradictingJ (Dand so the assumption ofInot being maximal. We conclude thatI must be maximal.
It is also possible to determine a PID by knowing a domain is Euclidean.
Theorem 4.3.4. Every Euclidean domain is a PID.
Proof. LetEbe a Euclidean domain, and letIbe an ideal inE. IfI ={0}thenI=h0i andIis principal. AssumeI6={0}. For alla∈Iwe have that1|a, and soφ(a)≥φ(1).
Then the setS ={φ(a)|a∈ I, a 6= 0}is a nonempty set withφ(1)as a lower bound.
Then there exists an elementb ∈ I such thatφ(b)is the smallest element in this set. If a ∈ I, we havea =qb+rforq, r ∈ E andφ(r) < φ(b). Butr = a−qb ∈ Iand φ(r)≥φ(b)by the choice ofb, so we must haver= 0, soa∈ hbi, henceI=hbi. Every ideal is principal and soEis a PID.
We dedicate the next and final section of this chapter to unique factorization domains, and we will see how these relate to our other types of domains.
4.4 Unique factorization domains
In a unique factorization domain, or UFD, elements can be factorized in a unique way.
In order to define what we mean by unique factorization precisely, we start by looking at elements in a domain being irreducible.
Definition. ForRa ring, a nonzero nonunit elementr ∈ Risirreducibleif forr= ab wherea, b∈R, eitheraorbis a unit.
Theorem 4.4.1. In an integral domainD, every prime element is irreducible.
Proof. Letp∈Dbe a prime element, and letp=ab, wherea, b∈D. Nowab=p·1, and sop|ab. Sincepis prime we must have thatp|aorp|b, i.e. a/p ∈ D orb/p ∈ D.
Since1 = (a/p)bor1 =a(b/p), we conclude that eitheraorbis a unit ofD, makingp an irreducible element ofD.
In a general ring, an element with the property of being irreducible is the equivalent of an integer in Z being prime. In the same way that every (nonzero nonunit) integer has a unique prime factorization, every (nonzero nonunit) element in a UFD has a unique factorization into irreducible elements.
Note that an element which is irreducible in one ring, may be reducible in another. It is therefore essential to specify in which ring the factorization of an element is to be carried out. Before giving the precise definition of a UFD, we take a closer look at irreducible elements.
Example 4.4.2. Look at the integral domainZ[√
−5], where an elementris of the form r={a+b√
−5|a, b∈Z}.
The norm is defined as earlier, that is
N(r) =a2+ 5b2.
From the properties of the norm we get that the only units arer=±1. Now consider the element9∈Z[√
−5], and look at the factorization 9 = (2 +√
−5)(2−√
−5).
We want to check that(2 +√
−5)and(2−√
−5)are irreducible factors. To reach our objective, let(2 +√
−5) =rs. Then we obtain N(2 +√
−5) =N(r)N(s),
which gives9 =N(r)N(s). Now we must have thatN(r)orN(s), let us chooseN(r), is equal to1,3or9. ForN(r) = 3we get
a2+ 5b2= 3,
which is not solvable for integersaandb. This leaves us with the two situations where {N(r) = 1,N(s) = 9}or{N(r) = 9,N(s) = 1}. In either case(2 +√
−5)is a product of two elements where one of them is a unit, hence(2 +√
−5)is irreducible. Doing the corresponding calculations for(2−√
−5)shows that it also is an irreducible factor.
While we for an integral domain showed that every prime element is also irreducible, we can prove the converse implication for a PID.
Theorem 4.4.3. An irreducible element in a principal ideal domain is always prime.
Proof. LetR be a PID, and considerp ∈ R to be an irreducible element withp|ab, for a, b∈R. Look athpi+hai. This is an ideal inR, so there exists an elementc∈Rwith
hpi+hai=hci.
Thenp∈ hci, i.e.p=cdfor somed∈R. Sincepis irreducible, eithercordis a unit. If cis a unit, thenhci=R, so
hpi+hai=R.
Then, for somex, y∈Rwe have that
1 =px+ay, and so
b=pbx+aby.
