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(√
We may write, without loss of generality that α=−a+√
Letnbe another squarefree integer such thatK=Q(√
n). ThenQ(√
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.
This impliesy2 = 1, asdis squarefree, and so d = n. This proves thatdis uniquely determined byK.
For any algebraic number field we may obtain the subset containing all its algebraic in-tegers. We will define this subset for a general algebraic number field, and then determine it for the quadratic extensions ofQ.
Definition. ForK an algebraic number field, the subset of all algebraic integers in K, denotedOK, is calledthe ring of integers of the algebraic number fieldK.
Note that since every element ofOK is an algebraic integer,OK is integral overZ. Also, in the case whereK =Q, we have thatOK =Z. Before giving several properties of the ring of integers for a general algebraic number fieldK, we look at the special case whereKis a quadratic field.
Theorem 5.3.4. LetK=Q(√
d)wheredis a squarefree integer. ThenOKis given by
OK= only if the coefficients of the minimal polynomial
x−a+b√ pdividesb, asdis squarefree. This contradicts with the assumption thata, b, chave no common prime factor. Hence, since2a/c∈Z, we must havec= 1orc= 2. Forc= 1,
and soa2≡1 (mod4)andb2≡1 (mod4), leading to the conclusion thatd≡1 (mod4). Example 5.3.5. Consider the quadratic fieldsK1 = Q(√
13)andK2 =Q(√
−1). We want to determine their ring of integers by using Theorem 5.3.4. Since13≡1(mod4) we have Euclidean domains with respect to the norm, as proven in [3, Theorem 4.17].
For the remainder of this chapter, we will prove the properties ofOKthat we will use in our next chapter, considering Dedekind domains.
Theorem 5.3.6. ForKan algebraic number field,OKis an integral domain.
Proof. AsKis a field, andOK⊆K, we conclude thatOK is an integral domain.
Theorem 5.3.7. ForKan algebraic number field we have thatQuot(OK) =K.
Proof. ForF = Quot(OK)andα∈F we haveα =b/cwhereb, c∈ OK andc 6= 0.
Theorem 5.3.8. ForKan algebraic number field,OKis integrally closed.
Proof. LetKbe an algebraic number field. By Theorem 5.3.7 we haveQuot(OK) =K.
An elementα∈Kthat is integral overOK, will also be integral overZby Theorem 5.1.6, asOK is integral overZ. Thenαis an algebraic integer inK, and soα∈ OK, making OKintegrally closed.
Theorem 5.3.9. ForKan algebraic number field,OKis a Noetherian domain.
Proof. LetIbe an ideal ofOK. ForI ={0}we have thatI =h0iis finitely generated.
ForI6={0}we have thatIis finitely generated by [4, Theorem 6.5.2]. Thus every ideal ofOKis finitely generated, and so, by Theorem 3.2.3,OK is a Noetherian domain.
Theorem 5.3.10. LetKbe an algebraic number field andPa nonzero prime ideal ofOK. ThenPis a maximal ideal ofOK.
Proof. We will prove this theorem by assuming the contrary of what it states, and find that the assumption does not hold. To that end, letKbe an algebraic number field, and assume there exists a prime idealP1ofOK that is not maximal. Define the set
S ={Iproper ideal ofOK|P1(I}
which is not empty asP1is not a maximal ideal. By Theorem 5.3.9OKis a Noetherian domain. Then, by Theorem 3.2.3, S contains a maximal element. This means that there exists a maximal idealP2such that
P1(P2(OK.
By Theorem 2.4.6P2is then also a prime ideal. As by [4, Theorem 6.1.7] every nonzero ideal inOK contains a rational integer, we must haveP1∩Z6={0}, and soP1∩Zis a prime ideal ofZby Theorem 2.4.8. But we saw in example 4.3.2 thatZis a PID, meaning P1∩Z=hpifor somep∈Z. This elementpis prime by Theorem 2.4.4, and so we obtain
hpi=P1∩Z⊆P2∩Z⊆Z.
P2is a proper ideal ofOK, and so1∈/ P2, makingP2∩Z6=Z. Also,hpiis a maximal ideal by Theorem 4.3.3, thus
hpi=P1∩Z=P2∩Z.
Now,P1(P2, so there exists an elementb∈P2that is not inP1. Sinceb∈ OK,bis an algebraic integer, and then integral overZ. Thus we have that
bn+an−1bn−1+· · ·+a1b+a0= 0
forai∈Zwherei= 0, ..., n−1, and so, since0∈P1
bn+an−1bn−1+· · ·+a1b+a0∈P1.
