If $\{f_i\}$ generate the unit ideal in a ring, so do $\{f_i^N\}$ for any positive $N$ [duplicate]

Let $R$ be a commutative ring, and let $\{f_i\}$ be a finite set of elements generating the unit ideal in R. Then $\{f_i^N\}$ also generate the unit ideal in $R$, for any positive $N$.

Why is this true?

• What is the unit ideal in a ring? Is it $(R1)$, i.e. the whole ring? – Robert Lewis Oct 1 '13 at 18:50
• mathoverflow.net/a/59087/450 – Georges Elencwajg Oct 1 '13 at 18:54
• I have the feeling this has been asked, or even duplicated already. Couldn't find the right search parameters though... – rschwieb Oct 1 '13 at 18:58
• Another way to think about this, is that the only ideal whose radical is the full ring is the full ring (this is obvious if you think about the radical as the intersection of prime ideals above the ideal). But, $\sqrt{(f_i^N)}\supseteq (f_i)=A$, so $\sqrt{(f_i^N)}=A$. From what I just said this implies that $(f_i^N)=A$ – Alex Youcis Oct 1 '13 at 19:03
• Although, now that I think about it, I think this fact is just couched inside of what @GeorgesElencwajg linked to. – Alex Youcis Oct 1 '13 at 19:05

If $\{f_i\}$ generates $R$, then write $$1=\sum_{k=1}^Ma_kf_k$$ where $a_k\in R$. Then raise this monstrosity to the power of $MN$. Then every term has some $f_k^N$ or a higher power in it. Appropriate rearragement shows that $1$ is in the span of $f_k^N$.
Any prime ideal containing $f^n$ contains $f$. Therefore, any prime ideal containing $I=\left<\{f_i^n\}\right>$ contains $R$. This is absurd, and therefore there is no prime ideal containing $I$, and this implies $I=R$.