Let $\alpha:R\rightarrow B$ be a $R$-Algebra, $f\in R$, $S=\{f^n\}$. If $\alpha(S)\subseteq B^\times$ must $f^{-1}\in B$?

Question

I'm missing just one step for a bigger proof. The problem and my current proof can be found here

Let $\alpha:R\rightarrow B$ be a $R$-Algebra, $f\in R$, $S=\{f^n\}$.

If $\alpha(S)\subseteq B^\times$ must $f^{-1}\in B$?

I can't come up with a counterexample, but I also don't see a way to prove this is in fact true.

Can we show that $f^{-1}\in B$, or does this in fact break my proof?

Answer 1

I think I don’t get the question. If $f\in R$ how can $f^{-1}$ be in $B$?

Clearly $\alpha(f)\in B^\times$ implies that $\alpha(f)^{-1}$ exists in $B$.

On the other hand $\alpha(f^n)\in B^\times$ for all $n\in \Bbb N_0$ does not imply that $f\in R^\times$. For example $\Bbb Q$ is a $\Bbb Z$-algebra via the inclusion. Take $2\in\Bbb Z$ then $2^n \in \Bbb Q^\times$ for all $n\in \Bbb N_0$, but $2\notin \Bbb Z^\times$.