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The Arithmetic of Elliptic Curves by Joseph H. Silverman defines a smooth (or non-singular) variety as follows:

Let $V$ be a variety, $P\in V$, and $f_1, \ldots, f_m \in \overline{K}[X]$ a set of generators for $I(V)$. Then, $V$ is non-singular (or smooth) at $P$ if the $m\times n$ matrix $$\left(\frac{\partial f_i(P)}{\partial X_j} \right)_{1\le i\le m\\ 1\le j\le n}$$ has rank $n - \dim V$. If $V$ is nonsingular at every point, then we say that $V$ is nonsingular (or smooth).

Note that the partial derivatives $\frac{\partial f_i(P)}{\partial X_j}$ have nothing to do with analysis; they are just formal expressions.

Where does the above definition come from? Does it have something to do with partial derivatives in analysis? I can't help but believe so due to the very suggestive notation used by the author. Thank you!

  • An affine algebraic set $V$ in $\Bbb A^n$, we define the ideal of $V$ as $$I(V) := \{f\in \overline{K}[x_1,x_2,\ldots,x_n]: f(P) = 0\quad \forall P\in V\}$$
  • $\dim V$ is the transcendence degree of $\overline{K}(V)$ over $\overline{K}$. Here, $K(V)$ is the quotient field of $K[V]$, the affine coordinate ring of $V/K$.
stoic-santiago
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    Over $\mathbb{C}$ you can define a functor from smooth varieties to complex manifolds called analytification. For example the analytification of a smooth projective curve of genus $g$ is a compact Riemann surface of genus $g$. It's not at all accurate to say that the partial derivatives "have nothing to do with analysis"; again, over $\mathbb{C}$ they are (complex) derivatives in the usual sense. – Qiaochu Yuan Aug 09 '22 at 17:01
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    "Have nothing to do with analysis" is a bit strong. They do coincide with the analytical derivatives when that makes sense. – Captain Lama Aug 09 '22 at 17:03
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    Have a look at these notes: https://math.mit.edu/~mckernan/Teaching/09-10/Spring/18.726/l_8.pdf. Being smooth means that the tangent space has the right dimension. This tangent space can be identified with the kernel of the Jacobian matrix you describe: https://math.stackexchange.com/questions/675773/how-to-think-of-the-zariski-tangent-space. – Elchanan Solomon Aug 09 '22 at 17:05
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    The key notion, which no one has mentioned, is the implicit function theorem. In the setting of multivariable calculus, this gives submanifolds of euclidean space, which are the smooth analogue of what you're thinking about. – Ted Shifrin Aug 09 '22 at 17:20

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