commutative algebra – Depth of tensor product of modules which are locally free on the punctured spectrum of regular local ring

Let $M,N$ be (non-zero) finitely generated modules over a regular local ring $(R, mathfrak m)$ of dimension $d$ such that $M_P, N_P$ are free (non-zero) $R_P$-modules for every prime ideal $Pne mathfrak m$ of $R$.

Then is there any elementary way to prove that

$$text {depth}(M otimes_R N)=max {0, text {depth}(M)+text {depth}(N)-d }$$ ?

I know that this is essentially contained in Proposition 3.1 of
https://doi.org/10.7146/math.scand.a-12871 , but I was wondering if there is a direct proof.

Thanks

ag.algebraic geometry – Commutative square of module of differential is cartesian?

Let $R$ be a regular local $mathbb{Q}$-algebra and $f$ be a normal crossing divisor(i.e. $f = x_{1}x_{2}…x_{r}$ such that $R/(x_{i})$ is regular for each $i$). Then we have commutative diagram

$require{AMScd}$
begin{CD}
R/fR @>{d}>> Omega^{1}_{R/fR} \
@VVV @VVV\
widehat{R}/fwidehat{R} @>{ widehat{d}}>> Omega^{1}_{widehat{R}/fwidehat{R}}
end{CD}

where $widehat{R}$ denotes the completion of local ring and maps are induced by completion map and d denote the differential map. Is the above square is cartesian?

I am thinking that this should be a cartesian square since there is similar square for $R$ and $widehat{R}$ namely $require{AMScd}$
begin{CD}
R @>{d}>> Omega^{1}_{R} \
@VVV @VVV\
widehat{R} @>{ widehat{d}}>> Omega^{1}_{widehat{R}}
end{CD}
above this and that seems to be cartesian but I am not sure. Even if second square is cartesian it does not mean that first square is cartesian but it is definitely necessary condition. Any help would be great.

ag.algebraic geometry – Any finite flat commutative group scheme of $p$-power order is etale if

This question is immediately related to Discriminant ideal in a member of Barsotti-Tate Group
dealing with Barsotti-Tate groups and here I
would like to clarify a proof presented by
Anonymous in the comments from linked thread which I not completly understand. Although meanwhile I found another proofs of the claim below I have a big interest on understanding this proof below.

Assume $G$ is a finite flat commutative group scheme
over a field $k$ of
order $p^N$. Assume $p$ prime and $p in k backslash {0}$,
equivalently invertible on the base.

Claim: Any finite flat commutative group scheme of $p$-power order
is etale if $p$
is invertible on the base.

Anonymous’ proof works as follows:
Firstly we reduce to case over a field (because a
finitely presented flat map is etale if it is
fibrewise etale). Since we assumed $G$ commutative
the multiplication by $p^N$ map $f_{p^N}: G to G$
is well defined and by Deligne’s theorem
$p^N$ kills $G$ since it’s the order of $G$.
That means that $f_{p^N}$ is the zero map: equivalently
it factorize over $Spec(k)$.

What now comes I not understand:

It is clamed that $f_{p^N}$ is unramified
“as the map on the tangent spaces is given
by $p^N$, which is invertible”.

Question I: why the induced by $f_{p^N}$
maps on tangent spaces is given
by $p^N$?

Question II: assume we understand Question 1.
Why this implies $G$ unramified?

when we know this we are done because unramified finite type schemes over a field are etale.

ct.category theory – adjunctions that capture the duality between ideals and saturated monoids in a commutative ring?

To let $ R $ be a commutative ring. A saturated monoid in $ R $ is a multiplicative submonoid $ S subset R $ which is closed among dividers, i.e. $ xy in S implies x in S $. This is the reverse of the analog axiom for ideals $ x in I implies xy in I $. Saturated monoids and ideals thus have a kind of duality, of which part swaps addition and multiplication and part reverses the direction of the implication.

