I have been trying to understand the proof of Gabriel's Theorem from Ralf Schiffler's 'Quiver Representations'. In chapter 8, he has defined, for a finite quiver $Q=(Q_0,Q_1)$ without oriented cycles and for a (fixed) dimension vector $\textbf{d}:= (d_i)_{i \in Q_0} \in \mathbb{Z}_{\geq 0}^n$, the space $E_\textbf{d}$ of representations $M$ of $Q$ having dimension vector equal to $\textbf{d}$, and the group $G_\textbf{d} := \prod_{i \in Q_0} GL_{d_i} (k)$, which acts on $E_\textbf{d}$ by conjugation: that is, for $g:=(g_i)_{i \in Q_0} \in G_\textbf{d}$ and $M := (M_i, \phi_\alpha)_{i \in Q_0, \alpha \in Q_1} \in E_\textbf{d}$, $$g \cdot M := (M_i, \hspace{1mm} g_{t(\alpha)} \hspace{0.5mm} \phi_\alpha \hspace{0.5mm} g_{s(\alpha)}^{-1})_{i \in Q_0, \alpha \in Q_1} \in E_\textbf{d}$$ where the source and target of arrow $\alpha \in Q_1$ are given by $s(\alpha)$ and $t(\alpha)$ respectively.
In the proof of Lemma 8.2, it is stated that $E_\textbf{d}$ is an irreducible algebraic variety and that a codimension zero orbit is open in $E_\textbf{d}$. Unfortunately, all I know is the definition of an algebraic variety (the set of common zeroes of a system of polynomial equations over the real or complex numbers) and of an irreducible algebraic variety, and while it seems that $E_\textbf{d}$ gets the structure of a variety, being the set of common zeroes in polynomials involving the $\phi_\alpha$'s, which arise out of commutativity conditions in the definition of a morphism (is this correct?), it is not clear to me why $E_\textbf{d}$ must be irreducible as an algebraic variety or why a codimension zero orbit should be open in $E_\textbf{d}$ (is $E_\textbf{d}$ getting the structure of a topological space?), and I would really appreciate some help in this regard.