Non-unital algebras in geometric algebra, smooth envelopes In Nestruev's (2000) Smooth Manifolds and Observables, the authors define an $\mathbb{R}$-algebra as a commutative, associative algebra with unit (p. 21). A natural generalization of this definition would drop the requirement of a unit. (For example, any self-adjoint, commutative, and non-unital C*-algebra defines such a "non-unital $\mathbb{R}$-algebra".) I am interested in how many of Nestruev's constructions carry over to the non-unital case. For example: is the notion of a smooth envelope of a geometric $\mathbb{R}$-algebra $\mathcal{F}$ well-defined if $\mathcal{F}$ lacks a unit?
P.S. Apologies if this question is too preliminary; I am new to posting on this site. Many thanks for reading :)
 A: 
For example: is the notion of a smooth envelope of a geometric R-algebra F well-defined if F lacks a unit?

Yes.  Recall the construction: $F$ is geometric if the Gelfand homomorphism
$$\def\Map{\mathop{\rm Map}}
\def\Spec{\mathop{\rm Spec}}
\def\R{{\bf R}}
\def\Hom{\mathop{\rm Hom}}
G\colon F → \Map(\Spec(F),\R)$$
is injective, where $\Spec(F)=\Hom(F,\R)$ is the set of homomorphisms $F→\R$
and $\Map$ denotes the set of maps of sets $\Spec(F)→\R$.
The smooth envelope of $F$ is then constructed as
the real algebra $E$ generated by the image of $G$
inside $\Map(\Spec(F),\R)$
and closed under smooth compositions:
if $f_1,…,f_n∈E$ and $g\colon \R^n→\R$ is smooth,
then $g(f_1,…,f_n)∈E$.
What changes if $F$ is nonunital?
Homomorphisms $F→\R$ are defined in the same manner as before,
and in fact coincide with unital homomorphisms $\def\hF{{\hat F}} \hF→\R$,
where $F→\hF$ is the unitization of $F$.
The same definition of the Gelfand homomorphism continues to work.
The resulting smooth envelope is a unital algebra even if $F$ is not:
indeed, the unit is produced by taking $n=0$ and $g=1$ in the construction of the smooth envelope of $F$.
Even if we exclude $n=0$, we can always take $g=1$ for any $n$,
which still gives us the unit.
