Ordinals which satisfy $\beta \cdot \alpha=\alpha$

Question

Let $\beta>1$ be a fixed ordinal. I want to find a nice characterization of the ordinals $\alpha>1$ which satisfy $\beta\cdot\alpha=\alpha$. I have already seen that $\beta+\alpha=\alpha$ if and only if $\beta\cdot\omega\leq\alpha$. Using this as a guide, I conjectured that $\beta\cdot\alpha=\alpha$ if and only if $\beta^{\omega}\leq\alpha$, but this turned out to be false (the forward implication is true but the reverse one is false). Are there any nice characterization of the ordinals $\alpha$?

Answer 1

The ordinals of interest are multiples of $\beta^\omega$ i.e. $\alpha=\beta^\omega\cdot\gamma$ for some ordinal $\gamma$. It is straightforward to see that these ordinals satisfy the equation. To prove these are the only ones, one may simply see that $\beta\cdot(\beta^\omega\cdot\gamma+\delta)=\beta^\omega\cdot\gamma+\beta\cdot\delta>\beta^\omega\cdot\gamma+\delta$ whenever $\beta>1$ and $0<\delta<\beta^\omega$.