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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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    You may want to specify that you want your category to be "Rings with 1" and morphisms to send $1$ to $1$; otherwise, it's unclear to me whether what you describe is a functor.2011-09-25
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    Sure, thanks for the suggestion.2011-09-25
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    It is actually unclear to me how this functor should act on morphisms. Since it outputs sets, you are forgetting any extra structure and hence it is basically a constant (except char 2) functor. Input any $R$, it outputs a set with two elements. If I have $f:R\to S$ where does the map go? Is it just the map $1\mapsto 1$ and $-1\mapsto -1$?2011-09-25
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    @Matt, if I have the base set $\{5,7,9,13,19,42\}$, then I can make it into a ring in various ways, simply by giving tables for the sum and product operations. One of these rings may have $7$ as its one-element and $13$ as its minus-one, another may have it the opposite way. If there's a homomorphism from one to another, its image in Set under the functor should be the map $\{7,13\}\to\{7,13\}$ that interchanges the two elements.2011-09-25
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    @Matt: Yes, that's right, this is usually constant at the 2-element set. I gave the background information so that anyone who cares to answer would know what I'm *actually* looking for. I'm pretty sure an answer to my title question would also be an answer to my real question.2011-09-25
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    In your edit: presumably, all your rings are also commutative :)2011-09-26
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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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    Thanks Dylan! Now apply Spec and say everything backwards... :P2011-09-26
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    Note that the functor in the statement does not map all rings to a $2$ element set.2011-09-26
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    That's true. Of course, all we need to adapt the proof is to find a pair of rings $R,S$ so that $F(R)$, $F(S)$, and $F(R\times S)$ are all supposed to have exactly 2 elements.2011-09-26
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    e.g. $R$ and $S$ not of characteristic 2 :)2011-09-26