Towards an algebraic understanding of jet bundles: $R/\mathfrak{m}^r \cong \bigoplus^r_{j=0}\text{Symm}^j(\mathfrak{m}/\mathfrak{m}^2)$?

Question

Let $R$ be a commutative ring and $\mathfrak{m}_p$ a maximal ideal. Does the following isomorphism hold?

$$R/\mathfrak{m}_p^{r+1}\cong \bigoplus^r_{j=0}\text{Symm}^j(\mathfrak{m}_p/\mathfrak{m}_p^2)$$

Plugging in $R=C^\infty(M)$ and $\mathfrak{m}_p= \{f:M \to \mathbb{C}: f(p)=0 \}$ we get the familiar identity:

$$C^\infty(M)/\mathfrak{m}_p^{r+1} \cong \bigoplus^r_{j=0} \text{Symm}^j(T_p^*M)$$

1. Does the first identity holds generally?

2. Is the $r$-th order jet bundle nothing but the $r$-th order symmetric algebra bundle on the cotangent bundle $J^r(M) \cong \bigoplus^r_{j=0} S^j(T^*M)$?

Answer 1

1. No. Take $R = \mathbb{Z}, m = (2), r = 1$.

2. I think the answer is "yes, but not naturally." That is, for a particular $M$, I think it's true that we have the desired isomorphism of vector bundles, but I also think this isomorphism cannot be made natural in $M$. The RHS is something like the associated graded of the LHS, so I think both sides have algebra structures that might differ.