How to represent the naive PGL functor?

Let $k$ be a field. Consider the functor

$F : \mathrm{Alg}(k) \to \mathrm{Set}, ~ R \mapsto \mathrm{GL}_n(R) / R^*$

[You might call this $\mathrm{PGL}_n^{\text{naive}}$, since it does not coincide on the correct $\mathrm{PGL}_n = \mathrm{GL}_n / \mathbb{G}_m$ - only on local $k$-algebras. See also Milne's script on algebraic groups, Example I.9.5. But this is not really relevant for this question]

I claim that $F$ is representable by Freyd's representability criterion (Mac Lane, Categories for the Working Mathematician, Theorem V.6.3): It is easy to verify that $F$ preserves all limits. Every solution set for $\mathrm{GL}_n$ is also one for $F$, but $\mathrm{GL}_n$ is even representable.

Question. Which specific $k$-algebra represents $F$?

Edit. I think I can spell out the proof of the Theorem in this special case: Consider the category $\int F$ of elements of $F$: Objects are pairs $(R,s)$, where $R$ is a $k$-algebra and $s \in F(R)$. A morphism $(R,s) \to (R',s')$ is a map of $k$-algebras $R \to R'$ which maps $s \mapsto s'$. Now $\int F$ is complete (since $F$ is continuous). Besides, it has a weakly initial object, namely the representation of $\mathrm{GL}_n$, that is $w=(k[\{X_{ij}\}]_{\mathrm{det}},\overline{(X_{ij})})$. I hope the notation is clear. Now the proof of Theorem V.6.1 tells us how to construct an initial object: We have to take the equalizer $v$ of all endomorphisms of $w$ in $\int F$. This means that $v=(R,\overline{(X_{ij})})$, where $R \subseteq k[\{X_{ij}\}]_{\mathrm{det}}$ is the subalgebra whose elements are fix under every $k$-algebra homomorphism $k[\{X_{ij}\}]_{\mathrm{det}} \to k[\{X_{ij}\}]_{\mathrm{det}}$, $X_{ij} \mapsto P_{ij}$, such that $\overline{(P_{ij})} = \overline{(X_{ij})}$ in $\mathrm{PGL}_n$, i.e. $P_{ij} = \lambda X_{ij}$ for some $\lambda \in (k[\{X_{ij}\}]_{\mathrm{det}})^* = k^* \cdot \mathrm{det}^{\mathbb{Z}}$ (here we use that $\mathrm{det}$ is an irreducible polynomial).

In other words, $R$ consists of those polynomials in the $X_{ij}$ localized at $\mathrm{det}$ such that the substitution $X_{ij} \mapsto \lambda \cdot X_{ij}$ for some $\lambda \in k^*$, and also $X_{ij} \mapsto \mathrm{det} \cdot X_{ij}$, doesn't change them. Is there any way to make this algebra more explicit? Can be describe it by generators and relations?

Solution 1:

Argh! The premise of my question

It is easy to verify that $F$ preserves all limits.

is wrong. It is correct that $F$ preserves products, but equalizers are not preserved; see below. And this is the reason why the whole search for a representation issue doesn't make sense.

Let $f,g : R \to S$ be two homomorphisms of algebras. Their equalizer is $E = \{r \in R : f(r)=g(r)\}$. The equalizer of $F(f),F(g) : F(R) \to F(S)$ consists of those $[M]$, $M \in \mathrm{GL}_n(R)$, such that $g(M)^{-1} f(M) \subseteq S^* \subseteq \mathrm{GL}_n(S)$. But the image of $F(E) \to F(R)$ consists of those $[M]$, $M \in \mathrm{GL}_n(R)$, such that there is some $r \in R^*$ such that $r M \in \mathrm{GL}_n(E)$, which is easily seen to be equivalent to $g(M)^{-1} f(M) = g(r)^{-1} f(r)$ for some $r \in R^*$. Thus, if $R=S$, $g=\mathrm{id}$, on the one hand we have $f(M) \subseteq S^*$ and on the other hand $f(M) \in f(R^*)$. These conditions are different for $n \geq 2$. Perhaps someone can add an example.