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Is the function $\mbox{Rings}\rightarrow\mbox{Sets}$ given by $R\mapsto \{\pm 1\in R\}$ corepresentable?

Of course this might be problematic in characteristic 2 since this set is then a singleton, but a char-2 ring can't map to anything but a char-2 ring anyways, so maybe it's alright. Actually what I was originally thinking about is the cogroupoid-in-$\mbox{Rings}$ $(A,\Gamma)$ corepresenting the groupoid whose objects are $\{x^2+bx+c:b,c\in R\}$, with morphisms Hom((b,c),(b',c')) = \{ r \in R : (x+r)^2+b(x+r)+c = x^2+b'x+c\}. Explicitly, $A=\mathbb{Z}[b,c]$ and $\Gamma = A[r]$ (where the copy of $A$ selects the source of a morphism). I'd like to extend this to allow my morphisms to take the form $x\mapsto \pm x + r$ (instead of just $x\mapsto x + r$, as it is now). The obvious guess is to set $\Gamma = A[e,r]/((e-1)(e+1)) = A[e,r]/(e^2-1)$, but of course this is going to allow our linear coefficient to be any order-2 element of $R^\times$.

I suspect there isn't such a ring, because I'd think if there were then it'd be easy to see what it should be. Either way, there's probably an algebro-geometric reason for the answer, which I'd love to see.

EDIT: To be completely clear: all my rings are commutative and have 1, and all my ring homomorphisms are unital.

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    Yup!!!!!!!!!!!!2011-09-26

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There is no such corepresentable functor. More is true: there is no corepresentable functor $F: \text{Ring} \rightarrow \text{Set}$ that takes every ring to a two-element set. Indeed, if $F$ is corepresentable then $F(R \times S) = F(R) \times F(S)$, for any rings $R$ and $S$ (this is just the universal property of the product). Certainly $R\times S$ is a ring if $R$ and $S$ are, and the product of a two element set with itself has cardinality 4.

The closest thing would be $\mathbb{Z} \times \mathbb{Z}$. Then we have that $\text{Hom}(\mathbb{Z}\times \mathbb{Z}, R)$ has two elements as long as 0 and 1 are the only idempotents.

(This was an exercise in Waterhouse's book on affine group schemes.)

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    e.g. $R$ and $S$ not of characteristic$2$:)2011-09-26