The only elements given to us suggest a pair of operations between ideals and saturated monoids
$$ begin {gathered} mathrm {Ideal} (R) rightleftarrows mathrm {SatMon} (R), \ I mapsto (1 + I) _ { text {sat}} text {and} S . mapsto left langle S-1 right rangle. end {gathered} $$

Take those from the powerset of induced poset structures $ R $, given by inclusions, both operations above are poset morphisms (take inclusions to inclusions). However, they are not additional functors:

$$ begin {assembled} left langle S-1 right rangle ⊂ I iff S-1 subset I iff S subset 1 + I \ S supset (1 + I) _ text { sat} iff S supset 1 + I end {gathered} $$

This is a strange situation for functors $ mathsf C rightleftarrows mathsf D $ say $ F, G $ and yet bijections $$ mathsf D (FA, B) cong mathsf C (GB, A) = mathsf C ^ text {op} (A, GB). $$

Question 1. Are the above operations based on a missing addition?

If not then:

Question 2. Is the formal duality between ideals and saturated monoids captured by another adjunction?

If not then:

Question 3. Is there more structure in the abstract environment I described above with interesting category theory?

linear algebra – product of a matrix with its cofactor matrix: commutative property

Let A be a quadratic ordering matrix $ n $. Then we have the following known result:

$$ textbf {A} big ( text {cof} ( textbf {A}) big) ^ T = big ( text {det} ( textbf {A}) big) textbf {I . } $$

Where $ text {cof} ( textbf {A}) $ is the cofactor matrix of $ textbf {A} $. Suppose the determinant of $ text {det} ( textbf {A}) neq 0 $. Then,

$$ frac {1} { text {det} ( textbf {A})} textbf {A} big ( text {cof} ( textbf {A}) big) ^ T = textbf { I} Rightarrow bigg ( big ( text {cof} ( textbf {A}) big) ^ T bigg) ^ {- 1} = frac {1} { text {det} ( textbf {A})} textbf {A} $$

So we showed that the transpose of the cofactor matrix is ​​not singular, so we can also write:

$$ frac {1} { text {det} ( textbf {A})} textbf {A} big ( text {cof} ( textbf {A}) big) ^ T = big ( text {cof} ( textbf {A}) big) ^ T frac {1} { text {det} ( textbf {A})} textbf {A} $$

or

$$ textbf {A} big ( text {cof} ( textbf {A}) big) ^ T = text {det} ( textbf {A}) textbf {I} = big ( text {cof} ( textbf {A}) big) ^ T textbf {A} $$

How do I prove this commutative property when the determinant is zero: $ text {det} ( textbf {A}) = 0 $?

commutative algebra – assumption to be affine in the proof of the algebraic Hartogs lemma

The following sentence is the sentence $ 6.45 in the book Algebraic Geometry I by Torsten Wedhorn and Ulrich Görtz

sentence To let $ X $ be a local noetheric normal scheme and let $ U subset X $ to be an open subset $ mathrm {codim} _X (X – U) geq 2 $. Then the restriction map $ Gamma (X, mathscr {O} _X) to Gamma (U, mathscr {O} _X) $ is an isomorphism.

The proof starts with: "We can accept that $ X = mathrm {Spec} (A) $ is affine … ". I wonder how you could assume that $ X $ be affine? I find out this lemma on StackProject, but it doesn't really help.

commutative algebra ring in which Jacobson Radical differs from Null Radical

Define $ Jac (R) = cap _ { mathfrak {m}, maximum} mathfrak {m} $ and $ Nil (R) = cap _ { mathfrak {p}, prime} mathfrak {p} $, Where $ R $ is a commutative ring. Can someone give me an example of a commutative ring? $ R $ so that $ Jac (R) ne Nil (R) $, I have ruled out $ mathbb {Z} _ {n}, mathbb {Z} $and every artinian ring. Should I look at polynomial rings?