$A \colon= {\Bbb C}[S_1,\ldots,S_n]$ 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 '17 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 '17 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 '17 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 '17 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 '17 at 14:58

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.