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This is Dummit & Foote. Appendix 2. Exercise 1.3. (Category Theory)

The map $\mathsf{Ring}$ to $\mathsf{Grp}$ by mapping a ring to its group of units defines a functor. Show by explicit examples that this functor is neither faithfull nor full.

It's a functor clearly, but I am not able to give examples about the second part. Any idea? Thanks!

Zev Chonoles
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Peter
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1 Answers1

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For fullness:

How many ring homomorphisms are there from $\mathbb{Z}/5\mathbb{Z}$ to $\mathbb{Z}/3\mathbb{Z}$?

How many group homomorphisms are there from $(\mathbb{Z}/5\mathbb{Z})^\times$ to $(\mathbb{Z}/3\mathbb{Z})^\times$?

For faithfulness:

How many ring homomorphisms are there from $(\mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/2\mathbb{Z})$ to $(\mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/2\mathbb{Z})$?

How many group homomorphisms are there from $(\mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/2\mathbb{Z})^\times$ to $(\mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/2\mathbb{Z})^\times$?

Zev Chonoles
  • 129,973
  • For faitfhulness we could also consider $\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$ to just $\mathbb{Z}/2\mathbb{Z}$. – Anakhand Jan 18 '24 at 23:09