仿射簇奇点证明解释：闭域下理想与Kⁿ同构性分析

Alright, let's walk through this proof step by step—this is a key result for understanding tangent spaces to affine varieties, so let's make sure every piece makes sense.

First, let's recap the setup to keep us grounded:

• $K$ is an algebraically closed field, $X \subseteq \mathbb{A}_K^n$ is an irreducible closed affine variety, and $P=(a_1,\dots,a_n)$ is a point on $X$.
• $\alpha_P = (x_1-a_1,\dots,x_n-a_n)$ is the maximal ideal of polynomials vanishing at $P$ (this follows directly from Hilbert's Nullstellensatz, since $K$ is algebraically closed).
• The map $\theta$ sends a coset $f + \alpha_P^2$ to the tuple of partial derivatives of $f$ evaluated at $P$: $\theta(f+\alpha_P^2) = \left(\frac{\partial f}{\partial x_1}(P),\dots,\frac{\partial f}{\partial x_n}(P)\right)$.

We need to show three things for $\theta$ to be an isomorphism: it's well-defined, injective, and surjective. Let's tackle each one.

1. $\theta$ is well-defined #

First, we need to confirm that if two polynomials $f,g \in \alpha_P$ are congruent modulo $\alpha_P^2$ (i.e., $f - g \in \alpha_P^2$), then $\theta(f) = \theta(g)$.

Elements of $\alpha_P^2$ are finite sums of products of two elements from $\alpha_P$. So any $h \in \alpha_P^2$ looks like $\sum_{k} u_k v_k$, where $u_k, v_k \in \alpha_P$. Now, take the partial derivative of $h$ with respect to $x_i$:
$$\frac{\partial h}{\partial x_i} = \sum_{k} \left( \frac{\partial u_k}{\partial x_i} v_k + u_k \frac{\partial v_k}{\partial x_i} \right)$$
Since $u_k(P) = 0$ and $v_k(P) = 0$ (they're in $\alpha_P$, so they vanish at $P$), evaluating this at $P$ gives $0 + 0 = 0$ for every term. So $\frac{\partial h}{\partial x_i}(P) = 0$ for all $i$.

If $f - g = h \in \alpha_P^2$, then $\frac{\partial f}{\partial x_i}(P) - \frac{\partial g}{\partial x_i}(P) = \frac{\partial h}{\partial x_i}(P) = 0$, so $\theta(f) = \theta(g)$. Perfect, $\theta$ is well-defined.

2. $\theta$ is surjective #

To show surjectivity, take any tuple $(c_1,\dots,c_n) \in K^n$. We need to find an $f \in \alpha_P$ such that $\theta(f) = (c_1,\dots,c_n)$.

Just take $f = c_1(x_1 - a_1) + c_2(x_2 - a_2) + \dots + c_n(x_n - a_n)$. Clearly, $f \in \alpha_P$ because each $(x_i - a_i)$ is in $\alpha_P$, and $\alpha_P$ is an ideal (closed under linear combinations).

Now compute the partial derivatives: for each $j$, $\frac{\partial f}{\partial x_j} = c_j$, so evaluating at $P$ gives $c_j$. That's exactly the tuple we wanted! So $\theta$ hits every element of $K^n$—surjectivity is done.

3. $\theta$ is injective #

Suppose $\theta(f) = 0$, meaning $\frac{\partial f}{\partial x_i}(P) = 0$ for all $i$. We need to show $f \in \alpha_P^2$.

Use the Taylor expansion of $f$ around $P$. For any polynomial $f \in K[x_1,\dots,x_n]$, we can write:
$$f = f(P) + \sum_{i=1}^n \frac{\partial f}{\partial x_i}(P) (x_i - a_i) + \text{terms in } \alpha_P^2$$
Since $f \in \alpha_P$, $f(P) = 0$ (by definition of $\alpha_P$). And we assumed all partial derivatives at $P$ are 0. So this simplifies to $f = \text{something in } \alpha_P^2$, which means $f \in \alpha_P^2$.

Therefore, if $\theta(f+\alpha_P^2) = 0$, then $f+\alpha_P^2 = 0+\alpha_P^2$—so $\theta$ is injective.

Conclusion #

$\theta$ is a well-defined $K$-linear map (partial derivatives are linear, and the quotient map preserves linearity) that's both injective and surjective. By definition, that makes it an isomorphism of $K$-vector spaces.

内容的提问来源于stack exchange，提问作者Minato