Sincep|ab, we get thatp|b. Now, ifdis a unit, sincep=cd, we havehpi=hci, so hpi+hai=hpi.
Thena∈ hpiand sop|a. We conclude thatpis a prime element.
Now we turn to unique factorization domains.
Definition. An integral domainRis called aunique factorization domainif the following hold:
i) Fora∈R\ {0}not a unit,a=p1p2· · ·pmfor some irreducible elementspi∈R.
ii) Ifp1, ..., pmandq1, ..., qnare irreducible elements inRandp1p2· · ·pm=q1q2· · ·qn, thenm=nand for all1≤i≤nthere exists a unitui∈Rwithpi =uiqi.
Remark. Elements of the formpi =uiqi, as in the definition of a UFD, are calledasso- ciates. These are elements that differ by multiplication of a unit. This means that since piandqi are associates, denotedpi ∼qi, we have thatpi|qiandqi|pi. In a commutative integral domain the converse is also true.
What the definition tells us is, as mentioned before, that every nonzero nonunit element in a UFD factorizesuniquelyinto irreducible elements. With that in mind we revisit our previous example.
Example 4.4.4. We have seen that the element9∈ Z[√
−5]factorizes into a product of irreducible factors, namely
9 = (2 +√
−5)(2−√
−5).
But we also have that9 = 3·3. As before, we now let3 =rs, wherer, s∈Z[√
−5]. We obtain
N(3) =N(r)N(s),
which gives9 =N(r)N(s), the same as in the previous example. Then we know that3is irreducible. This means that9can be factorized into two products of different irreducible factors, namely
9 = 3·3 = (2 +√
−5)(2−√
−5).
Since the only units inZ[√
−5]are1and−1, we deduce that the factors are not associates, and so the factorization from before is not unique. We conclude thatZ[√
−5]isnota UFD.
Before we eventually give an example of a ring that is in fact a UFD, we give a result connecting PIDs and UFDs.
Theorem 4.4.5. Every principal ideal domain is a unique factorization domain.
Proof. LetRbe a a PID. Then we know from Theorem 4.3.1 that Ris Noetherian, and so it does not have any infinite properly ascending chain of ideals. Now leta 6= 0be an element ofR that is not a unit. In order forRto be a UFD, amust factorize uniquely into a finite product of irreducible elements. Assumeais not irreducible, that is, it can be written asa=a1bwhere neithera1norbare units. Assume now thata1is not a product of irreducible elements, and thata1 = a2c where we leta2 be a product of reducible elements. Repeating this process gives an ascending chain of ideals
hai(ha1i(ha2i(· · ·
which in the case ofanot being a finite product of irreducible elements will be infinite.
But withRbeing Noetherian we deduce that the chain must be finite, henceais a finite product of irreducible elements.
Next we need to show that this product is unique. Assume there exists a nonzero ele- menta∈Rnot a unit, that factorizes into two different products of irreducible elements.
Suppose
a=p1p2· · ·pmanda=q1q2· · ·qn
wherepifori= 1, ..., mandqjforj= 1, ..., nare all irreducible inR, andn≥m. Then p1divides the productq1· · ·qn. Sincep1is irreducible it is also prime by Theorem 4.4.3, and sop1dividesqjfor somej. Without loss of generality we may supposep1|q1. Then, sincep1andq1are both irreducibles,q1=u1p1for some unitu1ofR. Thus
p1p2· · ·pm=u1p1q2· · ·qn
and
p2· · ·pm=u1q2· · ·qn. By continuing this process we reach
1 =u1u2· · ·umqm+1· · ·qn.
Asqjis not a unit for anyjwe havem=n, andp1, ..., pmare associates ofq1, ..., qnin some order. This contradicts with the assumption thatahas two different factorizations, and we may conclude that a factorizes into a finite and unique product of irreducible elements, making every PID a UFD, and thus the proof is complete.
Next we give the relation between prime elements and irreducible elements in a UFD, just as we did for integral domains and PIDs.
Theorem 4.4.6. For an elementp∈R, whereRis a UFD, we have that pis irreducible ⇐⇒ pis prime.
