# An ideal in the ring of infinitely differentiable functions

Well, I was just doing an elementary exercise, but am a tad bit skeptical about how I've gone about it. It goes as follows:

Let $R$ be the ring of infinitely differentiable functions defined on, say, the open interval $-1 < t <1$. Let $J_n$ be the set of functions $f \in R$ such that $D^k f (0) = 0$ for all $0 \le k \le n$. Here, $D$ is the differential operator. So, $J_n$ is the set of functions that whose derivatives up to order $n$ vanish at $0$. Show that $J_n$ is an ideal in $R$.

Now, what I did was:

Let $g \in R, f \in J_n$

Consider $D^k (g(f(x))) = D^{k-1} ((Dg)(f(t)) \times(Df)(t))$, from the chain-rule.

Now, as $Df(t) = 0$, we have $D^k (g(f(x))) = 0$ and $gf \in J_n$. Following which we can conclude that $J_n$ is an ideal.

Now, my question is: was I justified in taking composition rather than multiplication to be the product rule in the ring of functions? And, also, what about the case, $k=0$?

How would I deal with this, instead?

• "Now, my question is: was I justified in taking composition rather than multiplication to be the product rule in the ring of functions?" What is the addition in your ring? Jul 23, 2013 at 21:07
• Pointwise addition. And I realized that the multiplication must also be pointwise, rather than composition. But, thanks for the hint. Jul 23, 2013 at 21:22

I guess it should be clear that the addition in $R$ is that of pointwise addition of functions. Is composition of functions a suitable choice for multiplication? This would require among others that $h\circ(f+g)=h\circ f+h\circ g$, i.e. $h(f(x)+g(x))=h(f(x))+h(g(x))$ for all $f,g,h,x$. You can show by simple counterexamlpes that this does not hold in general (it holds with linear functions). Rather, the multiplication in $R$ is understood to be pointwise as well, i.e. $(gf)(x)=g(x)\cdot f(x)$.