Showing a Functor is not Representable 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. 
 A: 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.
A: A good technique is to compare to a similar functor which is representable, in this case $G(A)=\{(a,b,c,d)\in A^4:ac+bd=1\},$ which is represented by $B:=\mathbb{Z}[x,y,z,w]/(xz+yw-1)$. 
Given a putative representation $(A,(a,b))$ of $F$, we allegedly get a unique homomorphism $p:B\to A$ from any choice of $c,d\in A$ with $ac+bd=1$ from the fact that $B$ represents $G$. Now $p$ would induce a natural transformation between the functors represented by $A$ and $B$, witnessed for each $g:R\to S$ by commutative squares with one route 
$$\mathrm{Hom}(A,R)\to \mathrm{Hom}(B,R)\to \mathrm{Hom}(B,S)$$
and the other route
$$\mathrm{Hom}(A,R)\to \mathrm{Hom}(A,S)\to \mathrm{Hom}(B,S).$$
The morphisms are precomposition with $p$ or postcomposition with $g$. 
For any $f:A\to R$ with $f(a)=r_1$ and $f(b)=r_2$, denote $f(p(z))$ by $\hat{r_1}$ and $f(p(w))$ by $\hat{r_2}$. The naturality above says, in short, that for any $g:R\to S$, we have $(\widehat{g(r_1)},\widehat{g(r_2)})=(g(\hat{r_1}),g(\hat{r_2}))$. In particular, if $h:R\to S$ with $h(r_i)=g(r_i)$, we must have $h(\hat{r_i})=g(\hat{r_i})$. That is absurd. 
For instance, let $R=B$ and let $(r_1,r_2)=(x,y)$, and let $S=\mathbb{Z}\times\mathbb{Z}$, with $g,h:B\to  \mathbb{Z}\times\mathbb{Z}$ given by $g(x)=h(x)=(1,0),g(y)=h(y)=(0,1),g(z)=\widehat{(1,0)},g(w)=\widehat{(0,1)}$. Then we can define $h(z)=h(w)=(1,1)$ to get the desired counterexample, unless $\widehat{(1,0)}=\widehat{(0,1)}=(1,1)$, in which case we can let $h(z)=(1,0)$ or $h(w)=(0,1)$. 

What we have shown above is, in effect, that there exists no natural transformation $F\to G$ splitting the projection $G\to F$ that forgets about $c$ and $d$. However, it is a general fact that an epimorphism of representable functors must always admit a splitting, and indeed must always come from a split monomorphism in the domain category, as we demonstrated in the example above. In other words, there are very few epimorphisms between representable functors! This can be very useful-if a given functor is a quotient of a representable, an argument like this one will often work to blow things up. 
More generally still, it's extremely rare that a colimit of representable functors is still representable, though due to the idiosyncrasies of the UCLA curriculum you may be in the small population of people who studies representability without knowing the word colimit.
This is not, in fact, the most usual way of proving that a functor is not representable. For reasonable, familiar categories (except fields) it is a theorem that a (covariant) functor is representable if and only if it preserves products, equalizers, and satisfies something called the solution set condition, which would never fail in any example you're likely to see. Many functors (such as $F$) preserves products, as well, so your go-to technique in these problems is often to find an equalizer which is not preserved by the functor. However, actually computing equalizers involving rings like $B$ is pretty annoying, so I wasn't able even with a fair bit of work to get such an argument for this case.
