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? 



Generic Algebraic Hyperplane Section of a Prime Ideal
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](http://stacks.math.columbia.edu/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.


