Generic Algebraic Hyperplane Section of a Prime Ideal Let $P\subset k[x_0,\dots, x_n]=:S$ be a prime ideal such that
$\dim( \text{Proj}(S/P))\geq 2$.
Is it true that for a general linear form $H\in S_1$ we have that
the ideal $\langle P, H \rangle$ is again a prime ideal? 
If not, for what kind of projective schemes $\text{Proj}(S/P)$ we can expect this? 
 A: This type of theorem is called a Bertini theorem, and there are versions for smoothness, (geometric) irreducibility, and geometric reducedness. Over a perfect field, geometric reducedness is equivalent to reducedness [Tag 035X]. However, my last example below shows that this distinction is important.
Using part (1) and (2) of Cor. I.6.7 of Jouanolou's Théorèmes de Bertini et applications, one can prove that the result is true when $k$ is infinite and perfect. Note that I am citing the affine version, and not the projective one, because we want to know whether $S/I$ is a domain; not whether $\operatorname{Proj}(S/I)$ is an integral scheme. (I don't think these questions are equivalent.)
For counterexamples when $k$ is finite, see Poonen's papers on Bertini theorems over finite fields (these counterexamples are constructed using what he calls anti-Bertini theorems).
There are also counterexamples when $k$ is imperfect, related to the difference between reducedness and geometric reducedness in this case. A low-dimensional example is given by the $\mathbb F_2(t)$-algebra
$$\mathbb F_2(t)[x,y]/(x^2-ty^2).$$
It is a domain because $ty^2$ is not a square, but if we divide out $\lambda x + \mu y$ with $\lambda,\mu \in \mathbb F_p(t)$, we pick up nilpotents. Indeed, if $\lambda \neq 0$, then wlog $\lambda = 1$, so $x = -\mu y$, so $(\mu^2-t)y^2 = 0$, which forces $y^2 = 0$ (since $\mu^2 - t \neq 0$); yet $y \neq 0$. If $\lambda = 0$, then we get the algebra $\mathbb F_2(t)[x]/(x^2)$, which also has nilpotents. The analogous example over $\mathbb F_p(t)$ should be
$$\mathbb F_p(t)[x,y]/(x^p-ty^p),$$
but it's a bit harder to prove that this is a domain. There should be many high-dimensional examples as well. Note however that the reducedness statement (unlike the irreducibility statement) typically does not have a dimension assumption.
