TITLE: Finite number of elements generating the unit ideal of a commutative ring
QUESTION [8 upvotes]: 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.

REPLY [5 votes]: Alternatively, clearly $(f_1,\dots,f_r) \subset \sqrt{(f^{n_1}_1,\dots, f^{n_r}_r)}$. Thus, $\sqrt{(f^{n_1}_1,\dots, f^{n_r}_r)} = A$, and so $1 \in (f^{n_1}_1,\dots, f^{n_r}_r)$.