Consider the (non-unital) $\mathbb{C}$-algebra (point-wise multiplication) of $\mathcal{S}$ of Schwartz functions on $\mathbb{R}$.


**Question:** Does there exist some character (non-zero multiplicative functional to $\mathbb{C}$) $\omega$ of $\mathcal{S}$ that is *not* the evaluation map at any point in $\mathbb{R}$, i.e. does $\mathcal{S}$ as an algebra, admit characters other than the functionals given by Dirac measures?


Some remarks:


1. This can not happen if we require $\omega$ to be continuous with respect to the $L^\infty$ norm (one can extend $\omega$ to $C\_0(\mathbb{R})$, which is a commutative $C^\ast$-algebra, then use Gelfand-Naimark), my question does not pose any continuity restriction on $\omega$, and can be seen of an algebraic nature.
2. This amounts to the question whether the ideal $C\_c^\infty(\mathbb{R})$ of compactly supported smooth functions is contained in a codimension $1$ ideal of $\mathcal{S}$.



Characters of algebra of Schwartz functions
Let $m$ be a multiplicative functional. Let $A={\mathbb C}\oplus\mathcal S$ be the algebra $\mathcal S$ extended by the constant functions. This algebra is unital. Setting $m(f+\lambda)=m(f)+\lambda$ extends $m$ to a multiplicative functional on $A$.
For $f\in A$ let $s(f)=\sup\_{x\in\mathbb R}|f(x)|$. Assume $s(f)<1$. Then we claim that $\frac1{1-f}$ lies in $A$.
We have $\frac1{1-f}-1=\frac f{1-f}$ and any derivative of the latter is of a quotient of a polynomial in the derivatives of $f$ divided by a power of $1-f$. One concludes that $\frac f{1-f}$ lies in $\mathcal S$, hence $1-f$ is invertible in $A$.
For $\lambda\ne 0$ we have $f-\lambda=\lambda(\frac f\lambda -1)$, so if $|\lambda|>s(f)$, the latter is invertible. Now we have $m(f-m(f))=0$, hence $f-m(f)$ is not invertible, hence $|m(f)|\le s(f)$. This means that $m$ is continuous in the sup-norm. Hence it extends to a continuous functional on $C\_c({\mathbb R})$, i.e., a Radon measure. For this to be multiplicative, it must be a point measure.


