$C^{\infty}(U)$ is a ring. Where $U$ is open in $\mathbb{R}^n$.
Not $D_jf(a)$ denote the partial derivative of $f$ in the direction of $e_j$ at a.
My attempt:
Abelian group:
Let $f,g\in C^{\infty}(U)$. So, f and g are continuous, and have continuous partial derivatives of all order. Observe that for all $j$, $D_jf$ and $D_jg$ exist and are continuous. Then, since $D_j(f+g)(a)=D_jf(a)+D_jg(a)$, and the sum of continuous functions is continuous, $D_j(f+g)(a)$ exists and is continuous. So for any k, $D_k......D_{i_1}(f+g)(a)=$ $D_k.....D_{i_1}(a)f+D_k....D_{i_1}g(a)$, and so it is continuous and exists by assumption.
Clearly, if $f+g\in C^{\infty}(U)$ then $g+f\in C^{\infty}(U)$. Hence,
. 3. Consider $0: U\rightarrow \mathbb{R}$, defined by $0(x)=0$ for all x. Clearly it is continuous (since it is constant) and all derivatives are 0 (constant), hence they exist and are continuous.
Associativity holds clearly.
Let $f\in C^{\infty}(U)$. Put -f to be defined by $(-f)(x)=-(f(x))$ Note, that adding $f$ to $-f$ yields the zero map. Hence it suffices to show that $-f$ is smooth. It is continuous clearly. Observe that $D_j-f=-D_jf$ and so it exists and is continuous. Hence, $D_{k}D_{k-1}....D_{i_1}-f=-D_k.....D_{i_1}f$ which exists and is continuous for every k and every $a\in U$.
Product: If $f\in C^{\infty}(U)$ and $g\in C^{\infty}(U)$ then I claim that $fg\in C^{\infty}(U)$
Note that $D_jfg(a)=f(a)D_jg(a)+g(a)D_jf(a)$. Higher order derivatives are just a combination of the product rule and linearity, as shown to hold in 1.
Distributivity and associativity of multiplication: Clearly holds.
Identity: Define $1:U\rightarrow \mathbb{R}$ by $1(x)=1$ for all $x\in U$. It is smooth by a similar argument to 3.
Moreover it is clearly commutative.
Is this reasonable?