Why the Smooth Function Module Is Injective over Real Polynomials but the Polynomial Module Is Not

Almost 12 am, and a question has come to mind:

Proposition.

Let $M$$M$ be a $\mathbb{R}[D]$$\mathbb R[D]$ module where $D={\displaystyle \frac{d}{dx}}$$D=\dfrac{d}{dx}$. For example, $M=\mathbb{C}[[X]]$$M=\mathbb C[[X]]$.

Then $M$$M$ is injecitve iff every $p(D)y=g$$p(D)y=g$ has solution in $M$$M$, where $p(D)\ne 0$$p(D)\ne 0$.

Proof. Since $\mathbb{R}[X]$$\mathbb{R}[X]$ is a PID, this is equivalent to asking whether $M$$M$ is divisible. $◻$$\square$

Q:

View $C^{\mathrm{\infty }}(\mathbb{R})$$C^{\infty}(\mathbb{R})$ as an $\mathbb{R}[X]$$\mathbb{R}[X]$-module, where the module structure is given by the map

$$\mathbb{R}[X]⟶\mathbb{R}[D],p(X)\mapsto p(D)$$

with $D={\displaystyle \frac{d}{dx}}$$D = \dfrac{d}{dx}$. Is this an injective module?

The answer is yes—see the introduction of the paper for details.

https://link.springer.com/article/10.1007/s13163-018-0266-5

 

We have $\mathbb{R}$$\mathbb R$ is convex open set, hence $C^{\mathrm{\infty }}(\mathbb{R})$$C^\infty(\mathbb R)$ is divisible $\mathbb{R}[X]$$\mathbb R[X]$ module hence injecitve $\mathbb{R}[X]$$\mathbb R[X]$.

Proposition. $\mathbb{R}[x]\subset C^{\mathrm{\infty }}(\mathbb{R})$$\mathbb R[x]\subset C^{\infty}(\mathbb R)$ is not a injective submodule.

Proof. Consider $(D-1)y=0◻$$(D-1)y=0\quad\square$