Proof. The left implication follows from Theorem 4.4.1. For the right implication let p ∈ R be an irreducible element and supposep|ab for ab ∈ R. Then there exists an elementc∈Rsuch thatab=pc. SinceRis a UFD, we have that
a=p1· · ·pk, b=q1· · ·qm, c=r1· · ·rn,
wherep1· · ·pk,q1· · ·qmandr1· · ·rn are irreducible elements ofR, not necessarily distinct. We obtain
(p1· · ·pk)(q1· · ·qm) =p(r1· · ·rn).
Now, sinceRis a UFD,pmust be an associate of one of thepi’s orqj’s, implying thatp|a orp|b. Hencepis prime.
For the final result of this section, we combine Theorem 4.3.4 and Theorem 4.4.5 to obtain the following Corollary.
Corollary 4.4.7. Every Euclidean domain is a unique factorization domain.
In light of this corollary, we summarize as follows for a ringR:
REuclidean domain⇒Rprincipal ideal domain⇒Runique factorization domain.
Or as inclusions:
n
Euclidean domainso (
n PIDso
( n
UFDso . We note that the inclusions are proper, and look at an example.
Example 4.4.8. LetR be a polynomial ring over a fieldF in variablesxandy, that is R = F[x, y]. As shown in [1, Theorem 4.3, page 222-223],Ris then a UFD. Consider the idealI = hxi+hyiof R. I can not be of the formhf(x, y)i for any polynomial f(x, y)∈Rsince
hxi+hyi=hf(x, y)i ⇒x=cf(x, y), y=df(x, y)
for some nonzero elementsc, d∈F. This gives x c =y
d,
and sodx−cy= 0, which can not be the case asxandyare independent variables overF.
HenceR=F[x, y]is not a PID. This stresses the fact that a UFD need not be a PID. Also, in [2] there are given several PIDs that are not Euclidean domains, for exampleZ[√
−19].
We finish this chapter with two additional examples regarding UFDs.
Example 4.4.9. In section 4.2 we concluded thatZ[√
2]is a Euclidean domain. Then, by Corollary 4.4.7 it follows immediately thatZ[√
2]is also a UFD. We can emphasize this fact by taking associates into account. Look at the element(8−3√
2)∈Z[√
2]. We may factorize it into two products as follows:
(8−3√
2) = (5 +√
2)(2−√
2) = (11−7√
2)(2 +√ 2).
By the same procedure as in the previous examples in this section, we can use the norm to show that none of these four factors are units inZ[√
2], and that all of them are irreducible.
Then, as we knowZ[√
2]is a UFD, we must have that the two products differ only by a unit, in order to satisfy the second condition of the definition of a UFD. In other words, the factors have to be associates. Look atu= (3 + 2√
2)∈Z[√
2]. We have that N(u) = 32−2·22= 1,
henceuis a unit ofZ[√
2]. By usinguwe can verify that the factors are associates.
(2−√
2)u= (2 +√
2)⇒(2−√
2)∼(2 +√ 2), (11−7√
2)u= (5 +√
2)⇒(11−7√
2)∼(5 +√ 2), and so the conditions of a UFD are satisfied for(8−3√
2).
Example 4.4.10. Consider the integral domainZ[√
−p] ={a+b√
−p|a, b∈Z}where pis a prime number. Using the norm we can verify that its only units are1and−1, and that both elements(1 +√
−p)and(1−√
−p)are irreducible inZ[√
−p]. We can now show that this domain isnota UFD whenpis an odd prime number, that isp >2. Look at the element2∈Z[√
−p]. We can show that neither(1 +√
−p)nor(1−√
−p)divides 2. Assume the contrary, that is(1 +√
−p)|2, which gives 2 = (a+b√
−p)(1 +√
−p).
Calculating the norm on each side of the equation yields 4 = (a2+b2p)(1 +p).
