Proof: Suppose $P = Q = (0,0)$. Then necessarily $T_1$ and $T_2$ are of the form $$T_1 = ... + ax+by$$ $$T_2 = ... + cx+dy$$ with no constant terms. Write $F = F_m + ... + F_n$ in terms of its form decomposition; then we see $F^T = F_m^T + ... + F_n^T$, and for each $i$, $F_i^T$ contributes to the sum forms of degree $≥i$. This shows the nonzero forms in $F^T$ are all of degree $≥m$, so that $m_{(0,0)} F^T ≥ m_{(0,0)} F$. Now consider the general case for points $P,Q$. Let $S_1, S_2$ be the translations mapping the origin to $P,Q$ respectively. Then $$m_Q F^T = m_{(0,0)} F^{T∘S_2} = m_{(0,0)} F^{S_1∘S_1^{-1}∘T∘S_2}$$$$= m_{(0,0)} (F^{S_1})^{S_1^{-1}∘T∘S_2} ≥ m_{(0,0)} F^{S_1} = m_P F$$ Now to consider the properties of the Jacobian: Again suppose $P = Q = (0,0)$, and factor $F_m$ into a product of linear factors. Since $F_i^T$ for $i > m$ only contribute forms of degree $> m$, it will suffice to show $F_m^T$ contributes a nonzero form of degree $m$ when the Jacobian is invertible. When the Jacobian is invertible, that is to say its rows are linearly independent, so that if $ex+fy$ is a nonzero linear term, then $T(ex+fy) = eT_1(x) + fT_2(y)$ will have a nonzero form of degree $1$ in its decomposition. Since the smallest nonzero form of a product is the product of the smallest nonzero forms, we see that $m_{(0,0)} F^T = m_{(0,0)} F$. The argument for the general case of $P, Q$ is exactly as before, together with the observations that (1) the Jacobian of a composition of polynomial maps is the same as the matrix product of the Jacobian of polynomial maps at the appropriate points, and (2) that the Jacobians of translations are the identity. As for the demonstration that the converse is not in general true, we note that in the example given, $m_{(0,0)} F = 1 = m_{(0,0)} y - x^4 = m_{(0,0)} F^T$ despite the fact that the Jacobian of $T$ at $(0,0)$ is the noninvertible $$J_{(0,0)}T = \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix}~~\square$$
Friday, April 29, 2016
Plane Curve Multiplicities Under Polynomial Mappings (3.1.8)
William Fulton Algebraic Curves, chapter 3, section 1, exercise 8:
MathJax TeX Test Page
Let $k$ be an algebraically closed field, $F∈k[x,y]$, and $T : \mathbb{A}^2 → \mathbb{A}^2$ be a polynomial map with $T(P)=Q$. Show that $m_P F^T ≥ m_Q F$, with equality achieved if the Jacobian of $T$, defined as the matrix $(dT_i / dx_j (Q))$ where $T = (T_1,T_2)$, is invertible. Show that the converse is false in considering $T = (x^2,y)$ and $F=y-x^2$, $P=Q=(0,0)$.
