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

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. –  Arturo Magidin Sep 25 '11 at 21:28
Sure, thanks for the suggestion. –  Aaron Mazel-Gee Sep 25 '11 at 21:30
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 $-1\mapsto -1$? –  Matt Sep 25 '11 at 22:17
@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. –  Henning Makholm Sep 25 '11 at 22:51
@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. –  Aaron Mazel-Gee Sep 25 '11 at 22:56

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.
Note that the functor in the statement does not map all rings to a $2$ element set. –  Mariano Suárez-Alvarez Sep 26 '11 at 4:28
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. –  Aaron Mazel-Gee Sep 26 '11 at 5:58
e.g. $R$ and $S$ not of characteristic 2 :) –  Dylan Wilson Sep 26 '11 at 7:07