# Multiplicity under polynomial maps

I'm reading Fulton's book and I'm trying to solve the following exercise :

Let $T : \Bbb A^2 \rightarrow \Bbb A^2$ be a polynomial map, $T (Q) = P$.

(a) Show that $m_Q(F^T ) \geq m_P (F)$.

(b) Let $T = (T1,T2)$, and define $J_QT = (\partial T_i /\partial X_j (Q))$ to be the Jacobian matrix of T at Q.

Show that $m_Q(F^T ) = m_P (F)$ if $J_QT$ is invertible.

a) For the first question I can write $\partial^n F^T/ \partial X_j^n (Q)$ in terms of $\partial^n F / \partial X_j^n(P)$ (using for example Faà di Bruno's formula). For the first derivative for instance one can easily show that $F_X^T(Q) = F_X(P)T_X(Q)$ and thus : If $\partial^n F^T/ \partial X_j^n (Q) \neq 0$ Then there exists an $m \leq n$ such that $\partial^m F / \partial X_j^m(P) \neq 0$.

b) Do you have some suggestions for the second question ?

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a) Let $T : \Bbb{A}^2 \rightarrow \Bbb{A}^2$ be the polynomial map where $T(Q)=P$. We want to show that $m_Q(F^T) \geq m_P(F)$. We show this first for $Q = (0,0)$ and $P = (0,0)$. Let $T = (T_1, T_2)$. We have that $T(0,0) = (0,0)$. So $T_1$ and $T_2$ have no constant term. Now we have that $F^T = F(T_1(X,Y),T_2(X,Y)$ and $T_1$ and $T_2$ have no constant term. So $T$ does not decrease the lowest degree term of $F$. Therefore $m_Q(F^T) = m_P(F)$.
Now assume either $Q \neq (0,0)$ or $P \neq (0,0)$. Let $T'$ be the translation that takes $(0,0)$ to $Q$ and $T''$ be the translation that take $(0,0)$ to $P$. Then we have that $T \circ T'(0,0) = T''(0,0)$. Now we have that $m_Q(F^T) = m_{(0,0)}(F^{T\circ T'})$ and $m_P(F) = m_{(0,0)}(F^{T''})$. Also we know that $$(T'')^{-1}\circ T \circ T'(0,0) = (0,0).$$ Let $G = F^{T''}$. Then we have that $m_{(0,0)}(F^{T\circ T'}) = m_{(0,0)}(G^{(T'')^{-1}\circ T \circ T'})$. Now let $\bar{T} = (T'')^{-1} \circ T \circ T'$. So $\bar{T}(0,0) = (0,0)$. So by the ealier case, we have that $m_{(0,0)}(G^{\bar{T}}) \geq m_{(0,0)}(G)$. Therefore $m_Q(F^T) \geq m_P(F)$.
b) Let $T = (T_1,T_2)$ and $$J_Q T = \begin{pmatrix} \dfrac{\partial T_1}{\partial X}(Q) & \dfrac{\partial T_1}{\partial Y}(Q) \\ \dfrac{\partial T_2}{\partial X}(Q) & \dfrac{\partial T_2}{\partial Y}(Q) \end{pmatrix}.$$ As before we show the result for the case $P = (0,0)$ and $Q = (0,0)$.
Assume $J_Q T$ is invertible. Since $J_Q T$ is invertible, we can't have both $\dfrac{\partial T_1}{\partial X}(Q) = 0$ and $\dfrac{\partial T_1}{\partial Y}(Q) = 0$ or both $\dfrac{\partial T_2}{\partial X}(Q) = 0$ and $\dfrac{\partial T_2}{\partial Y}(Q) = 0$. Since $Q = (0,0)$, this implies that the decomposition of $T_1$ and $T_2$ into homogeneous polynomials are $$T_1 = G_1 + G_2 + G_3 + \ldots$$ and $$T_2 = H_1 + H_2 + H_3 + \ldots$$ where the $G_i$'s and the $H_i$'s are forms. By part (a), $T_1$ and $T_2$ have no constant terms. Thus, $$F^T = F(T_1,T_2) = F_m(T_1,T_2) + F_{m+1}(T_1,T_2) + \ldots$$ Since the lowest degree forms of $T_1$ and $T_2$ are of degree $1$, we have that $T$ does not decrease of the form $F_m(T_1,T_2)$. Similarly, $T$ does not decrease the degree of $F_{m+1}(T_1,T_2), F_{m+2}(T_1,T_2), \ldots$. Therefore we have that $m_{(0,0)}(F^T) = m_{(0,0)}(F)$.
Now assume that either $Q = (a_1,b_1) \neq (0,0)$ or $P = (a_2,b_2) \neq (0,0)$. Assume that $J_Q T$ is invertible. Let $T'$ be the translation that takes $(0,0)$ to $Q$ and $T''$ be the translation that takes $(0,0)$ to $P$. Then we have that $$\begin{pmatrix} \dfrac{\partial T_1}{\partial X}(X+a_1) & \dfrac{\partial T_1}{\partial Y}(X+a_1) \\ \dfrac{\partial T_2}{\partial X}(Y+b_1) & \dfrac{\partial T_2}{\partial Y}(Y+b_1) \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$$ and $$\begin{pmatrix} \dfrac{\partial T_1}{\partial X}(X+a_2) & \dfrac{\partial T_1}{\partial Y}(X+a_2) \\ \dfrac{\partial T_2}{\partial X}(Y+b_2) & \dfrac{\partial T_2}{\partial Y}(Y+b_2) \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}.$$ So by the chain rule, we have that $J_Q T = J_{(0,0)}((T'')^{-1} \circ T \circ T')$. Letting $\bar{T} = (T'')^{-1} \circ T \circ T'$, we have that $J_0 \,\bar{T}$ is invertible and the result follows from our ealier case.