Universit¨at Konstanz Dr. Andrew Dolphin Fachbereich Mathematik und Statistik Prof. Dr. Markus Schweighofer Wintersemester 2010/2011
Ubungsblatt 7 zur Einf¨ ¨ uhrung in die Algebra: Solutions
Aufgabe 1. Sei R ein kommutativer Ring. Sei p ⊆ R ein Ideal. Zeige, dass folgende Aussagen
¨aquivalent sind:
(i) p ist prim.
(ii) p echt und f¨ur alle IdealeI,J von RmitIJ ⊆pgiltI⊆p oderJ⊆p.
(iii) R\pist eine multiplikative Menge.
Solution
(i)⇒ (ii): p is a proper (echt) ideal by definition. Let I,J be ideals of R such that IJ ⊆ P.
If I ⊆p, nothing remains to be shown. Otherwise, let i ∈I\p. Then for all j ∈ J the product ij∈IJ, henceij ∈p. Sincep is prime, it follows thatj∈p, for allj∈J, and thereforeJ ⊆p.
(ii)⇒(i): Ifxy∈pthen (xy) = (x)(y)⊆p. Hence (x)∈por (y)∈p, and hencex∈pory∈p.
(i)⇔(iii): 1∈R\pif and only if 1∈/ pif and only ifp is a proper ideal.
We also have that
(xy∈p⇒x∈por y∈p)⇔(x /∈pand y /∈p⇒xy /∈p).
Together these results give thatp is prime if and only ifR\pis a multiplicative set.
Aufgabe 2.SeienA undB kommutative Ringe undϕ:A→B ein Epimorphismus. Zeige, dass die Zuordnungen
I7→ϕ(I) und J7→ϕ−1(J)
eine Bijektion zwischen der Menge der Ideale/Primideale/max. IdealeIvonAmit ker(ϕ)⊆Iund der Menge der Ideale/Primideale/max. Ideale vonB vermitteln.
Solution
We must show that following:
(a) If I is an ideal/prime ideal/max ideal of A with ker(ϕ) ⊆ I, then ϕ(I) is a ideal/prime ideal/max ideal ofB.
(b) IfJ is an ideal/prime ideal/max ideal ofB, thenϕ−1(J) is a ideal/prime ideal/max ideal of A with ker(ϕ)⊆I.
(c) I=ϕ−1(ϕ(I)) for all idealsIofA with ker(ϕ)⊆I.
(d) J =ϕ(ϕ−1(J)) for all idealsJ ofB.
We shall show (d) and (c) first.
(d): This is true even for all subsetsJ ofB asϕis surjective, as we know show.
”⊇” is trivial.
For”⊆”, sinceϕis surjective, for allx∈J there exists somea∈Asuch thatϕ(a) =x. Therefore a∈ϕ−1(J), and henceϕ(a) =x∈ϕ(ϕ−1(J)).
(c)”⊆” is trivial. For
”⊇”, takey∈ϕ−1(ϕ(I)). Thenϕ(y)∈ϕ(I). Ify /∈I, thenϕ(y) =ϕ(z) for somez∈I. Hencey−z∈ker(ϕ)⊆I. But sincez∈Iandy−z∈I, theny∈I, a contradiction.
Soy∈I andI=ϕ−1(ϕ(I)).
We now show (a) and (b) for ideals. (b) was shown in the lecture
(a) That the image of an ideal I is a subgroup has been shown in the lecture. We must only show that if j ∈ ϕ(I), then bj ∈ ϕ(I) for any b ∈ B. j = ϕ(i) for somei ∈ I, and since ϕ is
surjective, b =ϕ(a) for somea ∈A for all b ∈B. Therefore bj =ϕ(a)ϕ(ai) = ϕ(ai) ∈ϕ(I) as ai∈I for alli∈I anda∈AasI is an ideal.
For the rest of the question, we need only show that the image and preimage of a prime/max ideal is prime/max.
LetJ =ϕ(I) for some idealJ ofB andI an ideal ofA. Then consider the composition of the maps
A→ϕ B→B/J.
This gives a mapA→B/J which is surjective asϕis surjective, with kernelI (using (c) and (d)).
Hence, by the isomorphism theoremA/I ∼=B/J.
We know that for any ringRand idealKwe have thatR/Kis an integral domain/field if and only ifKis a prime ideal/ max ideal. Therefore, sinceA/I∼=B/J,B/J is an integral domain/field if and only ifA/I is an integral domain/field. This givesI is a prime/max ideal if and only ifJ is a prime/max ideal, which is what we wanted to show.