Proof: Suppose $P = Q = (0,0)$. Then necessarily $T_1$ and $T_2$ are of the form $$T_1 = ... + ax+by$$ $$T_2 = ... + cx+dy$$ with no constant terms. Write $F = F_m + ... + F_n$ in terms of its form decomposition; then we see $F^T = F_m^T + ... + F_n^T$, and for each $i$, $F_i^T$ contributes to the sum forms of degree $≥i$. This shows the nonzero forms in $F^T$ are all of degree $≥m$, so that $m_{(0,0)} F^T ≥ m_{(0,0)} F$. Now consider the general case for points $P,Q$. Let $S_1, S_2$ be the translations mapping the origin to $P,Q$ respectively. Then $$m_Q F^T = m_{(0,0)} F^{T∘S_2} = m_{(0,0)} F^{S_1∘S_1^{-1}∘T∘S_2}$$$$= m_{(0,0)} (F^{S_1})^{S_1^{-1}∘T∘S_2} ≥ m_{(0,0)} F^{S_1} = m_P F$$ Now to consider the properties of the Jacobian: Again suppose $P = Q = (0,0)$, and factor $F_m$ into a product of linear factors. Since $F_i^T$ for $i > m$ only contribute forms of degree $> m$, it will suffice to show $F_m^T$ contributes a nonzero form of degree $m$ when the Jacobian is invertible. When the Jacobian is invertible, that is to say its rows are linearly independent, so that if $ex+fy$ is a nonzero linear term, then $T(ex+fy) = eT_1(x) + fT_2(y)$ will have a nonzero form of degree $1$ in its decomposition. Since the smallest nonzero form of a product is the product of the smallest nonzero forms, we see that $m_{(0,0)} F^T = m_{(0,0)} F$. The argument for the general case of $P, Q$ is exactly as before, together with the observations that (1) the Jacobian of a composition of polynomial maps is the same as the matrix product of the Jacobian of polynomial maps at the appropriate points, and (2) that the Jacobians of translations are the identity. As for the demonstration that the converse is not in general true, we note that in the example given, $m_{(0,0)} F = 1 = m_{(0,0)} y - x^4 = m_{(0,0)} F^T$ despite the fact that the Jacobian of $T$ at $(0,0)$ is the noninvertible $$J_{(0,0)}T = \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix}~~\square$$
Proof: Suppose $P = Q = (0,0)$. Then necessarily $T_1$ and $T_2$ are of the form $$T_1 = ... + ax+by$$ $$T_2 = ... + cx+dy$$ with no constant terms. Write $F = F_m + ... + F_n$ in terms of its form decomposition; then we see $F^T = F_m^T + ... + F_n^T$, and for each $i$, $F_i^T$ contributes to the sum forms of degree $≥i$. This shows the nonzero forms in $F^T$ are all of degree $≥m$, so that $m_{(0,0)} F^T ≥ m_{(0,0)} F$. Now consider the general case for points $P,Q$. Let $S_1, S_2$ be the translations mapping the origin to $P,Q$ respectively. Then $$m_Q F^T = m_{(0,0)} F^{T∘S_2} = m_{(0,0)} F^{S_1∘S_1^{-1}∘T∘S_2}$$$$= m_{(0,0)} (F^{S_1})^{S_1^{-1}∘T∘S_2} ≥ m_{(0,0)} F^{S_1} = m_P F$$ Now to consider the properties of the Jacobian: Again suppose $P = Q = (0,0)$, and factor $F_m$ into a product of linear factors. Since $F_i^T$ for $i > m$ only contribute forms of degree $> m$, it will suffice to show $F_m^T$ contributes a nonzero form of degree $m$ when the Jacobian is invertible. When the Jacobian is invertible, that is to say its rows are linearly independent, so that if $ex+fy$ is a nonzero linear term, then $T(ex+fy) = eT_1(x) + fT_2(y)$ will have a nonzero form of degree $1$ in its decomposition. Since the smallest nonzero form of a product is the product of the smallest nonzero forms, we see that $m_{(0,0)} F^T = m_{(0,0)} F$. The argument for the general case of $P, Q$ is exactly as before, together with the observations that (1) the Jacobian of a composition of polynomial maps is the same as the matrix product of the Jacobian of polynomial maps at the appropriate points, and (2) that the Jacobians of translations are the identity. As for the demonstration that the converse is not in general true, we note that in the example given, $m_{(0,0)} F = 1 = m_{(0,0)} y - x^4 = m_{(0,0)} F^T$ despite the fact that the Jacobian of $T$ at $(0,0)$ is the noninvertible $$J_{(0,0)}T = \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix}~~\square$$
Subscribe to:
Post Comments (Atom)
No comments:
Post a Comment