Suppose $X, Y\subseteq \mathbb{P}_\mathbb{C}^n$ are complex projective varieties that meet transversally. That is, for each point $p\in X\cap Y,$ $X$ and $Y$ are both smooth at $p$, and their tangent spaces at $p$ satisfy $T_p(X) + T_p(Y) = T_p(\mathbb{P}^n).$ Denote by $I(X\cap Y), I(X), I(Y)$ the homogeneous ideals of the algebraic sets $X\cap Y, X, Y$, respectively. E.g., with coordinates $\mathbb{P}^n(x_0:\ldots :x_n)$, we have $$I(X) = (\{f\in \mathbb{C}[x_0,\ldots,x_n] \mid f \text { homogeneous and } f(p) = 0 \text{ for every } p\in X\}).$$ At the moment I am haunted by the following seemingly elementary question:
In the above situation, is it true that $I(X\cap Y) = I(X) + I(Y)$?
This is certainly not true without transversality, which motivates the definition of scheme theoretic intersection. My main interest in this question comes from an application of Bertini's theorem, wherein $X$ is smooth and $Y$ is a general hyperplane. I feel this must be true but I am struggling to come up with a proof or find a reference. Thanks in advance for any help!
This is in general false. Let $I_X$ denote the ideal sheaf of $X$ and $I_H=\mathcal{O}_{\mathbb{P}^n}(-1)$ that of a general hyperplane. Let $I_{X\cap H}$ the ideal sheaf of the intersection. Then, you have an exact sequence, $0\to I_X(-1)\to I_X\oplus I_H\to I_{X\cap H}\to 0$. Taking twists and global sections, we get a long exact sequence, $0\to I(X)\to I(X)\oplus I(H)\to I(X\cap H)\to H^1_*(I_X)\stackrel{H}{\to} H^1_*(I_X)$. So, if $H^1_*(I_X)\neq 0$, you get a counter example, since $H^1(I_X(r))=0$ for all large $r$, pick the largest $r$ for which this is non-zero. Then the map $H^1(I_X(r))\stackrel{H}{\to} H^1(I_X(r+1))$ is zero and so the previous map can not be surjective.
The simplest such example is a smooth rational curve embedded in $\mathbb{P}^3$ as a quartic.