Let $C$ be a smooth projective (irreducible) curve in $\mathbb{P}^n$ for some $n$. Denote by $I_C$ the ideal of $C$. Let $g \in I_C\backslash I_{C}^2$, an irreducible element. Is it true that for any positive integer $m$, $I_C^m \cap (g) \cong (g)I_C^{m1}$?

$\begingroup$ math.stackexchange.com/questions/1164842/… $\endgroup$– user26857Feb 27 '15 at 2:04
Denote by $J$ the ideal generated by $g$ and by $\mathcal{J}$ the associated coherent sheaf. Note first, if $I_C^m \cap J \not= I_C^{m1}J$ then there exists a closed point $x \in C$ and $h \in \mathcal{I}_{C,x}^m \cap \mathcal{J}_x$ not contained in $\mathcal{I}_{C,x}^{m1}\mathcal{J}_x$. Since $C$ is smooth, there exists a regular sequence consisting of $n1$ elements $f_1,..,f_{n1}$ in $\mathcal{O}_{\mathbb{P}^n,x}$ such that $\mathcal{I}_{C,x}$ is generated by these elements. Use Hartshorne's Algebraic Geometry, Theorem II.$8.21$A to see that there exists an isomorphism $$(\mathcal{O}_{\mathbb{P}^n,x}/\mathcal{I}_{C,x})[t_1,...,t_{n1}] \to \oplus_{n \ge 0} \mathcal{I}_{C,x}^n/\mathcal{I}_{C,x}^{n+1}$$ defined by sending $t_i$ to $f_i$. Furthermore, the symmetric power $S^n(\mathcal{I}_{C,x}/\mathcal{I}_{C,x}^{2}) \cong \mathcal{I}_{C,x}^n/\mathcal{I}_{C,x}^{n+1}$.
Denote by $g$ again the generator of $\mathcal{J}_x$. Then, $h=gh'$ for some $h' \in \mathcal{O}_{\mathbb{P}^n,x}$. Let $m_0$ be the maximal integer such that $h' \in \mathcal{I}_{C,x}^{m_0}\backslash \mathcal{I}_{C,x}^{m_0+1}$, i.e., $h'$ is nonzero in $\mathcal{I}_{C,x}^{m_0}/\mathcal{I}_{C,x}^{m_0+1}$. By the definition of $h'$, $m_0$ must be strictly less than $m1$.
Since $\mathcal{O}_{C,x}$ is an integral domain, so is $\mathcal{O}_{C,x}[t_1,...,t_{n1}]$. Using the isomorphism mentioned above $h=gh'$ is then nonzero in $\mathcal{I}_{C,x}^{m_0+1}/\mathcal{I}_{C,x}^{m_0+2}$. Using the direct sum decomposition above, $m_0+1$ must be greater than or equal to $m$. But this contradicts the bound on $m_0$ above. So, there does not exist any such $h$.
Is this correct?