By inspection we see that the only solutions to the last equation are{a = b =p = 1}, {a= 2, b=p= 0}and{a= 1, b= 0, p= 3}. Bothp= 1andp= 0contradicts withp being a prime number, and from the last solution we obtain
2 = 1 +√
−3 which is not true. Hence (1 +√
−p) - 2. By the same procedure we conclude that (1−√
−p)- 2. Now, the element(1 +p) ∈Z[√
−p]can be factorized into irreducible elements as
1 +p= (1 +√
−p)(1−√
−p),
and sincepis an odd prime, 1 +phas to be even, i.e. 2|(p+ 1). Then2is a factor of (p+ 1). As(1±√
−p)-2they are not associates, and so(1 +p)does not have a unique factorization, henceZ[√
−p]is not a UFD whenpis an odd prime.
Chapter 5
Algebraic number fields
In this chapter we will study algebraic number fields, focusing on the set of algebraic integers contained in these fields, and their properties. More specifically, we will look at the algebraic integers in a finite extension field ofQ. If we name this extension fieldK, the algebraic integers form a subring ofK, which we will define as the ring of integers of K. To that end, we start by defining the termintegral over a domain.
5.1 Integral elements
Definition. LetAandB be two integral domains such thatA ⊆ B. Then the element b∈Bisintegral overAif there areai∈Aandn≥1such that
bn+an−1bn−1+· · ·+a1b+a0= 0.
That is,bis a root of a monic polynomial with all its coefficients inA.
In the special case whereA = ZandB = C, makingb ∈ Ca complex number, b is called analgebraic integer. As a simple example of this case, letb = √
2. As√ 2is a solution to the equationx2−2 = 0, which is monic and has both its coefficients inZ,
√2is an algebraic integer. Leading up to the definition of algebraic number fields, we will later look at the case whereAis assumed to be a field, makingbnot only integral overA, but also what we will callalgebraic overA. For now, we focus on elements being integral over a domain, and prove the following results.
Theorem 5.1.1. LetA⊆B⊆Cbe a tower of integral domains. Ifc∈Cis integral over Athencis integral overB.
Proof. Sincec∈Cis integral overAthere exists a polynomial cn+an−1cn−1+· · ·+a1c+a0= 0
whereai ∈Afori= 0,1, ..., n−1. Now, sinceA⊆Bwe have thatai ∈B, makingc integral overB.
Before our next theorem we note thatA[b]denotes the polynomial ring inboverA, that is, the set of polynomials of the form
anbn+ab−1bn−1+· · ·+a1b+a0
whereai ∈Afori= 0, ..., n.
Theorem 5.1.2. For integral domainsAandBwhereA⊆Bandb∈B, we have that bis integral overA ⇐⇒ A[b]is a finitely generatedA-module.
Proof. Assume thatbis integral overA. Then we have that bn−an−1bn−1− · · · −a1b−a0= 0 forai∈Awherei= 0,1, ..., n−1. From here we obtain
bn=an−1bn−1+an−2bn−2+· · ·+a1b+a0, bn+1=an−1bn+an−2bn−1+· · ·+a1b2+a0b.
Now,
bn∈Abn−1+Abn−2+· · ·+Ab+A while
bn+1∈Abn+Abn−1+· · ·+Ab2+Ab⊆Abn−1+· · ·Ab+A.
From here we obtain by induction onn, that for all integersk≥0we have bk∈Abn−1+· · ·+Ab+A.
HenceA[b]is a finitely generatedA-module, as it is generated bybn−1+· · ·+b+ 1.
For the converse, assume thatA[b]is a finitely generatedA-module. Then we have A[b] =Au1+· · ·+Aun
forui ∈A[b]wherei= 1,2, ..., n, which can not all be zero. We have thatbui ∈A[b], implying that there existaij ∈Asuch that
bu1=a11u1+· · ·+a1nun, ...
bun=an1u1+· · ·+annun. We rewrite the equations and obtain for unknownsx1, ..., xn
(b−a11)x1−a12x2− · · · −a1nxn= 0,
−a21x1+ (b−a22)x2− · · · −a2nxn = 0, ...
−an1x1−an2x2− · · ·+ (b−ann)xn= 0.
