# Extending effective Cartier divisors

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.

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

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 $$D=D_1$$.
• I do not understand why $f \in (x^d, y)$? It should be a degree d polynomial in $(x,y)$. Commented May 3, 2022 at 18:30
• 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 $d-1$. Commented May 4, 2022 at 10:06
• 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. Commented May 4, 2022 at 11:55
• 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 $x^{d-1}$. Commented May 5, 2022 at 11:47