Prove the functor is not representable Let $F$ be the functor from rings to sets that takes a ring R to the set of all pairs $(a,b)\in R \times R$ such that $aR + bR = R$.  Show that $F$ is not representable.
This is a 2019 UCLA qualifying exam question.  I have tried using that if $(a,b)$ is a universal element in a ring $R$, then so is $(b,a)$.
 A: For convenience, let $k$ be a field.

Lemma As subrings of $k(x,y)$, we have $k[x,y] = k[x,y,x^{-1}] \cap k[x,y,y^{-1}]$.
Proof. Clearly, we have $k[x,y] \subseteq k[x,y,x^{-1}] \cap k[x,y,y^{-1}]$. Next, let $p \in k[x,y,x^{-1}] \cap k[x,y,y^{-1}]$ be arbitrary. Then we have $p = f/x^n$ and $p = g/y^m$ for some $f,g \in k[x,y]$ and $n,m \geq 0$. This yields $y^m f = x^n g$, so $y^m$ divides $x^n g$, whence $y^m$ divides $g$, so $p = g/y^m \in k[x,y]$. $\square$
Corollary The following is a pullback square in the category of (commutative) rings:
$\require{AMScd}$
\begin{CD}
k[x,y] @>>> k[x,y,y^{-1}]\\
@VVV @VVV \\
k[x,y,x^{-1}] @>>> k(x,y)
\end{CD}

Now, suppose for contradiction that $F$ is representable. Then $F$ preserves limits, so in particular the following would be a pullback square in the category  of sets:
\begin{CD}
F(k[x,y]) @>>> F(k[x,y,y^{-1}]) \\
@VVV @VVV \\
F(k[x,y,x^{-1}]) @>>> F(k(x,y))
\end{CD}
Moreover, we see directly that $F$ preserves inclusions (if $f : R \to S$ is an inclusion, then $F(f) = (a,b) \mapsto (f(a),f(b)) : F(R) \to F(S)$ is an inclusion). Thus, we have
$$F(k[x,y]) = F(k[x,y,x^{-1}]) \cap F(k[x,y,y^{-1}])$$
However, this is not true, since $(x,y) \in F(k[x,y,y^{-1}]) \cap F(k[x,y,x^{-1}])$ but $(x,y) \notin F(k[x,y])$.