Now, from linear algebra we know that this system of equations has a solution if and only if its coefficient matrix is not invertible, meaning its determinant is equal to zero, i.e.
b−a11 −a12 · · · −a1n
−a21 b−a22 · · · −a2n
... ... ...
−an1 −an2 · · · b−ann
= 0.
By computation of the determinant we obtain an equation bn+an−1bn−1+· · ·+a1b+a0= 0 for somea0, a1, ..., an−1∈A, makingbintegral overA.
The proof of the next theorem follows the last one closely, and we abstain from show- ing all the details.
Theorem 5.1.3. LetAandBbe two integral domains such thatA ⊆B andb ∈ B. If there exists an integral domainCsuch that
A[b]⊆C⊆B
and C is a finitely generatedA-module, thenbis integral over Aand A[b] is a finitely generatedA-module.
Proof. SinceCis a finitely generatedA-module we have thatC =Ac1+· · ·+Acn for ci ∈ Cnonzero, wherei = 1,2, ..., n. We haveb ∈ A[b]andA[b] ⊆ C, which gives b ∈ C. SinceC is an integral domain we havebci ∈ C. Then, foraij ∈ Awe obtain a linear system as the one in the proof of Theorem 5.1.2. Following the same procedure, computing the determinant of the coefficient matrix, we reach the conclusion that b is integral overA, makingA[b]a finitely generatedA-module by Theorem 5.1.2.
Theorem 5.1.4. Let A ⊆ B ⊆ C be a tower of integral domains. If B is a finitely generatedA-module andCa finitely generatedB-module, thenCis a finitely generated A-module.
Proof. By assumption we have thatB =Ab1+· · ·+AbmandC=Bc1+· · ·+Bcnfor bi∈Bandcj∈Cwherei= 1, ..., mandj= 1, ..., n. Letc∈C. Then we have
c=
n
X
j=1
xjcj
wherexj ∈B. Foraij ∈Awe have xj =
m
X
i=1
aijbi.
Combining the two sums yields c=
n
X
j=1 m
X
i=1
aijbicj
makingC=Ab1c1+· · ·+Abmcna finitely generatedA-module.
Theorem 5.1.5. LetAandBbe two integral domains such thatA⊆B, and letbi ∈B be integral overAfori= 1, ..., n. ThenA[b1, ..., bn]is a finitely generatedA−module.
Proof. We prove the statement by induction onn. First, ifb1∈Bis integral overAthen A[b1]is a finitely generatedA-module by Theorem 5.1.2, and so the theorem is true for n= 1, completing the base case.
Next, assume thatA[b1, ..., bn−1]is a finitely generatedA-module, wherebi ∈ B is integral overA, andi = 1, ..., n−1for n ≥ 2. Also, letbn ∈ B be integral overA.
Then, by Theorem 5.1.1, bn is integral over A[b1, ..., bn−1], and so by Theorem 5.1.2, we have that(A[b1, ..., bn−1])[bn] =A[b1, ..., bn]is a finitely generatedA-module, which completes the inductive step, thus verifying the theorem by induction.
There are not only elements that may have the property of being integral over a domain, but also domains themselves.
Definition. LetAandBbe two integral domains such thatA⊆B. Then, if every element b∈Bis integral overA,Bis integral overA.
We go on by presenting a result combining both elements and domains being integral.
Theorem 5.1.6. LetA⊆B⊆Cbe a tower of integral domains. IfBis integral overA andc∈Cis integral overB, thencis integral overA.
Proof. Sincec∈Cis integral overB, we have that
cn+bn−1cn−1+· · ·+b1c+b0= 0
forbi∈B, wherei= 0, ..., n−1, thuscis integral overA[b0, ..., bn−1]. Now, since every bi∈BandBis integral overA, everybiis also integral overA, and soA[b0, ..., bn−1]is a finitely generatedA-module by Theorem 5.1.5. By Theorem 5.1.2 and the fact thatcis integral overA[b0, ..., bn−1], we deduce that(A[b0, ..., bn−1])[c] =A[b0, ..., bn−1, c]is a finitely generatedA-module. Then finally, by Theorem 5.1.3cis integral overA.
