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The following question is from the Fall 2016 UCLA algebra qualifying exam:

Let $F$ be a field and $a\in F$. Show that the functor that takes $R$, commutative $F$-algebras to the invertible elements of $R[X]/(X^2-a)$ is representable.

What I have so far: If $a\in F$, then then we have that $R[X]/(X^2-a)\cong R\times R$. Hence we get that $$ Hom_{F-\text{alg}}(F[t,t^{-1}]\otimes_FF[t,t^{-1}],R)\cong Hom_{F-\text{alg}}(F,R\times R)\cong U(R)$$ where $U$ is the functor that take $R$ to units of $R[X]/(X^2-a)$. Hence in this case, the functor is representable.

I'm unsure how to extend this to the general case.

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  • $\begingroup$ Do you mean if $a\in F^2$? $\endgroup$ – Arnaud D. Mar 21 at 20:37
  • $\begingroup$ The question says $a\in F$. $\endgroup$ – Leon Sot Mar 21 at 21:47
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    $\begingroup$ I meant after "What I have so far". $\endgroup$ – Arnaud D. Mar 21 at 22:01
  • $\begingroup$ Sorry, I meant that $a$ has a square root in $F$. $\endgroup$ – Leon Sot Mar 22 at 1:45
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First of all observe that if $A$ is a ring then naturally $ A^\times \cong \{(x,y)\in A^2: xy=1\}. $

One always has $R^2\cong R[X]/(X^2-a)$ as $R$-modules via $(a_1,a_0)\mapsto a_1X+a_0$, and an element corresponding to a pair $(a_1,a_0)$ is invertible if and only if there exists $(b_1,b_0)\in R^2$ such that $$ (a_1b_0+a_0b_1,a_0b_0+aa_1b_1-1)=(0,0) \hspace{1cm} (\ast) $$ The $F$-algebra $F[A_0,A_1,B_0,B_1]/(A_1B_0+A_0B_1,A_0B_0+aA_1B_1-1)$ represents the desired functor on $F$-algebras $$ R\mapsto \{(a_1,a_0,b_1,b_0)\in R^4 \text{ satisfying } (\ast)\}\cong \{(\gamma,\delta)\in (R[X]/(X^2-a))^2 : \gamma \delta=1\}. $$

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