# $(x_1,\ldots x_n)=(1)\implies (x_1^{k_1},\ldots, x^{k_n})=(1)$ [duplicate]

Let $R$ be a commutative ring with unit, I'm trying to prove why in this ring

$$(x_1,\ldots x_n)=(1)\implies (x_1^{k_1},\ldots, x^{k_n})=(1)$$

It seems an easy question, but I couldn't prove it, I need a hint or something.

Thanks

• Hint: Write $1$ as a linear combination of the $x_i$ and raise both sides of the equation to a suitable large power (try starting with just $x_1$ and $x_2$ and see what happens). – Tobias Kildetoft Mar 6 '14 at 14:26
• @TobiasKildetoft I've already thought about this hint, it seems a little bit hard to write it down, but it seems the only way to solve this question. – user108205 Mar 6 '14 at 14:31

Let $k = \sup(k_1, \dots, k_n)$. We can write $1 = \sum a_i x_i$ by hypothesis. Therefore $$1 = 1^{nk} = (\sum a_i x_i)^{nk}$$

By the binomial theorem this sum can be written as a sum of elements of the form $b_{a_1, \dots, a_n} x_1^{a_1} \dots x_n^{a_n}$ (where the $b_?$ are some coefficients we don't care about) with $a_1 + \dots + a_n = nk$. So necessarily one of the $a_i \ge k \ge k_i$, and so $x_i^{a_i} \in (x_1^{k_1}, \dots, x_n^{k_n})$, so the whole expression is in the ideal and therefore $1 \in (x_1^{k_1}, \dots, x_n^{k_n})$.