Quotient by a subring and not an ideal

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?

• 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 $\mathcal{O}^{(k)}_{M,p} / \sim$? – terrygarcia Mar 21 at 19:40