Problem 1150

4.15. We will prove that for a finite set $M$ of positive numbers, there exists a basis $B$. We will call a subset $S$ of positive numbers a superbasis for $M$ if each number in $M$ can be represented as a product

$$\alpha_{1}^{i_{1}} \ldots \alpha_{m}^{i_{m}}, \text { where } \alpha_{i}, \ldots, \alpha_{m} \in S, i_{i}, \ldots, i_{m} \in \mathbf{Z}$$

For example, the set $M$ itself is a superbasis for $M$. Among all superbases for $M$, we choose the set

$$S_{0}=\left\{\beta_{i} ; \ldots ; \beta_{n}\right\}$$

containing the minimum number of elements. We will prove that if $n \geqslant 2$, then $S_{0}$ is a basis for $M$. Suppose that some element $u \in M$ admits different representations as a product of integer powers of elements from $S_{0}$:

$$u=\beta_{1}^{i_{1}} \ldots \beta_{n}^{i_{n}}=\beta_{1}^{f_{1}} \ldots \beta_{n}^{i_{n}}, \quad \text { i.e. } \quad \beta_{1}^{k_{1}} \ldots \beta_{n}^{k_{n}}=1$$

for integers $k_{l}=i_{l}-l_{l}$, not all zero for $l=1, \ldots, n$. Without loss of generality, we can assume that $k_{n} \neq 0$. Let

$$S_{1}=\left\{\gamma_{1}, \ldots, \gamma_{n-1}\right\}, \quad \text { where } \gamma_{l}=\beta_{l}^{1 / k_{n}} \text { for } l=1_{2} \ldots, n-1 .$$

Then each element of the set $S_{0}$ can be represented as a product of integer powers of elements from $S_{1}$:

$$\beta_{l}=\gamma_{l}^{k_{n}} \text { for } l=1, \ldots, n-1 ; \beta_{n}=\gamma_{1}^{-k_{1}} \ldots \gamma_{n-1}^{-k n-1}$$

Therefore, the set $S_{1}$ is a superbasis for $M$, containing $n-1$ elements, which contradicts the choice of $S_{0}$. Hence, $S_{0}$ is a basis for $M$. If there exists a superbasis $S_{0}$ for $M$ containing a single element $\beta \neq 1$, then $S_{0}$ is also a basis for $M$, since the equality $\beta^{i}=\beta^{\prime}$ is impossible for $i \neq j$. Finally, if the set $S_{0}=\{1\}$ is a superbasis for $M$, then $M=\{1\}$ and the set $S_{1}=\{2\}$ will be a basis for $M$.