@EmilJeřábek I edit to "algebraic independent"

@OlegEroshkin Thank you for your helpful comment

@EmilJeřábek Thank you for your helpful comments

@EmilJeřábek There is no polynomial $F(x,y)$ with integer coefficients with $F(\alpha,\beta)=0$

@RobertBryant is the question obvious in the particular case :foliation of TM by verticql space?

If i remember correctly the Lima paper contains an answ3r to Smale question:what is the rank of S^3 answrwd by Lima(=1)

@SpencerKraisler My apology for my late repley. To be honest the answer I provided in the post was based on immediate computation. It was not based on a reference on Stein equation. at the time of answering I was not aware of the special terminology, i.e stein equation, for such equation. So I do not know any resource on the topic. I will come back you if I find some resources

So your answer is actually $A\mapsto A\otimes C_0(X)$. The tensor product of a C^* algebra and a Z^* algebra is again Z^* algebra. X being uncountable is not an approximately sigma compact. After all I am really curious about this category, the category of Z^* algebras. Please see my conversation with David Roberts about a precise structure for the category of C^* algebras(what kind of morphisms we are consideringt, etc). Any way it would be interesting that the two categories would be isomorphic in a reasonable sense

I gave a comment on to OP which justifies the word algebra provided we let A to be a non commutative algebra