Let $F, G \subset \mathbb{A}^2$ be affine plane curves that share no common tangents at a point $P \in \mathbb{A}^2$. Let $I$ be the ideal of $k[X, Y]$ generated by $X$ and $Y$. By a change of coordinates, we may assume $P = (0, 0)$. Write $m$ (resp. $n$) for the multiplicity of $F$ (resp. $G$) at $P$. It is a lemma in Fulton's Algebraic Curves (chapter 3, section 3) that if $t \geq m + n - 1$, then $I^t \subset (F, G)\mathcal{O}_P(\mathbb{A}^2)$. I am having trouble understanding the proof of this lemma, which I paraphrase here:
To simplify notation, let $\mathcal{O} = \mathcal{O}_P(\mathbb{A}^2)$. Let $L_1, \dots, L_m$ be the tangents to $F$ at $P$, and $M_1, \dots, M_n$ be the tangents to $G$ at $P$. Furthermore, let $L_i = L_m$ if $i \geq m$, and $M_j = M_n$ if $j \geq n$. Set $A_{ij} = L_1 \cdots L_i M_1 \cdots M_j$. Because no $L_i = M_j$ for any $i, j$, it is known that $\{A_{ij} | i + j = t\}$ forms a basis for the vector space of all forms of degree $t$ in $k[X, Y]$. It thus suffices to show that $A_{ij} \in (F, G)\mathcal{O}$ for all $i + j \geq m + n - 1$. But $i + j \geq m + n - 1$ means $i \geq m$ or $j \geq n$, so say for example $ i \geq m$. Then $A_{ij} = A_{m0}B,$ where $B$ is a form of degree $t = i + j - m$. Write $F = A_{m0} + F'$, where all terms of $F'$ are of degree $\geq m + 1$. Then $A_{ij} = BF - BF'$, where each term of $BF'$ has degree $\geq i + j + 1$.
The claim is that at this point, it suffices to show that $I^t \subset (F, G)\mathcal{O}$ for all sufficiently large $t$ (i.e., there is some $N$ for which $I^t \subset (F, G) \mathcal{O}$ whenever $t \geq N$). This is a consequence of the Nullstellensatz, but I fail to see how that implies the lemma. Any help is appreciated.
Edit: I believe I understand it now. This is a descending argument by induction: suppose that $I^{t+1} \subset (F, G) \mathcal{O}$. Then $BF' \in (F, G) \mathcal{O}$, so $I^t \subset (F, G)\mathcal{O}$. The base case is handled by the "sufficiently large" part.