Letkbe the least positive integer for which there existci∈Zwherei= 0, ..., k−1, such that
bk+ck−1bk−1+· · ·+c1b+c0∈P1. Sinceb∈P2we have that
bk+ck−1bk−1+· · ·+c1b=b(bk−1+ck−1bk−2+· · ·+c1)∈P2.
Now, asP1(P2andP2is an ideal ofOK
c0= (bk+· · ·+c1b+c0)−(bk+· · ·+c1b)∈P2,
butc0∈Z, and soc0∈P2∩Z=P1∩Z, makingc0an element ofP1. Then bk+ck−1bk−1+· · ·+c1b= (bk+ck−1bk−1+· · ·+c1b+c0)−c0∈P1.
In the case wherek= 1we will haveb∈P1, clearly contradicting withb /∈P1, sok≥2 and
b(bk−1+· · ·+c1)∈P1.
Finally, sinceP1is a prime ideal andb /∈P1, we must have bk−1+· · ·+c1∈P1
which contradicts withkbeing the least positive integer, as(k−1)is positive whenk≥2.
Hence the assumption ofP1not being maximal does not hold, and we conclude that every prime ideal ofOKis maximal.
Chapter 6
Ideal factorization in Dedekind domains
We have arrived at the final chapter of this thesis. Here we will define aDedekind domain, and at the end we will prove our main theorem.
6.1 Dedekind domains
Definition. An integral domainDis called aDedekind domainif it satisfies the following conditions:
i) Dis a Noetherian domain.
ii) Dis integrally closed.
iii) Every nonzero prime ideal inDis a maximal ideal.
Throughout this thesis we have shown several properties of the domainZ. It will also be our first example of a Dedekind domain.
Example 6.1.1. Consider the integral domainZ. We check for the properties of a Dedekind domain.
i) Zis Noetherian by example 3.2.6.
ii) Zis integrally closed by example 5.2.2.
iii) Zis a PID by example 4.3.2, and so every nonzero prime ideal inZis a maximal ideal by Theorem 4.3.3.
HenceZis a Dedekind domain.
We now know thatZis both a principal ideal domain and a Dedekind domain. In fact, this is an example of a general rule.
Theorem 6.1.2. Every PID is a Dedekind domain.
Proof. AssumeDto be a PID. Then by Theorem 4.3.1Dis Noetherian. D is integrally closed by Corollary 5.2.3, and finally, by Theorem 4.3.3 every nonzero prime ideal inD is a maximal ideal. HenceDis a Dedekind domain.
We now turn to the ring of integers of an algebraic number field. As mentioned in the previous chapter, we will make use of the properties ofOKthat we have already proven.
Theorem 6.1.3. ForKan algebraic number field,OKis a Dedekind domain.
Proof. LetK be an algebraic number field andOK its ring of integers. Then we know that OK is Noetherian by Theorem 5.3.9, integrally closed by Theorem 5.3.8 and that every nonzero prime ideal ofOK is a maximal ideal by Theorem 5.3.10. HenceOKis a Dedekind domain.
We have seen that the quadratic domainZ[√
n]wherenis a squarefree integer, is the ring of integers of the algebraic number fieldK =Q(√
n)whenn 6≡1 (mod4). Then, by our last theorem, we conclude thatZ[√
n]is a Dedekind domain whenn6≡1 (mod4).
Forn≡1 (mod4)however, we know thatOK 6=Z[√
n], so we can not as easily jump to any conclusions. Actually, in this case,Z[√
n]is not a Dedekind domain.
Proposition 6.1.4. Ifnis a squarefree integer such thatn≡1 (mod4), thenZ[√ n]is not a Dedekind domain.
Proof. We want to show thatZ[√
n]is not integrally closed whennis a squarefree integer such thatn≡1 (mod4). For such values we may expressnasn= 4k+ 1fork∈Z. Let which has all its coefficients inZ[√
n]. Forx=αwe obtain
n], and so we have an integral element in the quotient field ofZ[√
n], not inZ[√
n]. Hence, whenn≡1 (mod4),Z[√
n]is not integrally closed, thus not a Dedekind domain. Since every UFD is integrally closed, it is not a UFD either.
In order to provide the proof of our main theorem, we must first consider ideals in Dedekind domains, and look at how they contain a product of prime ideals.