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.