Aufgabe 3.SeiAein kommutativer Ring undS⊆Amultiplikativ. Zeige, dass die Zuordnungen p7→S−1ιS(p) :=
a s
a∈p, s∈S
und q7→ι−1S (q)
eine Bijektion zwischen der Menge der Primideale p von A mit p∩S = ∅ und der Menge der Primideale vonAS vermitteln.
Solution
Similarly as in the previous question, we must show that following:
(a) Ifp is a prime ideal ideal ofAwithS∩p=∅, thenS−1ιS(p) is a prime ideal ideal ofAS. (b) Ifq is a prime ideal ofAS, thenι−1S (q) is a prime ideal of AwithS∩ι−1S (q) =∅.
(c) p=ι−1S (S−1ιS(p)) for all prime idealsp ofAwithS∩p=∅.
(d) q=S−1ιS(ι−1S (q)) for all idealsqofAS. Again, we show (c) and (d) first
(c) First we show that p⊆ι−1S (S−1ιS(p)). Clearly p⊆ι−1S (ιS(p)). Since 1∈ S we have that ιS(p)⊆S−1ιS(p) and hence p⊆ι−1S (ιS(p))⊆ι−1S (S−1ιS(p)).
Now we showι−1S (S−1ιS(p))⊆p. Takex∈ι−1S (S−1ιS(p)), then x= as ∈AS, for some a∈p, s ∈ S. This gives that s0(sx−a) = 0 for some s0 ∈ S, and hence s0sx ∈ p. Since p is prime, this implies that ss0 ∈ p or x ∈ p. But ss0 ∈ S and S ∩p = ∅. Hence x ∈ p and therefore ι−1S (S−1ιS(p))⊆p.
Hence p=ι−1S (S−1ιS(p)).
(d) First we show thatS−1ιS(ι−1S (q))⊆q. Take as ∈S−1ιS(ι−1S (q)), for somea∈ι−1S (q),s∈S.
Thena∈q, butqis an ideal inAS, therefore as =1s·a∈q, and henceS−1ιS(ι−1S (q))⊆q.
Now we show that q⊆S−1ιS(ι−1S (q)). Take as ∈q for somea∈A, s∈S. Then s· as =a∈q as qis an ideal in AS and hence we have that a∈ι−1S (a) = a, therefore as ∈S−1ιS(ι−1S (q)), and thereforeq⊆S−1ιS(ι−1S (q)).
Hence q=S−1ιS(ι−1S (q)).
(a) Letp be a prime ideal that has an empty intersection withS . Then it is easy to see that S−1ιS(p) is an ideal. We now show that it is prime.
First we show that it is a proper ideal. If 1∈S−1ιS(p), then there is ans∈S anda∈p such that s=a, i.e. there is a t∈S such that ts=ta∈pButts∈S and S∩p=∅, a contradiction.
HenceS−1ιS(p) is a proper ideal.
Assume now that as·bt ∈S−1ιS(p), for some a,b∈Aands,t∈S. Then abst = p
s0 forp∈pand some s0 ∈S. Hences00(abs0−pst) = 0 for some s00∈S. Hences00abs0 =s00pst∈p. But s00s0 ∈/ p, hence ab∈p, and hencea∈p orb ∈p, and therefore as ∈S−1ιS(p) or bt ∈S−1ιS(p), and so we conclude thatS−1ιS(p) is prime.
(b) Letqbe a prime ideal ofAS. Take the composition of the maps A→ιS AS →AS/q.
This map has kernelι−1S (q) by (d), hence there exists an injectionA/ι−1S (q),→AS/q. Ifqis prime, then AS/q is an integral domain, and hence A/ι−1S (q) must be too. Hence ι−1S (q) must also be prime.
Aufgabe 4.SeiRein faktorieller Ring. Sein∈N0undx∈R mit x=pα11. . . pαnn
mitpi∈R prim und paarweise nicht assoziiert undαi ∈N. Seiy∈mit y=p1. . . pn.
Zeige, dass
p(x) = (y).
Solution
Without loss of generality, we may assume thatn6= 0. We will use thatp
(x) ={a∈R| ∃n∈ N0:an∈(x)}.
”⊇” Let α= max{α1, . . . ,αn}. Then we have
yα= (p1· · ·pn)α=pα1· · ·pαn =pα−α1 1· · ·pα−αn npα11· · ·pαnn =yx and thusyα∈(x). This shows the first inclusion.
”⊆” Assume that z ∈p
(x). Then there is n∈N such thatzn ∈(x). Thusx divides zn. Of course for anyi∈ {1, . . . ,n} we have that pi divides x. Thuspi divides zn and sincepi is prime, we obtain thatpi divides z. Now fori 6=j elementspi and pj are non-associated primes, thus y dividesz asR is factorial, and thereforez∈(y).
