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Let $k$ be a field, $R := k[x_1, \cdots , x_n]$ the polynomial ring in $n$ indeterminates over $k$ and $f$ a nonzero element of $R$. The following paper of Lyubeznik which I have been recently reading, gives a "characteristic-free" proof of the finiteness of the length of the localized ring $R_f$, viewed as a $\mathfrak D$-module over the ring $\mathfrak D$ of $k$-linear differential operators of $R$: https://www.sciencedirect.com/science/article/pii/S002240491000263X

I seem to be having the following problem with the proof of Proposition 3.1:

In the first paragraph of the proof (the one that basically shows the existence of a separating transcendence basis of a certain transcendental extension of a perfect field invoking the machinery of K"ahler differentials) the author seems to be considering the field of fractions $K$ of the domain $R/P$ where $P$ is a certain prime ideal of the polynomial ring $R$ and towards the end of the paragraph, he obtains (by relabeling) elements $x_{h+1},\cdots , x_n$ such that the ring of K"ahler differentials $\Omega_{K/k}$ is spanned as a $K$-vector space by $\{d(x_{h+1}),\cdots , d(x_n)\}$, where $d$ is the natural map $R/P \rightarrow \Omega_{K/k}$. Thereafter, he considers the polynomial subring $\mathscr R := k[x_{h+1},\cdots , x_n]$ of $R$ and denoting by, $\mathscr K$ its field of fractions, he speaks of the "fundamental exact sequence" $$\Omega_{\mathscr K / k} \otimes_{\mathscr K} K \longrightarrow \Omega_{K/k} \longrightarrow \Omega_{K/\mathscr K} \longrightarrow 0.$$

Now, I have recently read very little about K"{a}hler differentials and from what I know, if $B$ and $C$ are algebras over a commutative ring $A$ such that there is an $A$-algebra homomorphism from $B$ to $C$ (giving $C$ the structure of a $B$-module), then we have an exact sequence of $C$-modules $$\Omega_{B/A} \otimes_{B} C \longrightarrow \Omega_{C/A} \longrightarrow \Omega_{C/B} \longrightarrow 0.$$ If this is indeed the result being used by Lyubeznik in the part quoted above, I seem to be having trouble figuring out what the $k$-algebra map from $$\mathscr K := \text{Frac }k[x_{h+1},\cdots , x_n] = k(x_{h+1},\cdots , x_n) \hspace{5mm} \text{(which seems to play the role of $B$)}$$ to the ring $$K:= \text{Frac}(R/P) = \text{Frac}\left(k[x_1,\cdots , x_n] \big/ P \right) \hspace{5mm} \text{(which seems to play the role of }C\text{)}$$ should be.

I started with the $k$-algebra map $$\eta: \mathscr R = k[x_{h+1},\cdots , x_n] \hookrightarrow R \twoheadrightarrow R/P.$$ But now by the universal property of the total quotient ring, in order for this to factor through $\mathscr K := \text{Frac}(\mathscr R)$, every nonzero element of $\mathscr R$ must be a unit in $R/P$. This is clearly false if $P\mathscr R \neq 0$, for any nonzero element of $P\mathscr R \subset \mathscr R$ maps to zero under $\eta$. I apologize if I am missing something obvious but I would really appreciate some help in this regard. Thank you.

Addendum: It did occur to me that the perhaps the provided definition of $\mathscr R$ is a typo (it is just defined in one place after all and there are other definitions building upon it) and perhaps what was really meant was that $$\mathscr R:= k[x_{h+1},\cdots , x_n]/Pk[x_{h+1},\cdots , x_n],$$ which is still a domain. While it is intuitively clear that most of the other parts of the proof seem to go through with this choice of $\mathscr R$, I have not yet been able to rigorously verify the same. Is this guess correct?

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    $\begingroup$ Why don't you ask him? $\endgroup$ May 3, 2021 at 22:15
  • $\begingroup$ Thank you Sir, I will try doing that then. I have never had the opportunity to email authors of papers just for clarification before, hence I didn't consider it at first. $\endgroup$
    – AK12N1
    May 3, 2021 at 22:33

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