rank of jacobian matrix of generators of ideal of affine variety is the dimension as a subspace of k^n

Question

The question comes from the proof of Theorem I. 5.1 of Hartshorne's AG. Here I gave the theorem and the confusing part of the proof:
Theorem 5.1 Let $Y\subseteq \mathbb{A}^n$ be an affine variety. Let $P\in Y$ be a point. Then $Y$ is nonsingular at $P$ if and only if the local ring $\mathcal{O}_{P,Y}$ is a regular local ring.
Proof Let $P=(a_1,\cdots,a_n)\in\mathbb{A}^n$, and let $\mathfrak{a}_p:=(x_1-a_1,\cdots,x_n-a_n)$. We define a $k$-linear map $\theta:k[x_1,\cdots,x_n]\to k^n$ by $$\theta(f):=(\frac{\partial f}{\partial x_1}(P),\cdots,\frac{\partial f}{\partial x_n}(P))$$ for all $f\in k[x_1,\cdots,x_n]$. Then it induces an isomorphism $$\theta':\mathfrak{a}_p/\mathfrak{a}_P^2\cong k^n$$ Now let $\mathfrak{b}=I(Y)=(f_1,\cdots,f_t)$.
Here comes my question Then the rank of the Jacobian matrix $J=(\frac{\partial f_i}{\partial x_j}(P))$ is just the dimension of $\theta(\mathfrak{b})$ as a subspace of $k^n$.
Unfortunately, I cannot see how these two concepts are related. Any help is appreciated!

Answer 1

By definition, the rank of a matrix is equal to the dimension of the span of its columns.

The columns of the $n \times n$ matrix $J$ are precisely $\theta(f_1), \dots , \theta(f_t)$, where each $\theta(f_i)$ is viewed as a vector in $k^n$. Therefore, the rank of $J$ is equal to the dimension of $\text{Span}(\theta(f_1), \dots , \theta(f_t)) \subset k^n$.

I claim that $\theta(\mathfrak b) = \text{Span}(\theta(f_1), \dots , \theta(f_t))$.

• Proving $\theta(\mathfrak b) \subset \text{Span}(\theta(f_1), \dots , \theta(f_t)) $: Observe that if $g \in \mathfrak b = (f_1, \dots , f_t)$, then there exist $a_1, \dots, a_t \in k[x_1, \dots , x_t]$ such that $g = a_1f_1 + \dots + a_t f_t$. Then $\frac{\partial g}{\partial x_i} (p) = a_1(p)\frac{\partial f_1}{\partial x_i} (p) + \dots a_t(p)\frac{\partial f_t}{\partial x_i} (p)$, by the "product rule" for differentiation. (Note that the product rule also produces terms like $f_i(p)\frac{\partial a_i}{\partial x_i} (p)$, but these vanish because $f_i(p) = 0$.) Thus $\theta(g) = a_1 \theta(f_1) + \dots a_t \theta(f_t)$. Hence $\theta(g) \in \text{Span}(\theta(f_1), \dots , \theta(f_t))$.
• Proving $\text{Span}(\theta(f_1), \dots , \theta(f_t)) \subset \theta(\mathfrak b) $: If $v \in \text{Span}(\theta(f_1), \dots , \theta(f_t))$, then there exist $c_1, \dots, c_t \in k$ such that $v = c_1 \theta(f_1) + \dots + c_t \theta(f_t)$. Then $v = \theta(h)$, where $h = cf_1 + \dots + cf_t \in k[x_1, \dots, x_n]$.

To summarise, the rank of $J$ is equal to the dimension of $\text{Span}(\theta(f_1), \dots , \theta(f_t))$ as a vector subspace of $k^n$, and $\text{Span}(\theta(f_1), \dots , \theta(f_t)) = \theta(\mathfrak b)$. Hence the rank of $J$ is equal to the dimension of $\theta(\mathfrak b)$ as a vector subspace of $k^n$.