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Let $I$ be an $R$-ideal in a commutative algebra $B$ over a commutative ring $R.$ Given $b\in I$ I want to prove that $b\not \in I^2$. Are there any sufficient conditions for showing that $b\not\in I^ …

We have the adjunction $$\hom_{CDGA}(\Omega^\bullet(A),B^\bullet)\cong\hom_{CA}(A,B^0)$$ where $CDGA$ is the category of commutative diffferential graded algebras and $CA$ is the category of commutati …

Since on any commutative algebra $R$ over ring $S$ we have module of Kahler differentials $(\Omega_{R/S},d)$ which extends to the algebraic de-Rham complex $(\Omega^\bullet,d),$ it is natural to defin …

Let $M$ be a smooth manifold and let $Q$ be an arbitrary $C^\infty(M)$-module. $Q$ is called geometric if $$\bigcap_{p\in M}\mu_pQ=0,$$ where $\mu_p$ is an ideal in $C^\infty(M)$ of functions vanishin …

Let $M$ be a smooth manifold and let $A=C^\infty(M).$ We consider module of Kahler differentials $\Omega_k(A)$ and module of 1-forms $\Omega^1(M).$ Denote Kahler differential by $d_k$ and classical …

The relation between the exterior derivative $d:C^\infty(\mathbb{R})\to\Omega^1\mathbb(\mathbb{R})$ and the Kähler differential $d_{C^\infty(\mathbb{R})/\mathbb{R}}:C^\infty(\mathbb{R})\to\Omega_{C^\i …

$\DeclareMathOperator{\Id}{Id} \require{cancel}$ Jet Nestruev's "Smooth Manifolds and Observables" contains following exercise: Exercise. Show that $P$ is geometric if and only if the two modules …

The exercise is fine. I misread the definition of $\Gamma(P).$ Jet Nestruev defines $\Gamma(P)$ as an image $\phi(P).$ Hence surjectivity follows obviously from the definition of $\Gamma(P)$.

Cotangent space appears in both differential geometry (DG) and algebraic geometry (AG). In DG, given a smooth manifold $M$ and $x\in M$ one has an isomorphism $I_x/I_x^2 \cong T^*_xM$, where $I_x$ is …