$2$ Questions in the proof of Theorem $7.18$ of Clader and Ross Begining in Algebraic Geometry
问题内容
I am self studying algebraic Geometry from the textbook of Clader and Ross Beginning in Algebraic Geometry and I have $2$ in the proof of the theorem $7.18$ given on the page $201-203$.
Background: define $I_a= \{F\in K[X]\mid F(a)=0\} \subseteq K[X]$.
On page $201$ we see that $I_a$ can be written as equal to $I_a=\langle [x_1-a_1],...,[x_n-a_n] \rangle$ and ${I_a}^2=\langle[(x_i-a_i)(x_j-a_j)\mid 1\leq i,j\leq n]\rangle\subseteq I_a$
Theorem $7.18$: $I_a /I_a^2 =(T_a X)^{\vee}$
Proof: We define a canonical surjective linear map $\phi : I_a \to (T_aX)^{\vee}$ and prove that $ker(\phi)= {I_a}^2.$ To define $\phi$ suppose that $F\in I_a$ and write that $F=[f]$ for some $f\in K[x_1,...,x_n]$. Define $\phi(F) $ to be the linear map $\phi(F):T_a(X)\to K$ by $v \to \nabla {f(a)}. v$.
It can be shown that $\phi(F)$ is independent of the choice of representation $f$ and $\phi$ is linear. (I understand this part.)
I understand why $\phi$ is surjective. ( We define a linear polynomial (to be used later) \begin{equation} f= \overline{\rho}(e_1)(x_1-a_1)+\ldots +\overline{\rho}(e_n)(x_n-e_n) \in K[x_1,...,x_n]).\tag{1} \end{equation}
To prove $ker(\phi)= {I_a}^2$ first suppose that $F\in {I_a}^2$. $F=[f]$ and $f$ has the form \begin{equation} f=\sum_{i,j=1}^n (x_i-a_i)(x_j-a_j) f_{i,j}\text{ for some } f_{i,j} \in K[x_1,...,x_n].\tag{2} \end{equation}
I am not able to show that $\nabla{ f(a)}=0$. Can you please help me with that?
Rest of the proof to show the inclusion ${I_a}^2 \subseteq ker(\phi)$ is clear to me.
To prove the other inclusion suppose that $F\in ker(\phi)$ and write $F=[f]$ for some $f \in K[x_1,...,x_n]$.
Let $I(X)= \langle g_1,...,g_m \rangle$. It is clear to me how author writes $\nabla {f(a)}= \sum_{i=1}^m a_i \nabla g_i(a)$ for some $a_1,...,a_m\in K$. So, we get $\nabla{f(a)}=\nabla{g(a)}$.
Using that $F=[f]$ is in the domain of $\phi$ and thus $[f]\in I_a,$ we can write I don't understand how
$$ f=\sum_{i=1}^n b_i (x_i-a_i) +\sum_{i,j=1}^n f_{i,j} (x_i-a_i) (x_j-a_j). $$
Does $(1)$ and $(2) $ have some role to play in proving $f$ equals Right Hand side?
Please let me know how to prove these $2$ assertions.
Rest of the proof is clear to me.
回答 (1)
To show $\nabla f(a)=0$ the point is that $\nabla$ satisfies the following two properties for $f,g\in K[x_1,\dots,x_n]$:
- $\nabla(f+g)(a)=\nabla f(a)+\nabla g(a)$
- $\nabla(fg)(a)=f(a)\nabla g(a)+g(a)\nabla f(a)$
Using (1) you just need to show that $(\nabla (x_i-a_i)(x_j-a_j)f_{i,j})(a)=0$ for each pair $(i,j)$. And (2) implies that if $f(a)=g(a)=0$ then $\nabla(fg)(a)=0$. You can apply this e.g. for $f=(x_i-a_i)f_{ij}$ and $g=x_j-a_j$.
For the second question (I am going to drop all of the quotient notation because it was getting messy): if $f\in I_a=\langle x_1-a_1,\dots,x_n-a_n\rangle$ then you can write $f=(x_1-a_1)f_1+\cdots+(x_n-a_n)f_n$.
But notice that for an arbitrary polynomial $g$, one has $g-g(a)\in I_a$, and so $g=b+g'$ for some $b\in K$ (where $b=g(a)$ here) and $g'\in I_a$. Doing this for $g=f_i$ for each $i$, you write $f_i=b_i+f_i'$ with $f_i'\in\langle x_1-a_1,\dots,x_n-a_n\rangle$. Writing $f_i'$ as a combination of the $x_i-a_i$ and expanding $f$ above for these new representations of $f_i$ will yield the formula you're seeing