$A \colon= {\Bbb C}[S_1,\ldots,S_n]$ with $1 \leq n < \infty$

$B \colon= A[X_1,\ldots,X_d]$ with $2 \leq d < \infty$.

$O \colon= (0,\ldots,0)$ be the origin of ${\mathrm{Spec}}\,B$.

Suppose ${\frak P}$ be a prime ideal of $B$ such that ${\frak P} \cap A = 0$.

Scheme-theoretically, this is equivalent to the following condition$\colon$ ${\mathrm{Spec}}\,B/{\frak P} \to {\mathrm{Spec}}\,A$ is dominant, i.e. surjective after taking the closure of the image.

Let $H$ be a hyperplane of ${\mathrm{Spec}}\,B$ whose defining equation $h$ satisfies the following condition$\colon$ $h = \Sigma_{i=1}^{i=d}a_iX_i\, ;\, a_i \in B, \phantom{i}^{\exists}a_i \not= 0 \,\,{\mathrm{s.t.}}\,\, a_i \in {\Bbb C}$.

and define the intersection $D \colon= ({\mathrm{Spec}}\,B/{\frak P}) \cap H$. There is a natural morphism $\phi \colon D \to {\mathrm{Spec}}\,A$.

Q. Is it possible to choose $H$ such that the following three conditions are satisfied?

(i) $D$ passes through $O$.

(ii) $D$ is irreducible.

(iii) $\phi$ is dominant.

  • $\begingroup$ What is a hyperplane section passing through the origin means for you? Is it of the form $\sum a_iX_i$? Are $a_i$s allowed to be in $A$ or just $\mathbb{C}$? $\endgroup$
    – Mohan
    Oct 22, 2017 at 17:36
  • $\begingroup$ The condition of the coordinate of H is as follows: $a_i \in B$ but at least one of $a_i$ belongs to ${\Bbb C}$. That is at least one linear term of variables $X_1,\ldots,X_d$ must appear. $\endgroup$
    – Rinmyaku
    Oct 22, 2017 at 22:29
  • 1
    $\begingroup$ I presume when you said $a_i\in\mathbb{C}$, you meant $a_i\neq 0$ too. In which case, in general you can not expect (iii) to hold. $\endgroup$
    – Mohan
    Oct 23, 2017 at 0:37
  • $\begingroup$ What is the meaning of ``in general, (iii) cannot hold''? Does that mean that for a certain ${\frak P}$, it can happen that there is NO hyperplane H under the above conditions? Or does that mean that there ALWAYS exist such H but NOT generaically in original Bertini theorem? $\endgroup$
    – Rinmyaku
    Oct 23, 2017 at 14:32
  • $\begingroup$ Can you also explain what you mean when you say $D$ passes through the origin? This is not a closed point, so do you mean the prime ideal is contained in $(X_1,\ldots, X_d)$? $\endgroup$
    – Mohan
    Oct 23, 2017 at 14:58

1 Answer 1


The answer to the question Q is "no". Take $n=d=2$. Let $\frak P$ be the ideal generated by $X_1^2+X_2^2+S_1$ and $X_1^3+X_2^3+S_2$. Substituting any two complex numbers $s_1$ and $s_2$ for $S_1$ and $S_2$, respectively, we obtain an equation in the unknowns $X_1$ and $X_2$ which has a solution. Namely, we have $X_1=\sqrt{s_1-X_2^2}$ and $(s_1-X_2^2)^\frac32+X_2^3=s_2$, which is easy to solve. Thus the map ${\mathrm{Spec}}\,B/{\frak P} \to {\mathrm{Spec}}\,A$ is surjective, so ${\frak P}\cap A=(0)$.

Next, consider a ``hyperplane section'' $H$ of the form given in the question. The ideal $\frak P$ is prime of height 2 since it can be transformed into the ideal $(S_1,S_2)$ by an automorphism of $B$. Since $h\notin\frak P$, we have $ht({\frak P}+(h))=3$. Thus $\dim\frac B{{\frak P}+(h)}=4-3=1$, so the homomorphism $A\longrightarrow\frac B{{\frak P}+(h)}$ cannot be injective. We obtain that $\phi$ cannot be dominant.


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