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Strukturen und Algebra » Moduln » Existenz von Modul zeigen
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Universität/Hochschule J Existenz von Modul zeigen
LukasNiessen
Aktiv Letzter Besuch: in der letzten Woche
Dabei seit: 30.09.2019
Mitteilungen: 77
Aus: Rheinland-Pfalz, Asbach
Zum letzten BeitragZum nächsten BeitragZum vorigen BeitragZum erstem Beitrag  Themenstart: 2019-11-04

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Guten Abend,

ich hänge schon länger an folgender Aufgabe:

Sei $f: L \to M$ ein A-Homomorphismus zwischen zwei endlich erzeugten freien A-Moduln. Man zeige:
(i) Es existiert ein freier Untermoduln F von L mit $L = \text{Ker}(f) \oplus F$.
(ii) Es existieren Basen $x_1,...,x_n \text{ bzw } y_1,...,y_n$ von L bzw M sowie Elemente $\alpha_1,...,\alpha_r \in A-{0}, r \leq \min{(m,n)}$, sodass $f(x_i) = \alpha_i y_i$ für i Element {1,...,r} und für i > r gilt $f(x_i) = 0$. Zusätzlich kann man $\alpha_i \mid \alpha_{i+1}$ für $1 \leq i < r$ erreichen.

Ich habe versucht den Elementarteilersatz zu nutzen, sowie ein Lemma, dass wenn M endlich erzeugt ist und den Torsionsuntermodul T hat, dass es dann ein freien Untermoduln F von M gibt, mit $T \oplus F = M$.
Aber es hat weder bei (i) noch bei (ii) geholfen.


Vielen Dank für jede Hilfe! 😄



-----------------
Beste Grüße, Lukas Nießen
PS: Schreibt mir gerne bei Fragen. 😄
\(\endgroup\)


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Triceratops
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(i) Die Sequenz $0 \to \ker(f) \to L \to \mathrm{im}(f) \to 0$ spaltet, weil $\mathrm{im}(f)$ frei ist (als Untermodul eines freien Moduls).

(ii) Wende den Elementarteilersatz auf den injektiven(!) Homomorphismus $f|_F : F \to M$ an.



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xiao_shi_tou_
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2019-11-04 19:42 - Triceratops in Beitrag No. 1 schreibt:
(i) Die Sequenz $0 \to \ker(f) \to L \to \mathrm{im}(f) \to 0$ spaltet, weil $\mathrm{im}(f)$ frei ist (als Untermodul eines freien Moduls).
@Lukas hier


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”己所不欲,勿施于人“(Konfuzius)
PS: Falls ich plötzlich aufhöre in einem Thread zu antworten, dann kann es sein, dass ich es vergessen habe. Ihr könnt mir in diesem Fall eine Private Nachricht schicken.
\(\endgroup\)


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LukasNiessen
Aktiv Letzter Besuch: in der letzten Woche
Dabei seit: 30.09.2019
Mitteilungen: 77
Aus: Rheinland-Pfalz, Asbach
Zum letzten BeitragZum nächsten BeitragZum vorigen BeitragZum erstem Beitrag  Beitrag No.3, vom Themenstarter, eingetragen 2019-11-04


Danke!


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Beste Grüße, Lukas Nießen
PS: Schreibt mir gerne bei Fragen. 😄



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