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I'm working towards understanding the Zariski tangent space of a $C^k$ manifold, using this pdf.

The author defines $\mathcal{O}^{(k)}_{M,p}$ as the set of germs of $C^k$ functions at $p$, which forms a ring as well as an algebra over $\mathbb{R}$ (and itself, of course) with operations defined pointwise.

They further define $\mathcal{S}^{(k)}_{M,p}$ as the set of germs of $C^k$ functions stationary at $p$, i.e. $\mathbf{f} \in \mathcal{S}^{(k)}_{M,p} \iff \forall(f \in \mathbf{f}).(f \circ \varphi^{-1})'(\varphi(p))=0$, for some (and any) chart $\varphi$ at $p$. This is clearly a subring (and subalgebra) of $\mathcal{O}^{(k)}_{M,p}$, but not an ideal (as the author points out).

The author then asserts that $\mathcal{O}^{(k)}_{M,p} / \mathcal{S}^{(k)}_{M,p}$ is (1) a vector space that is (2) isomorphic to the space of linear derivations on $\mathcal{O}^{(k)}_{M,p}$ that vanish on $\mathcal{S}^{(k)}_{M,p}$, in turn (3) isomorphic to $T_{p}^{*}M$ the (usual) cotangent space. However, I do not understand the quotient $\mathcal{O}^{(k)}_{M,p} / \mathcal{S}^{(k)}_{M,p}$, since $\mathcal{S}^{(k)}_{M,p}$ is not an ideal. Is this a quotient of algebras/vector spaces? What is the implied equivalence relation?

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Since it is a subalgebra, it is also a vector subspace. Do you know how to take the quotient by a vector subspace? It preserves the vector space structure, but the ring structure disappears.

You're doing the same thing when you take the quotient of any ring by an ideal: primarily it is a quotient of an abelian group by a subgroup. The ring structure is a "bonus" that you get when the subgroup is an ideal.

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  • $\begingroup$ I see, so then the relation is the usual $a \sim b \iff a - b \in \mathcal{S}^{(k)}_{M,p}$ and the quotient is $\mathcal{O}^{(k)}_{M,p} / \sim$? $\endgroup$ – terrygarcia Mar 21 at 19:40
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    $\begingroup$ @terry Yes, that's right. $\endgroup$ – Matt Samuel Mar 21 at 19:41

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