Now we will circle back to the special case of the definition of integral elements where we letA=ZandB =C.
Theorem 5.1.7. LetAandB be integral domains such thatA ⊆ B. Ifb1, b2 ∈ B are integral overA, thenb1+b2,b1−b2andb1b2are also integral overA.
Proof. Sinceb1is integral overAwe have by Theorem 5.1.2 thatA[b1]is a finitely gen- eratedA-module. We haveA⊆A[b1] ⊆B, and sinceb2is integral overAwe have by Theorem 5.1.1 thatb2is integral overA[b1]. Then(A[b1])[b2] =A[b1, b2]is a finitely gen- eratedA[b1]-module by Theorem 5.1.2, makingA[b1, b2]a finitely generatedA-module by Theorem 5.1.4. Next, letxdenote any one of the elementsb1+b2,b1−b2,b1b2. Hence A ⊆ A[x] ⊆ A[b1, b2] ⊆ B where the integral domainA[b1, b2]is a finitely generated A-module. Then, by Theorem 5.1.3,xis integral overA.
This theorem allows us to conclude that in the situation whereA ⊆B, the set of all elements ofBthat are integral overAis a subdomain ofBcontainingA. ForA=Zand B =Cwe are left with the fact thatthe set of all algebraic integers is an integral domain.
5.2 Integral closure
As we just mentioned, in the situationA ⊆B whereAandB are integral domains, the set of all elements inB that are integral overAis a subdomain ofBcontainingA. We provide notation for this domain in the next definition.
Definition. LetAandB be two integral domains such thatA ⊆ B. Then theintegral closureofAinB, denotedABis the subdomain ofBconsisting of all elements ofBthat are integral overA.
Theorem 5.2.1. ForRa UFD letF be its field of quotients, that isF = Quot(R). Then for an elementf ∈Fwe have that
f is integral overR ⇐⇒ f ∈R.
Proof. In the case wheref ∈R,f satisfies the equationx−f = 0and sof is integral overR, proving the left implication.
For the converse, assume thatf ∈Fis integral overR, hence satisfying an equation fn+an−1fn−1+· · ·+a1f +a0= 0
forai ∈ Rwherei = 0, ..., n−1. Sincef ∈ F we writef = rs−1 fors 6= 0where r, s∈Rand gcd(r, s) = 1. We insert the new expression forfin the equation above and obtain
rns−n+an−1rn−1s1−n+· · ·+a1rs−1+a0= 0.
Multiplying bysnon both sides yields
rn+an−1rn−1s+· · ·+a1rsn−1+a0sn= 0.
We need to show thatsis a unit inR. To that end, assume thatsis not a unit. Then there exists an irreducible elementp∈Rsuch thatp|s. We have that
rn =−an−1rn−1s− · · · −a1rsn−1−a0sn
where s is a factor in every term, and so we deduce that p|rn. Since pis prime (by Theorem 4.4.6)p|r. This contradicts with the fact that gcd(r, s) = 1, hencesis a unit in Randf =rs−1∈R.
In light of this theorem, we look at the situation whereR =Z. We know thatZis in fact a UFD, as we showed in example 4.2.1 that it is a Euclidean domain. In example 2.2.2 we concluded thatQuot(Z) = Q. Now, for the integral domainsZandC, we get that ZCdenotes the set of all algebraic integers. ThenQ∩ZCdenotes all rational algebraic integers. By our last theorem, an elementais inQ∩ZC, i.e. ais integral overZ, if and only ifa∈Z, makingQ∩ZC=Z. In other words:a rational algebraic integer must be an ordinary integer.
To move on we look at the situation where the integral closure of an integral domain R, makesRintegrally closed.
Definition. LetRbe an integral domain. ThenRis said to beintegrally closedif for all elementsa∈Quot(R)that are integral overR, we havea∈R.
We observe the connection between our last definition and Theorem 5.2.1, and con- clude thatevery UFD is integrally closed.
Example 5.2.2. Consider the ringZ. We have seen thatZis a UFD, henceZis integrally closed.
As we know every PID is also a UFD we get the following corollary.
