A question in proof 4.8 of Chapter -1 of Daniel Perrin's Algebraic Geometry( Page 17)
问题内容
$k$ is an algebraically closed field. $\Gamma(V)$ is image of $r: k[X_1,...,X_n]\to F(V,k)$ and $V$ is affine algebraic set.
I am self studying Algebraic Geometry from Daniel Perrin's Algebraic Geometry and have a question in proof of Proposition $4.8 $ of Chapter $1$ on Page $17$.
Proposition $4.8:$ The following are equivalent. $V $ is finite iff $\Gamma(V)$ is a finite dimensional $k-$ vector space.
Proof: The side assuming $V$ is finite is clear to me.
For converse, let $\Gamma (V)$ is finite dimensional. Let $ \overline{X_i}$ be the image of $X-i$ in $\Gamma(V)$. So, the element $1,\overline{X_i} , {\overline{X_i}}^s$ is are not independent and hence in $\Gamma(V)$, there exists an identity ${a_s\overline{X_i}}^s+ a_1 \overline{X_i} +a_0=0$ such that $a_j \in k\;\text{and}\; $a_s \neq 0$.
I am not able to understand the next line of the proof, which is: If $u=(x_1,...,x_n)$ is an arbitrary point of $V$, it follows that we also have $a_s {x_i}^s+ ...+ a_1 x_i +a_0 =0$
Question:Why an arbitrary point of $V$ should also satisfy $a_s{x_i}^s +...+a_1x_i +a_0=0?$
Please let me know this.
回答 (1)
Regarding the question, $\overline{X}_i$ is the function that maps $(x_1,\cdots, x_n)$ to $x_i$ so if $a_0+\dots +a_s \overline{X_i}^s$ vanishes in $\Gamma(V)$, then evaluating $a_0+\dots +a_s \overline{X_i}^s=0$ at $u$ gives you $a_0+\dots +a_s x_i^s=0$