Let us say that a Hamel basis $H$ in an algebra $A$ is *closed under multiplication*, if $ab\in H$ whenever $a,b\in H$. It is an easy observation that if $A$ has such a basis then there it also has a character (a linear-multiplicative functional; see *Amer. Math. Monthly* **124** (2017), no. 7, 651–653.)


All characters on Banach algebras are automatically continuous. When $A=c\_0$, then they are just point evaluations.



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> Does $c\_0$ (with the pointwise product) have a Hamel basis closed under multiplication?
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One could also ask:



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> Is there an infinite-dimensional (complex) abelian Banach algebra that has a Hamel basis closed under multiplication?
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Bases closed under multiplication
Assume that a Banach algebra $B$ has such a basis $H$. Take any $h\in H$. Consider the element $x=ah+a^2h^2+a^3h^3+\dots$ where $a>0$ is chosen small enough to make the series converge in $B$. Assume $x=\sum\_{j=1}^n c\_jh\_j$ for some $h\_j\in H$. Write
$$
x=ah+a^2h^2+\dots+a^mh^m+a^mh^mx=ah+a^2h^2+\dots+a^mh^m+\sum\_j c\_ja^m(h^mh\_j)
$$
If all the basis elements $h^k\in H$, $k=1,\dots,m$ are distinct, then the elements $h^mh\_j\in H$ in the last sum can cancel only $n$ of them and we get a representation for $x$ of length $\ge m-n>n$ if $m>2n$. Thus some two of the powers must coincide and, thereby, $h^k$, $k\ge 1$ span only a finite dimensional space. Since it is true for every $h\in H$, we conclude that the powers of *every* element of $B$ span only a finite-dimensional space. This excludes any "decent" infinite-dimensional Banach algebra ($c\_0$ in particular). 


However *something* survives. The most degenerate example is the following. Take some Banach algebra, say $B=c\_0$, an idempotent element $y\in B$ (of norm $1$, say, though it doesn't matter), say $y=(1,0,0,0,\dots)$, some linear multiplicative functional $\psi$ on $B$ such that $\psi(y)=1$, which would be the evaluation of the first coordinate in that particular example, and redefine the product in $B$ to
$$
ab=\psi(a)\psi(b)y
$$  

(this is not a very exciting product, of course, but it is a product nevertheless).
Now just take any maximal linearly independent set containing $y$ in $B\setminus\operatorname{ker}\psi$ and normalize its elements $h$ by replacing them by $\frac 1{\psi(h)}h$. You have your basis.


