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Let $R$ be an arbitrary unital associative ring. In the category of $R$-algebras $\mathfrak{Alg_R}$, if we consider $R$ as an $R$-algebra over itself (trivially), what type of object is it then in $\mathfrak{Alg_R}$?

Is it initial, terminal or something else? (My intuition expects it to be initial).

Also, it is obviously free, but is it injective?

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2 Answers 2

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Hints:

An $R$-linear homomorphism has to take $1\mapsto 1$ and then be $R$-linear, so that $r\mapsto r$. All $R$-linear functions leaving $R$ are therefore completely determined.

Consider $R=\Bbb C$: Notice that evaluation mappings out of $\Bbb C[x]$ produce distinct mappings to $\Bbb C$.

You started the context in Alg(R) and then asked about injectivity and projectivity, but we most often talk about injectivity and projectivity in the category of modules. In the context of modules, $R$ is as you noted always a free module over itself, but it is not always injective module. Such rings $R$ are called self-injective rings.

In a principal ideal domain, like $\Bbb Z$, each ideal is module isomorphic to $\Bbb Z$. If $\Bbb Z$ is injective, so would these submodules be injective. But injective submodules are always summands of their supermodules. Can you see why this isn't possible in a domain like $\Bbb Z$?

If you really want to talk about free, projective and injective objects in the category of Alg(R), let me know.

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  • $\begingroup$ You are not working in the right category. There is only one $\mathbb{C}$-algebra homomorphism $\mathbb{C} \to \mathbb{C}$. $\endgroup$
    – Zhen Lin
    Commented Oct 31, 2013 at 17:23
  • $\begingroup$ Dear @ZhenLin : Right... I don't know how I talked myself into that one. But was your rather sweeping comment meant to apply to more than just the third line, or did you just stop reading there and write the comment? I can't really tell... $\endgroup$
    – rschwieb
    Commented Oct 31, 2013 at 17:37
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Hints:

  1. Let $A$ be an $R$-algebra. Can you define a map $R\to A$? Note that if $R$ is any ring then there are as many algebra homomorphisms $R[x]\to R$ as there are elements of $R$---this shows that in general $R$ cannot be terminal.
  2. Consider the category of $\mathbb{Z}$-algebras. Is $\mathbb{Z}$ injective as a $\mathbb{Z}$-module?
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  • $\begingroup$ True... But is R initial and would it generate $\mathfrak{Alg_R}$? $\endgroup$
    – ABIM
    Commented Oct 29, 2013 at 17:08

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