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.

  • $\begingroup$ Do you mean if $a\in F^2$? $\endgroup$ – Arnaud D. Mar 21 '19 at 20:37
  • $\begingroup$ The question says $a\in F$. $\endgroup$ – Leon Sot Mar 21 '19 at 21:47
  • 1
    $\begingroup$ I meant after "What I have so far". $\endgroup$ – Arnaud D. Mar 21 '19 at 22:01
  • $\begingroup$ Sorry, I meant that $a$ has a square root in $F$. $\endgroup$ – Leon Sot Mar 22 '19 at 1:45

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

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.