Corollary 5.2.3. Every PID is integrally closed.
Up till now we have defined what it means for an element in an integral domain to be integral over a domain, and to be an algebraic integer. In the upcoming section we will start by looking at elements being algebraic over a domain and algebraic numbers, leading up to the definition of an algebraic number field and its ring of integers.
5.3 The ring of integers
We have earlier considered integral domainsAandBwhereA⊆B. Now, we will look at the case whereAis not only an integral domain, but also a field.
Definition. LetAandBbe two integral domains such thatA⊆B. IfAis a field and an elementb∈Bis integral overA, thenbisalgebraic over A.
Just as we in the previous section obtained algebraic integers by looking atZandC, we know look at the situation where, in our definition,A=QandB =C. Then we name an elementb ∈ Cthat is integral over the fieldQ, analgebraic number. AsZ⊆ Q, an elementc ∈Cthat is integral overZ, will also be integral overQby Theorem 5.1.1. In other words:every algebraic integer is an algebraic number.
Theorem 5.3.1. Every algebraic numberbis of the formr/swherer∈Cis an algebraic integer ands6= 0is an integer.
Proof. Letbbe an algebraic number. Then forai∈Qwherei= 0, ..., n−1we have bn+an−1bn−1+· · ·+a1b+a0= 0.
Letsbe the least common multiple of the denominatorsai. Then06=s∈Zandsai∈Z. Now, multiplying the above equation bysnyields
(sb)n+ (san−1)(sb)n−1+· · ·+ (sn−1a1)(sb) + (sna0) = 0,
which is a monic polynomial with coefficients in Zand sb as a root. Hence sb is an algebraic integer which we denoter. Thenb=r/swhereris an algebraic integer ands is a nonzero integer.
As we now know what it means for an element to be an algebraic number, we define an algebraic number field. Afterwards, we will look at what we callthe ring of integers of an algebraic number field, and study it further.
Definition. Analgebraic number field K is a subfield ofCof the formQ(α1, ..., αn), whereαifori= 1, ..., n, are algebraic numbers.
Example 5.3.2. K =Q(√
2)is an algebraic number field, as we have shown that√ 2is an algebraic integer, hence an algebraic number.
We may also express an algebraic number field by saying thatK =Q(α1, ..., αn)is the smallest subfield ofCcontaining the whole ofQand all the elementsα1, ..., αn. Now, in the case whereK =Q(α)andα∈Cis a root of an irreducible quadratic polynomial x2+ax+b∈Q[x], we nameQ(α)aquadratic field, or aquadratic field extensionofQ. We want to be able to determine these fields in a unique way.
Theorem 5.3.3. IfKis a quadratic field, then there exists a unique squarefree integerd such thatK=Q(√
d).
Proof. LetK=Q(α)whereαis a root of the irreducible polynomialx2+ax+b∈Q[x].
Thus
α=−a±√ a2−4b
2 .
We may write, without loss of generality that α=−a+√
a2−4b 2 and so
K=Q(α) =Q
−a+√ a2−4b 2
=Q(√ c)
wherec=a2−4b∈Q. Sincex2+ax+bis irreducible inQ[x],cis not the square of a rational number. Letc=p/qwherep, q∈Zare such thatq >0and gcd(p, q) = 1. Now forpqletm2be the biggest square such thatm2|pq. Then we have thatpq =m2dfor a squarefree integerd6= 1. Furthermore
K=Q(√ c) =Q
rp q
=Q(√
pq) =Q(√
m2d) =Q(m√
d) =Q(√ d).
Letnbe another squarefree integer such thatK=Q(√
n). ThenQ(√
d) =Q(√ n), and
so √
d=x+y√ n for somex, y∈Q. After squaring we obtain
d=x2+ny2+ 2xy√ n.
Assumexy6= 0. Then
√n=d−x2−ny2 2xy which contradicts with √
n /∈ Qsincen is chosen to be a squarefree integer. Hence xy= 0. Fory = 0we have√
d=x, but asdis squarefree this contradicts with√ d /∈Q, and so we must havex= 0, making√
d=y√
nand finally d=y2n.
