Let $X$ be a non-singular, quasi-projective variety (over $\mathbb{C}$) of dimension at least $3$, $D_1, D_2$ are integral effective divisors in $X$ with $D_1 \cap D_2$ of codimension $2$ in $X$. Let $C \subset D_1$ be an integral closed subscheme in $D_1$ of codimension $1$. Note that, as $C$ is irreducible and contained in $D_1$, we have $C \cap (D_2 \backslash D_1 \cap D_2)=\emptyset$. Does there exist an effective Cartier divisor $D \subset X$ containing $C$ such that $D \cap (D_2 \backslash D_1 \cap D_2)=\emptyset$ and $D \cap D_1$ is of codimension $2$ in $X$? Moreover, if $C$ is non-singular can we get such a $D$ which is also non-singular?

If necessary, assume that $D_1, D_2$ and $D_1 \cap D_2$ are non-singular as well.

  • 1
    $\begingroup$ Any other conditions on $D$? E.g. why doesn't $D = D_1$ work? $\endgroup$
    – pinaki
    Commented Apr 19, 2022 at 13:00
  • $\begingroup$ @pinaki Thanks, I have made an edit. I have added the condition $D \cap D_1$ is of codimension $2$ in $X$. $\endgroup$
    – user45397
    Commented Apr 19, 2022 at 13:06

1 Answer 1


Take two planes $D_1 = V(x),D_2=V(y)\subset \mathbb{P}^3_{(x:y:z:t)}$ and consider the line $C=V(x,z)\subset D_1$. The equation of any hypersurface $D=V(f)$ of degree $d$ which only meets $D_2$ in the line $D_1\cap D_2$ must be contained in the ideal $f\in(x^d,y)$. Similarly $C\subset D$ implies that $f\in (x,z)$, and so we have $f\in (x^d,xy,yz)$. The only way that $D$ can be nonsingular at the point $C\cap D_2=(0:0:0:1)$ is if $d=1$, but this implies that $D$ is a plane, and thus $D=D_1$.

  • $\begingroup$ I do not understand why $f \in (x^d, y)$? It should be a degree d polynomial in $(x,y)$. $\endgroup$
    – user45397
    Commented May 3, 2022 at 18:30
  • $\begingroup$ It means that it is a polynomial of the form $f = Ax^d + B_{d-1}y$ for $A\in \mathbb C$ constant and $B_{d-1}\in \mathbb C[x,y,z,t]$ a homogeneous polynomial of degree $d-1$. $\endgroup$
    – Tom Ducat
    Commented May 4, 2022 at 10:06
  • $\begingroup$ But this is not true. A general hypersurface that meets $D_2$ in the line $D_1 \cap D_2$ is of the form $f=Ax+By$, where $A,B$ are degree $d-1$ polynomials in $\mathbb{C}[x,y,z,t]$. It need not have an $x^d$-term. $\endgroup$
    – user45397
    Commented May 4, 2022 at 11:55
  • $\begingroup$ Yes, but you asked to satisfy the condition $D\cap (D_2\setminus(D_1\cap D_2)) = \emptyset$, which means that $D$ can only meet $D_2$ in the line $D_1\cap D_2$. If $f=Ax+By$ then $D\cap D_2 = V(x,y) \cup V(A,y)$ and the condition $V(A,y)\subset V(x,y)$ implies that $A$ is a scalar multiple of $x^{d-1}$. $\endgroup$
    – Tom Ducat
    Commented May 5, 2022 at 11:47
  • $\begingroup$ Thanks for the explanation. I understand. $\endgroup$
    – user45397
    Commented May 5, 2022 at 21:57

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