I apologize as I am certain this is not research-level, but several days have gone by without an answer on stackexchange (<https://math.stackexchange.com/questions/4724245/jacobian-criterion-for-zariski-cotangent-space-over-arbitrary-field>).


Section 7.2 of the notes on etale cohomology (<https://people.dm.unipi.it/tdn/CoomologiaEtale/Note.pdf>) by Lombardo and Maffei states the following (up to a small typo):


Let $k$ be any field, let $A = \frac{k[x\_1,...,x\_n]}{\langle f\_1,...,f\_m \rangle}$, and let $Jf = [\frac{\partial f\_j}{\partial x\_i}]$ be the Jacobian matrix of the $f\_j$. Let $\mathfrak{m}$ be a maximal ideal of $A$, and $k(\mathfrak{m}) = \frac{A}{\mathfrak{m}}$. Then $\dim\_{k(\mathfrak{m})} \frac{\mathfrak{m}}{\mathfrak{m}^2} \le n - $ rank $Jf(\mathfrak{m})$, where $Jf(\mathfrak{m})$ is obtained from $Jf$ by reducing all entries mod $\mathfrak{m}$.


The proof is omitted in the notes. Can anyone please supply a proof or a reference?


I emphasize that there is no further assumption on $k$ or the extension $k(\mathfrak{m})/k$. If $k$ is algebraically closed this is standard and if the extension is separable it is easy to prove from the notes.



Jacobian criterion for Zariski cotangent space over arbitrary field (X-post from SE)
Here's an attempt at solution.


Let $I=(f\_i)$ be the ideal defining the scheme $Spec(A)$ and $\mathtt{n} \subset k[\underline{x}\_i]$ be the maximal ideal s.t. $\mathtt{n}/I=\mathtt{m}$. Then we have the following commutative diagrams


$\require{AMScd}$
\begin{CD}
I/(I \cap \mathtt{n}^2) @>\alpha>> \mathtt{n}/\mathtt{n}^2 @>>>\mathtt{m}/\mathtt{m}^2 @>>> 0\\
@AAA @V \theta VV @VV V @.\\
k(\mathtt{m}) \otimes I/I^2=I/\mathtt{n}I @>\delta>>\oplus^{n}\_{i=1} k(\mathtt{m}) dx\_i @>>> \Omega^1\_{A/k}(\mathtt{m}) @>>> 0
\end{CD}


where the first vertical map is the obvious surjection and the next two vertical maps are maps defined naturally(as in Prop'n 6.13 of the notes alluded to in the question). The top row is a short exact sequence and the bottom row is exact. The bottom is also obtained as in Prop'n 6.13 cited before, but only after tensoring by $k(\mathtt{m})$.


By commutativity, $\theta(im(\alpha))=im(\delta)$. Since $dim\_{k(\mathtt{m})}im(\delta)=$ rank$Jf(\mathtt{m})$, so $dim\_{k(\mathtt{m})}im(\alpha) \geq rankJf(\mathtt{m})$ or equivalently $dim\_{k(\mathtt{m})}\mathtt{m}/\mathtt{m}^2 \leq n-rankJf(\mathtt{m})$(since $k[\underline{x}\_i]$ is regular $dim\_{k(\mathtt{m})}\mathtt{n}/\mathtt{n}^2=n$, noting that $k[\underline{x}\_i]/\mathtt{n}=A/\mathtt{m}$).


