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Let $R$ be a (commutative) ring with identity. A covering of R is a subset $\{r_1,...,r_n\}$ of elements of $R$ such that $R$ is generated by $r_1,...,r_n$. I think that if $\{r_1,...,r_n\}$ is a covering family and $k>0$ is an integer, the family $\{r_1^k,...,r_n^k\}$ should also be a covering of $R$.

To see this you just see that the products $\{r_ir_j\}_{i,j}$ form also a covering of $R$ (taking the square of the relation $1=\sum_{i=1}^n\alpha_i r_i$ given by some $\alpha_i$ in $R$). Iterating the process $s$ times you get that the $\{r_{i_1}^{t_1}...r_{i_s}^{t_S}\}_{i_j,t_1+...+t_s=s}$ also form a covering and for $r$ sufficiently large you can regroup those terms to get a formula looking like $1=\sum_{i=1}^n\beta_i r_i^k$.

However the proof seems a bit rude to me. Do you think there could be a more elegant proof ?

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