Let $A$ be a commutative ring with $1$. Let $f_1,\dots,f_r$ be elements of $A$. Suppose $A = (f_1,\dots,f_r)$. Let $n > 1$ be an integer. Can we prove that $A = (f_1^n,\dots,f_r^n)$ without using axiom of choice?
EDIT It is easy to prove $A = (f_1^n,\dots,f_r^n)$ if we use axiom of choice. Suppose $A≠(f_1^n,…,f_r^n)$. By Zorn's lemma(which is equivalent to axiom of choice), there exists a prime ideal $P$ such that $(f_1^n,…,f_r^n) \subset P$. Since $f_i^n \in P$, $f_i \in P$ for all $i$. Hence $A = (f_1, \dots,f_n) \subset P$. This is a contradiction.