For ordinal $\alpha > 1$, what's the smallest $\beta > 0$ such that $\alpha \beta = \beta$?

Question

For finite $\alpha$, it's fairly obvious $\beta = \omega$. If, on the other hand, $\alpha$ is infinite, then I noticed $\beta = \sum_{i = 0}^\infty \alpha^i$ satisfies $\alpha \beta = \beta$, but is it the minimal one?

Answer 1

If $\alpha\beta=\beta$ then it follows by induction that $\alpha^n\beta=\beta$ for every $n\lt\omega.$ If we assume $\beta\gt0,$ it follows that $\beta=\alpha^n\beta\ge\alpha^n\cdot1=\alpha^n$ for every $n\lt\omega,$ whence $\beta\ge\sup_{n\lt\omega}\alpha^n=\alpha^\omega.$

On the other hand, as you observed, $\alpha\alpha^\omega=\alpha^{1+\omega}=\alpha^\omega.$ More generally, the solutions of the equation $\alpha\beta=\beta$ are all ordinals of the form $\beta=\alpha^\omega\gamma.$