# 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?

• Doesn't this follow from the Bertini theorems for irreducibility and reducedness, cf. e.g. Cor. I.6.11 of Jouanolou's book Théorèmes de Bertini et applications? You have to assume the field is infinite; for counterexamples over finite fields see for example Poonen's papers. – R. van Dobben de Bruyn Jul 17 '17 at 7:40
• Actually I'm not sure if the Bertini theorems are saying the exact thing that you want. It's clear that if $S$ is a domain, then $\operatorname{Proj}(S)$ is an integral scheme, but the converse sounds false. Perhaps throwing in words like 'projectively normal' can help here, but I don't know how exactly. – R. van Dobben de Bruyn Jul 17 '17 at 16:49
• Another footnote is that Bertini reducedness theorems only consider geometric reducedness. This is equivalent to reducedness when the field is perfect, but it seems that there should actually be counterexamples to a Bertini reducedness theorem (without the adjective geometric) when the field is imperfect. – R. van Dobben de Bruyn Jul 17 '17 at 16:50

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.