-
[ Pobierz całość w formacie PDF ]
r > 2, then (À) is properly contained in Q1Q2 = (±) and so ± is
a proper divisor of À, a contradiction. Therefore, it follows that a
principal ideal generated by a non-prime irreducible element factors
into the product of two prime ideals. Now what?)
5. Let R be as above, i.e., a Dedekind domain with ideal class group
of order at most 2. Let À1, À2 " R be irreducible elements. As we
seen in Exercise 4 above, any factorization of À1À2 will involve exactly
two irreducibles. Show that, up to associates, there can be at most
three distinct factorizations of À1À2 into irreducibles. a simple
"(As
illustration, it turns out that the Dedekind domain Z[ -5] has class
group of order 2; correspondingly we have distinct factorizations: 21 =
" " " "
3 · 7 = (1 + 2 -5)(1 - 2 -5) = (4 + -5)(4 - -5).)
4.6 A Characterization of Dedekind Domains
In this final section we ll prove the converse of Theorem 4.4.6, thereby
giving a characterization of Dedekind domains.
To begin with, let R be an arbitrary integral domain, with fraction field
E. In analogy with the preceeding section, if I †" R is an ideal, we set
I-1 = {± " E| ±I †" R}.
We say that I is invertible if I-1I = R.
Lemma 4.6.1 Assume that I †" R and admits factorizations
P1P2 · · · Pr = I = Q1Q2 · · · Qs,
1
See L. Carlitz, A characterization of algebraic number fields with class number two,
Proc. Amer. Math. Soc. 11 (1960), 391-392. In case R is the ring of integers in a finite
extension of the rational field, Carlitz also proves the converse.
102 CHAPTER 4. DEDEKIND DOMAINS
where the Pi s and the Qj s are invertible prime ideals. Then r = s, and
(possibly after re-indexing) Pi = Qi, i = 1, 2, · · · , r.
Lemma 4.6.2 Let R be an integral domain.
(i) Any non-zero principal ideal is invertible.
(ii) If 0 = x " R, and if the principal ideal (x) factors into prime ideals as
(x) = P1P2 · · · Pr, then each Pi is invertible.
Now assume that R is an integral domain satisying the following condi-
tion:
(*) If I †" R is an ideal of R, then there exist prime ideals P1, P2, · · · , Pr †"
R such that
I = P1P2 · · · Pr.
Note that no assumption is made regarding the uniqueness of the above
factorization. We shall show that uniqueness automatically follows. (See
Corollary 4.6.9.2 , below.) Of course, this is exactly analogous with what
happens in unique factorization domains.
Our goal is to show that R is a Dedekind domain.
Proposition 4.6.3 Any invertible prime ideal of R is maximal.
Proposition 4.6.4 Any prime ideal is invertible, hence maximal.
Corollary 4.6.4.1 Any ideal is invertible.
Corollary 4.6.4.2 Any ideal of R factors uniquely into prime ideals.
Proposition 4.6.5 R is Noetherian.
Our task of showing that R is a Dedekind domain will be complete as
soon as we can show that R is integrally closed. To do this it is convenient
to introduct certain overrings of R, described as below.
Let R be an arbitrary integral domain and let E = F(R). If P †" R is a
prime ideal of R we set
RP = {±/² " E| ±, ² " R, ² " P }.
It should be clear (using the fact that P is a prime ideal) that RP is a
subring of E containing R. It should also be clear that F(RP ) = E. RP is
called the localization of R at the prime ideal P .
4.6. A CHARACTERIZATION OF DEDEKIND DOMAINS 103
Lemma 4.6.6 Let I be an ideal of R, and let P be a prime ideal of R.
(i) If I †" P then RP I = RP .
-1
(ii) RP P properly contains RP .
Lemma 4.6.7 If ± " E then either ± " RP or ±-1 " RP .
The following is now really quite trivial.
Lemma 4.6.8 RP is integrally closed.
Proposition 4.6.9 R = )"RP , the intersection taken over all prime ideals
P †" R.
As an immediate result, we get
Corollary 4.6.9.1 R is integrally closed.
Combining all of the above we get the desired characterization of Dedekind
domains:
Corollary 4.6.9.2 R is a Dedekind domain if and only if every ideal of R
can be factored into prime ideals.
Exercises
1. A valuation ring is an integral domain R such that if I and J are ideals
of R, then either I †" J or J †" I. Prove that for an integral domain
R, the following three conditions are equivalent:
(i) R is a valuation ring.
(ii) if a, b " R, then either (a) †" (b) or (b) †" (a).
(iii) If ± " E := F(R), then either ± " R or ±-1 " R.
(Thus, we see that the rings RP , defined above, are valuation rings.)
2. Let R be a Noetherian valuation ring.
(i) Prove that R is a p.i.d.
104 CHAPTER 4. DEDEKIND DOMAINS
(ii) Prove that R contains a unique maximal ideal. (This is true even
if R isn t Noetherian.)
(iii) Conclude that, up to units, R contains a unique prime element.
(A ring satisfying the above is often called a discrete valuation ring .)
3. Let R be a discrete valuation ring, as in Exercise 2, above, and let
À be the prime, unique up to associates. Define ½(a) = r, where
a = Àrb, À / b. Prove that ½ is an algorithm for R, giving R the
structure of a Euclidean domain.
4. Let R be a Noetherian domain and let P be a prime ideal. Show that
the localization RP is Noetherian.
5. Let R be a ring in which every ideal I †" R is invertible. Prove that R is
a Dedekind domain. (Hint: First, as in the proof of Proposition 4.6.5,
R is Noetherian. Now let C be the set of all ideals that are not products
of prime ideals. Since R is Noetherian, C = " implies that C has a
maximal member J. Let J †" P , where P is a maximal ideal. Clearly
-1 -1 -1
J = P . Then JP †" P P = R and so JP is an ideal of R;
-1 -1 -1
clearly J †" JP . If J = JP , then JP = P1P2 · · · Pr so J =
-1
P P1P2 · · · Pr. Thus J = JP so JP = J. This is a contradition,
why?)
6. Here is an example of a non-invertible ideal in an integral domain R.
Let
"
R = {a + 3b -5| a, b " Z},
" "
and let I = (3, 3 -5), i.e., I is the ideal generated by 3 and 3 -5.
Show that I is not invertible. (An easy way to do this is to let J = (3),
the principal ideal generated by 3, and observe that despite the fact
that I = J, we have I2 = IJ.)
Chapter 5
Module Theory
5.1 The Basic Homomorphism Theorems
In Section 4.1 we introduced some of the basics of module theory, as they
were indespensible to our study of Dedekind domains. In the present chap-
ter, we embark on a more systematic study of module theory; one very im-
portant difference here is that unless otherwise stated, the rings in question
need not be commutative.
There are two basic homomorphism theorems worth mentioning here.
The proofs are entirely routine and mimick the corresponding proofs for
abelian groups (i.e.,Z-modules). [ Pobierz całość w formacie PDF ] - zanotowane.pl
- doc.pisz.pl
- pdf.pisz.pl
- zambezia2013.opx.pl