I have this old qual problem that I don't know how to do: So let $ F: Rings \to Sets$ given by $R \mapsto \{(a,b) \in R^2, aR + bR = R\}$. The action on the morphism is the obvious one. Now the problem asks to show that the functor is not representable.
So here is what I found on the internet: According to the method on the second page of this pdf http://pi.math.cornell.edu/~zbnorwood/ucla/files/repfunctors.pdf,
suppose $F = Hom(A, \_)$, then there is $(a, b) \in A^2, aA + bA =A$ such that for $B$ a ring and $(x, y) \in B^2, xB+ yB = B$ then there exists unique ring map $f: A \to B$ such that $f(a) = x, f(b) = y$. I don't know how to reach a contradiction from there.
Or does anyone have any great method for proving a not representable functor is not representable in general? I find this type of problems really hard.
Best Answer
Some Motivation
Recall that a (covariant) functor $F \colon \mathscr{C} \to \mathsf{Set}$ is representable by an object $A$ of $\mathscr{C}$ if and only if there exists a universal element $a \in F(A)$. This means that there exists for every other object $X$ of $\mathscr{C}$ and every element $x \in F(X)$ a unique morphism $f \colon A \to X$ in $\mathscr{C}$ with $F(f)(a) = x$. The natural bijection $\operatorname{Hom}_{\mathscr{C}}(A, -) \to F$ is then given by $f \mapsto F(f)(a)$.
A useful example to keep in mind in the functor $F \colon \mathsf{Ring} \to \mathsf{Set}$ given by $F(R) = R^n$. This functor is represented by the ring $A = \mathbb{Z}\langle t_1, \dotsc, t_n \rangle$, the (non-commutative) polynomial ring in the variables $t_1, \dotsc, t_n$. The (usual) universal element $a \in F(A)$ is given by $a = (t_1, \dotsc, t_n)$. And indeed, that this is a universal element means precisely that there exists for every other ring $R$ and every element $x \in F(R)$ with $x = (x_1, \dotsc, x_n)$ a unique ring homomorphism $f \colon A \to R$ with $F(f)(a) = x$, i.e. with $f(t_i) = x_i$ for every $i = 1, \dotsc, n$. And this is precisely how the ring $A$ represents the functor $F$.
The Problem
We show more generally that for every number of elements $n \geq 2$ the functor \begin{align*} F \colon \mathsf{Ring} &\to \mathsf{Set}, \\ R &\mapsto \{ (x_1, \dotsc, x_n) \in R^n \mid x_1 R + \dotsb + x_n R = R \} \end{align*} is not representable. Assume otherwise that the functor $F$ is representable by a ring $A$ and let $(a_1, \dotsc, a_n) \in F(A)$ be the universal element (corresponding to some choice of isomorphism $F \cong \operatorname{Hom}_{\mathsf{Ring}}(A,-)$).
We will consider some auxilary functors which are representable: We can consider for every index $i = 1, \dotsc, n$ the functor \begin{align*} E_i \colon \mathsf{Ring} &\to \mathsf{Set}, \\ R &\mapsto \{ (x_1, \dotsc, x_n) \in R^n \mid \text{$x_i$ is a unit in $R$} \} \,. \end{align*} This functor is representable by the ring $U_i := \mathbb{Z}\langle t_1, \dotsc, t_n, t_i^{-1} \rangle$. We can also consider for any two indices $i$, $j$ with $1 \leq i \neq j \leq n$ the functor \begin{align*} E_i \colon \mathsf{Ring} &\to \mathsf{Set}, \\ R &\mapsto \{ (x_1, \dotsc, x_n) \in R^n \mid \text{$x_i$ and $x_j$ are units in $R$} \} \,. \end{align*} This functor is representable by the ring $U_{ij} := \mathbb{Z}\langle t_1, \dotsc, t_n, t_i^{-1}, t_j^{-1} \rangle$. We fix for the rest of this argumentation two such indices $i$, $j$. (This is where we use that $n \geq 2$.)
We have inclusions of functors as follows:
These inclusions correspond to ring homomorphism between their representing objects:
The inclusion $E_i \subseteq E_{ij}$ correspond to the canonical ring homomorphisms $U_i \to U_{ij}$, and the inclusion $E_i \subseteq F$ correspond to the ring homomorphisms $f_i \colon A \to U_i$ with $f(a_k) = t_k$ for every index $k = 1, \dotsc, n$. (Such a homomorphism exists because it is the unique homorphism $f \colon A \to U_i$ with $F(f)( (a_1, \dotsc, a_n) ) = (t_1, \dotsc, t_n)$. So here we use that $(a_1, \dotsc, a_n)$ is a universal element and that $(t_1, \dotsc, t_n)$ is an element of $F(U_i)$.) Similar for $U_j$ instead of $U_i$.
The commutativity of the above diagram gives (by the faithfulness of the Yoneda embedding) the commutativity of the corresponding diagram:
This means that $f_i(a) = f_j(a)$ for every element $a \in A$. It follows that $f_i$ and $f_j$ restrict to the same ring homomorphism $f \colon A \to U_i \cap U_j$, with the intersection $U_i \cap U_j$ taken in $U_{ij}$. This intersection is precisely the usual polynomial ring $U := \mathbb{Z}\langle t_1, \dotsc, t_n \rangle$. We have seen in the above explicit description of the homomorphism $f_i$ (and $f_j$) that the homomorphism $f \colon A \to U$ is given by $f(a_k) = t_k$ for every index $k = 1, \dotsc, n$. The existence of such a homorphism means that $(t_1, \dotsc, t_n) \in F(U)$ bescause the ring $A$ represents the functor $F$ via the universal element $(a_1, \dotsc, a_n) \in F(A)$. But this would means that $U = t_1 U + \dotsb + t_n U$, which is not the case.
We hence see that such a representing object $A$ cannot exist.