The Stacks project

As stated, this definition is not clear. In the presentation of $S$, there are $n$ variables and $c$ polynomials, so one should end up with a $c\times n$ matrix. Is the definition assuming that $n\geqslant c$ and then truncating the matrix $(\partial f_i/\partial x_j)$, or is it saying that all $d\times d$ minors are invertible (where $d=\min\{c,n\}$)? Or is there some convention on what the determinant of a non-square matrix means? Either way, it should be clarified.

Hi, I agree that this is a bit vague, but I think it is not as bad as you make it out to be. Maybe a better formulation would be:

Let $R$ be a ring. Given integers $n \geq c \geq 0$ and $f_1, \ldots, f_c \in R[x_1, \ldots, x_n]$ we say $$S = R[x_1, \ldots, x_n]/(f_1, \ldots, f_c)$$ is a standard smooth algebra over $R$ if the polynomial $$g = \det \left( \begin{matrix} \partial f_1/\partial x_1 & \partial f_2/\partial x_1 & \ldots & \partial f_c/\partial x_1 \\ \partial f_1/\partial x_2 & \partial f_2/\partial x_2 & \ldots & \partial f_c/\partial x_2 \\ \ldots & \ldots & \ldots & \ldots \\ \partial f_1/\partial x_c & \partial f_2/\partial x_c & \ldots & \partial f_c/\partial x_c \end{matrix} \right)$$ maps to an invertible element in $S$.

What do you think?

OK, I went ahead and made the change. Thanks!

This definition as given applies to a presentation, not to R -> S itself. Shouldn't it have "for some presentation .." or "there exists .."? This makes more sense but would of course have minor effects elsewhere.

Would it not be useful later to also define locally standard smooth?

Hello! I've decided to leave this as is for now. If more people chime in, then I'll reconsider.

Definition is not clear. Maybe it is better to define 'standard smooth presentation' or use terms like 'there exists a presentation such that ...' in the definition. As Peter Johnson mentioned.

OK, I have changed this here in such a way that we can both talk about a given presentation being standard smooth and about an abstract $R$-algebra being standard smooth over $R$.