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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?

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    $\begingroup$ I think that "irreducible" here is used in the sense of "reduced and irreducible". $\endgroup$ Jan 8, 2015 at 14:05
  • $\begingroup$ @FrancescoPolizzi Yes, probably, a curve is a variety hence reduced, Thanks! Also, in his notes Vakil claims that even if we take an algebraically closed field, there is a difference between geometric irreducibility and irreducibility. I suspect that it is not true. $\endgroup$
    – peter
    Jan 8, 2015 at 20:12

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