Let $X$ be a $k$-scheme. We say that $X$ is geometrically irreducible if $X\times_k \mathrm{spec}{K}$ is irreducible for all algebraically closed extensions $K$ of $k$.
Ravi Vakil states in his notes, page 259 ex. 9.5.G that the set of points in $\mathbb{P}^{\binom{d+2}2-1}_k$ (which parametrizes closed subschemes of degree $d$ in $\mathbb{P}^2_k$) corresponding to geometrically irreducible curves is open. However in the case where $k$ is algebraically closed, geometric irreducibility is the same as plain irreducibility, and this set need not be open (curves given by polynomials $P^l=0$ where $\deg P=\frac{d}l$ and $P$ is irreducible are also irreducible).
Where is my mistake?