
Definition 10.135.6. 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 $S$.

Comment #235 by on

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.

Comment #240 by on

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 is a standard smooth algebra over $R$ if the polynomial maps to an invertible element in $S$.

What do you think?

Comment #1845 by Peter Johnson on

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?

Comment #1883 by on

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

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