Show their is only one ring morphism from Zn to Zk if k|n

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SUMMARY

The discussion focuses on proving that there is only one ring morphism from Zn to Zk when k divides n. The function f: Zn to Zk defined by f([x]n) = [x]k for all x in Zn is established as a ring morphism by verifying the necessary properties. The challenge lies in demonstrating that this is the sole morphism, which can be approached by contradiction, particularly by analyzing the implications of defining f(1) and its constraints.

PREREQUISITES
  • Understanding of ring theory and morphisms
  • Familiarity with modular arithmetic, specifically Zn and Zk
  • Knowledge of properties of ring homomorphisms
  • Experience with proof techniques, including proof by contradiction
NEXT STEPS
  • Study the properties of ring homomorphisms in detail
  • Explore the implications of defining morphisms in modular arithmetic
  • Learn about proof techniques, specifically proof by contradiction
  • Investigate the structure of finite rings and their morphisms
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This discussion is beneficial for mathematics students, particularly those studying abstract algebra, as well as educators and researchers interested in ring theory and its applications.

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Homework Statement


For the rings Zn and Zk show that if k|n, then the function f: Zn to Zk
s.t [x]n --->[x]k for all x in Zn
is a ring morphism. Show this is the only ring morphism from Zn to Zk.

The attempt at a solution

So I showed it is a ring morphism by just verifying the properties, no big deal. I have no idea how to show that it is the only one though.

I want to start out by contradiction and assume that their is another one... but I don't know what to do with that. Is there a way to show it directly?
 
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Once you know what f(1) is, you pretty much know what the morphism is, right? Is there any choice but to make f(1)=1?
 

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