The Stacks project

the zero prime ideal of $A$ doesn't correspond to any ideal of $\Gamma$ because by the definition here we don't allow zero ideal in $\Gamma$, unless you allow $\emptyset$ to be an ideal of $\Gamma$.

Typo in the statement of 0BUL: in the displayed formula, "$f^!\mathcal{O}_Y$" should be "$f^!\mathcal{O}_X$".

Is there a reason that $\mathcal O_{X,\overline{x}}$ is decorated with $sh$? It is already strictly Henselian (tag 04HX).

Typo: $h_{total} : (\mathop{\mathit{Sh}}\nolimits (\mathcal{C}_ {total}), \mathcal{O}) \to (\mathop{\mathit{Sh}}\nolimits (\mathcal{C}'_{total}), \mathcal{O}')$

In the first sentence, I believe the $(1, -1)$ should be $(f, -g)$

In the proof of (3), to apply 0315, we need to show that $K$ is finite. This seems true because $R^{\oplus n}$ is a Noetherian module ($R$ is Noetherian). Am I correct?

Should (2) not also include the possibility of $R_{\mathfrak m}$ being a field? In Definition 10.50.13. (tag 00IE) fields are excluded from being DVRs, yet they are Dedekind domains as defined in this section.

Pedantic point, but do you want to allow fields to be discrete valuation rings? For example, Tag 034X seems to consider fields as discrete valuation rings, but the value group will trivial.

I was confused by this too. I believe the following clarifies the argument: given $P$ we may choose $f \in k[x_1, \ldots, x_r, T]$ to be irreducible (for the full polynomial ring) by defining $f = P / c(P)$ where $c(P)$ is the gcd of the coefficients (well-defined because $k[x_1, \ldots, x_r]$ is a UFD). This subsequently shows that the given polynomial for $x_i$ is irreducible and therefore that $x_i$ is separable.

Don't we need to assume that $R$ is Noetherian?

Why the image of $f\times g$ is $f(A)g(B)$ in Lemma 04Q0 ?

I beleive the statement is supposed to be $$\mathop{\mathrm{Hom}}\nolimits _ {S^{-1}R} (S^{-1}M, N) \longrightarrow \mathop{\mathrm{Hom}}\nolimits _ R(M, N)$$ and not $\mathop{\mathrm{Hom}}\nolimits _ {R} (S^{-1}M, N)$.

Also, I think it's useful to add an explanation about what does "all the elements of S act as automorphisms on N" mean - with respect to what action? I find it a bit confusing.

If "$f$ is left inverse to $g$" means that $f \circ g$ is the identity (where $f \circ g$ means "apply $g$ first"), then I think "left" and "right" should be switched in this lemma.

Someone pointed out to me that this claim holds in more generality: finitely generated projective modules are dualizable objects in RMod, and tensoring with dualizable objects commutes with limits.

I believe the lemma is wrong in the case $q=n=0,d<0$. Unwrapping the definition we should have $H^0(\mathbb{P}^0,\mathcal{O}(d))=R\cdot T_0^d$, which is compatitable with $(\frac{1}{T_0}R[\frac{1}{T_0}])_d\neq 0$ but not compatitable with $(R[T_0])_d=0$ when $d<0$.

I have found the criterion given in claim 1 of the following link for commuting tensor with an inverse limit of modules useful, but have not seen it in any reference: https://math.stackexchange.com/a/4577954/304290 Perhaps it would be worth mentioning here, either on this tag or wherever else it would fit.

In the definition of Koszul complex $K_n$, should the differential be defined by $d(\xi_i)=f^n$?

In the second part of the proof for 10.160.8 (the case of positive characteristic), the terminology is slightly confusing. I suggest rewording the condition required of $\varphi_n$ to the commutative square $\pi \varphi_n = f \pi$, where $\pi$ are the projections $\Lambda/p^n \Lambda \to \Lambda/p\Lambda$ and $R/\mathfrak{m}^n \to R/\mathfrak{m}$, and $f: \Lambda/p\Lambda \to R/\mathfrak{m}$ is a chosen isomorphism. Moreover, the map to lift using formal smoothness is the composition $\Lambda/p^n\Lambda \rightarrow \Lambda/p^{n-1} \Lambda \overset{\varphi_{n-1}}{\longrightarrow} R/\mathfrak{m}^{n-1}$.

It could also be helpful for readers to mention that this proof shows that the map $A \mapsto A/pA$ from Cohen rings (up to isomorphism) to fields (up to isomorphism) is injective.

Here what is the ideal $I$ that defines the $I$-adic topology/completion?

I guess that $V$ in the first sentence of the last paragraph should be only "geometrically connected"?