## On Ext-duals of injective modules for commutative rings

Let $$R$$ be a commutative noetherian ring and $$I=E(R/p)$$ the injective hull of the module $$R/p$$ for a prime ideal $$p$$.

Question: Is there a (more) explicit description of the $$R$$-modules $$Ext_R^i(I,R)$$ for $$i geq 0$$ (at least in special cases such as $$R$$ being Gorenstein)? When are those modules finitely generated?

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## 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

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## 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.

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## 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.

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## faithful modules over a finite dimensional commutative algebra

Let $$A$$ be a commutative algebra over a field $$k$$ which is finite dimensional as a vector space over $$k$$. Let $$M$$ be a faithful $$A$$-module. Does it follow that $$dim_k(M)geq dim_k(A)$$?

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## 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?

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## 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: $$text {det} ( textbf {A}) = 0$$?

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## Simple \$ C ^ * \$ algebras are not commutative

Is it true that for simple $$C ^ *$$-Algebras, which means that they have no non-trivial bilateral ideals, it means that they are necessarily not commutative? And why?

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## 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.

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## 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?

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