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Probably this is easiest, but as I am somehow stuck I would be pleased about some comments.

What I give myself is a rule $f$ which does the following:

To every commutative ring $A$ with $1$ the rule $f$ assigns a unit $f(A)\in A^*$, and this assignment shall satisfy the following property:

If $\varphi: A\rightarrow B$ is any ring homomorphism, then $\varphi(f(A)) =f(B)$.

I am convinced that the only rules which can do this are the following two:

(i) $f(A)=1$ for each ring $A$ simultaneously,

(ii) $f(A)=-1$ for each ring $A$ simultaneously.

How could one show this (if it is right)?

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Should the rule say "$f(A) \in A^*$" or "$f(A) \in A$?" I am confused as $\phi$ is not a homomorphism with domain $A^*$. – Kris Williams Sep 19 '12 at 13:54
One could generalize and ask the analogous question for $R$-algebras instead or rings. Manny's nice proof carries over to this setting and shows that the set of those rules is in canonical bijection to the set of units in $R$. By the way, such a rule is the same as a natural transformation from $K_{\{\star\}}$ to $(-)^\times$, where $K_{\{\star\}}$ is the functor which associates to any ring the set $\{\star\}$ and $(-)^\times$ is the functor which associates to any ring its set of units. – Ingo Blechschmidt May 2 at 21:55
up vote 8 down vote accepted

Given the context of the problem, I am assuming that you want to impose the following conditions:

  • "Commutative ring" means commutative ring with identity.
  • "Ring homomorphism" means a homomorphism that preserves identity elements.

In this case, your guess is correct. Here is a proof.

First, consider that $f(\mathbb{Z}) \in \mathbb{Z}^* = \{1,-1\}$.

Next, let $A$ be any commutative ring. Then there is a unique homomorphism $\phi \colon \mathbb{Z} \to A$, determined by $\phi(1) = 1$. Given your requirements for the assignment $f$, it follows that $$f(A) = \phi(f(\mathbb{Z})).$$ If $f(\mathbb{Z}) = 1$, we conclude from the equation above that $f(A) = 1$ also. If $f(\mathbb{Z}) = -1$, it follows similarly that $f(A) = -1$. Thus $f$ falls into either case (i) or case (ii), to use your notation.

(How did I produce this answer? I looked for a universal example where I could use concrete reasoning. The commutative ring $\mathbb{Z}$ is "universal" in the sense that it has a unique homomorphism to every ring. The problem became much easier after understanding this specific example!)

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+1 Very nice explanation. – Michael Joyce Sep 19 '12 at 16:13

One program that often works for these sorts of questions is:

  • Select one or more rings for which it is easy to solve for all possibilities
  • Study if and how these solutions extend to all rings.
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