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Tag 00T6

Chapter 10: Commutative Algebra > Section 10.135: Smooth ring maps

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 $S$.

    The code snippet corresponding to this tag is a part of the file algebra.tex and is located in lines 35254–35284 (see updates for more information).

    \begin{definition}
    \label{definition-standard-smooth}
    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 {\it 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 $S$.
    \end{definition}

    Comments (5)

    Comment #235 by Daniel Miller (site) on July 7, 2013 a 10:02 pm UTC

    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 Johan (site) on July 16, 2013 a 2:17 pm UTC

    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 <i>standard smooth algebra</i> 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?

    Comment #241 by Johan (site) on July 16, 2013 a 8:29 pm UTC

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

    Comment #1845 by Peter Johnson on February 23, 2016 a 1:09 pm UTC

    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 Johan (site) on April 1, 2016 a 6:43 pm UTC

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

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