# Show that the functor that takes $R$ to the set of invertible elements of $R[X]/(X^2-a)$ is representable.

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.

• Do you mean if $a\in F^2$? – Arnaud D. Mar 21 at 20:37
• The question says $a\in F$. – Leon Sot Mar 21 at 21:47
• I meant after "What I have so far". – Arnaud D. Mar 21 at 22:01
• Sorry, I meant that $a$ has a square root in $F$. – Leon Sot Mar 22 at 1:45

## 1 Answer

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 $$R\mapsto \{(a_1,a_0,b_1,b_0)\in R^4 \text{ satisfying } (\ast)\}\cong \{(\gamma,\delta)\in (R[X]/(X^2-a))^2 : \gamma \delta=1\}.$$