I´m studying Mumford's geometric invariant theory and I came to the following question. Let us suppose that $k$ is an algebraically closed field of characteristic 0, and $G$ is a reductive linear algebraic group acting trivially on a scheme $X$. Let us consider a quasi-coherent sheaf of algebras, $\mathcal{A}$, equipped with a G-linearization. We can form its relative spectrum $Y=\text{Spec}(\mathcal{A})$ and we have a natural $G$ action on $Y$ given by the $G$-linearization of $\mathcal{A}$. On the other hand, we have a canonical map $\mathcal{A}^{G}\hookrightarrow\mathcal{A}$ given by the inclusion so, there is an induced map between their spectrums $\text{Spec}(\mathcal{A})\rightarrow\text{Spec}(\mathcal{A}^{G})$. It is true that the above morphism is a good universal quotient.
If $X$ is just the spectrum of $k$ we have the affine GIT quotient. In the general construction of quotients, we consider an ample $G$-linearized sheaf and we compute the semistable points so, in the case of affine schemes we are considering the structural sheaf (which is ample) equipped with the trivial linearization. Can we just prove directly without any references to linearizations that $\text{Spec}(\mathcal{A})\rightarrow\text{Spec}(\mathcal{A}^{G})$ is a GIT quotient as happens in the affine case?
Edit: Is the structure sheaf of $\text{Spec}(\mathcal{A})$ ample?
Thank you for your time.