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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?

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    Since $\textrm{PGL}_n(R)$ is the quotient of $\textrm{GL}_n(R)$ by the action of $R^*$, one expects the corresponding $k$-algebra to be the invariants under $R^*$? But that can't possibly be right. Could you perhaps explain in more detail why the conditions of the special adjoint functor theorem are satisfied? It is not obvious to me that $\textrm{Alg}(k)$ has a coseparating set.2012-03-16
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    Thanks for the comment about the coseparating set; now I've proven it with Freyd's representability criterion. // Meanwhile I've realized that the proof can be made mor explicit and indeed we get the $k^*$-invariants of $k[\{X_{ij}\}]$ as an representing object.2012-03-18
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    Glad I could help. I guess, when $k$ is infinite, they must be the homogeneous polynomials of degree a multiple of $n$, divided by an appropriate power of the determinant.2012-03-18

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