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Let $R=\frac{k[[x_1,\ldots,x_n]]}{(f_1,\ldots,f_c)}$ (or $\frac{k[x_1,\ldots,x_n]}{(f_1,\ldots,f_c)}$), where $k$ is a field with characteristic $0$ and $0\ne(f_1,\ldots,f_c)\subseteq (x_1,\ldots,x_c)$.

In the paper On the fitting ideal in free resolution (see also Jacobian ideal reference), there is a definition of the Jacobian ideal of $R$. Assume that the height of $(f_1,\ldots,f_c)$ in $k[[x_1,\ldots,x_n]]$ is $h$, i.e. $h=\operatorname{inf}\{\operatorname{height}(p)\mid p\in \operatorname{Spec}(k[[x_1,\ldots,x_n]]), (f_1,\ldots,f_c)\subseteq p\}$. Then the Jacobian ideal of $R$ is defined to be the ideal of $R$ generated by $h\times h$ minors of the Jacobian matrix $\frac{\partial(f_1,\ldots,f_c)}{\partial(x_1,\ldots,x_n)}$.

I have two questions about the definition.

Must the Jacobian ideal of $R$ be non-zero? I am confused that why the definition considers $h\times h$ minors of the Jacobian matrix. Why not higher minors?

Consider the example $R=\frac{k[[x,y,z]]}{(xy,xz)}$, the height of $(xy,xz)$ in $k[[x,y,z]]$ is $1$. We can check easily that there is a $2\times 2$ minor of the Jacobian matrix that is non-zero in $R$. So it seems that it is reasonable to consider higher minors.

Thank you in advance.

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The ideal can be zero, but only if $R$ is non-reduced: take for example $(x^2,xy,y^2)$ which has Jacobian ideal $(2x^2,2y^2,4xy)=0$. If $R$ is reduced, then $\operatorname{Spec} R$ is either a reduced variety (in the case you're taking $k[x_1,\cdots,x_n]$) or the germ of a reduced variety (in the case you're taking $k[[x_1,\cdots,x_n]]$) and such things are generically smooth if the characteristic of $k$ is zero. This is covered in most introductory algebraic geometry texts if you're looking for a reference.

The reason for considering $h\times h$ minors is that you're trying to look at nonsingularity of $V(f_1,\cdots,f_c)$, which happens when the Jacobian matrix is of rank $n-\dim_p V(f_1,\cdots,f_c)$ at each point of $p\in V(f_1,\cdots,f_c)$.

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  • $\begingroup$ Nice. Thank you. Cool, I found the last sentence you said in Eisenbud's book Corollary 16.20. Under the last condition you said, the Jacobian ideal defines the singular locus of $R$. $\endgroup$
    – Jian
    Nov 24, 2021 at 1:07
  • $\begingroup$ If $R$ is reduced, the ideal is non-zero. Can you give me a reference? I didn't find it. Also, do you think the ideal is also non-zero when $f_1,\ldots,f_c$ is a regular sequence in $k[[x_1,\ldots,x_n]]$ or $k[x_1,\ldots,x_n]$? $\endgroup$
    – Jian
    Nov 24, 2021 at 1:31
  • $\begingroup$ It's not as direct as that: the vanishing locus of the Jacobian ideal is the singular locus, which is known to be a proper subvariety when the characteristic is zero (see Hartshorne I.5 for a proof, for instance), hence the Jacobian ideal cannot be zero. The regular sequence idea can't work: if $f_1,\cdots,f_c$ is a regular sequence, $f_1^r,\cdots,f_c^r$ is also a regular sequence. $\endgroup$
    – KReiser
    Nov 24, 2021 at 1:32
  • $\begingroup$ Yes, the power of a regular sequence is still a regular sequence. In the regular sequence case, do you mean the $c\times c$ minor can be zero? Can you give an example here? Thank you. I don't find any example that the Jacobian ideal is zero when $f_1,\ldots,f_c$ is a regular sequence. $\endgroup$
    – Jian
    Nov 24, 2021 at 1:45
  • $\begingroup$ Sorry, perhaps I was hasty - at present I'm not totally sure about the claim either way. $\endgroup$
    – KReiser
    Nov 24, 2021 at 2:52

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