Sum of ideals equals R implies the sum of their powers is R

Question

So given $I_1,I_2,\dots, I_n$ are ideals in a commutative ring $R$ with unity, suppose that $$I_1+I_2+\dots+I_n=R$$ I want to show that given arbitrary $c_1, c_2, \dots, c_n\geq1$, we have $$I_1^{c_1}+I_2^{c_2}+\dots + I_n^{c_n}=R.$$

$\textbf{Here's what I know so far:}$

The arbitrary sum of ideals is an ideal by a simple lemma, products of ideals are ideals by another lemma, and if an ideal contains 1 then it is equal to the whole ring. Furthermore, if an ideal is equal to the whole ring, then it's sum or product with another ideal is also the whole ring.

Hence, there should be a "smallest ideal sum" $I_1 + \dots + I_j$ such that it contains 1, but that for all indices $1\leq k<j$, none of $I_1+\dots + I_k$ contains unity. Then there are elements $i_1 \in I_1, \dots ,i_j \in I_j$ such that $$i_1+\dots + i_j=1.$$

Now, I know that $i_k^{e_k}\in I_k^{e_k}$, but this doesn't actually seem relevant since I have no reason to believe they would sum to unity. The whole sum-of-ideals-equals-the-ring thing screams Chinese Remainder theorem, but I'm not sure how quotients of $R$ by $I_1, \dots, I_n$ are supposed to help me find elements in $I_1^{e_1},\dots,I_k^{e_k}$ that sum to unity. I feel like I have a few of the major pieces of a solution, but it's not clear how I should combine them. Can anyone point me in the right direction?

Answer 1

If $I_1+\cdots+I_n=R$ then there are $a_j\in I_j$ with $$a_1+\cdots+a_n=1.\tag{1}$$ Write $C=c_1+\cdots+c_n$. Take the $C$-th power of $(1)$, and we get $$(a_1+\cdots+a_n)^C=1^C=1.\tag{2}$$ Expanding out the left side of $(2)$, each term has a factor of $a_j^{c_j}$ for some $j$, so that $1\in I_1^{c_1}+\cdots+I_n^{c_n}$.

Answer 2

Here is a solution inspired by algebraic geometry: we first prove that for ideals $I_1, \ldots, I_n$, we have $I_1 + \cdots + I_n = \langle 1 \rangle = R$ if and only if for all prime ideals $\mathfrak{p}$ of $R$, then $I_i \not\subseteq \mathfrak{p}$ for some $i$. ($\Rightarrow$) If, on the contrary, we had $I_i \subseteq \mathfrak{p}$ for all $i$, then we would also have $I_1 + \cdots + I_n \subseteq \mathfrak{p}$ which would imply $\mathfrak{p} = R$, giving a contradiction since the unit ideal is not considered a prime ideal. ($\Leftarrow$) Suppose that $I_1 + \cdots + I_n \ne \langle 1 \rangle$. Then there exists a maximal ideal $\mathfrak{m}$ containing $I_1 + \cdots + I_n$, and this maximal ideal would also be prime.

Now, to apply this fact to conclude the proof, you just need to prove that for prime ideal $\mathfrak{p}$, ideal $I$ and $c \ge 1$, $I^c \subseteq \mathfrak{p}$ if and only if $I \subseteq \mathfrak{p}$. I'll leave the proof of this as an exercise.

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The connection to algebraic geometry is: given a ring $R$, you can form a space $\mathop{\operatorname{Spec}} R$ whose elements are exactly the prime ideals of $R$. It turns out that it is possible to put a topology (called the "Zariski topology") on $\mathop{\operatorname{Spec}} R$ such that the closed sets are exactly sets of the form $$ Z(\mathfrak{a}) := \{ \mathfrak{p} \in \mathop{\operatorname{Spec}} R \mid \mathfrak{a} \subseteq \mathfrak{p} \} $$ for ideals $\mathfrak{a}$ of $R$.

Now, we have identities such as: $Z(\mathfrak{a}_1 + \cdots + \mathfrak{a}_n) = \bigcap_{i=1}^n Z(\mathfrak{a}_i)$; $Z(\langle 1 \rangle) = \emptyset$; and $Z(\mathfrak{a}) = Z(\sqrt{\mathfrak{a}})$ where $\mathfrak{a}$ is the radical ideal $\{ x \in R \mid \exists n \ge 1, x^n \in \mathfrak{a} \}$. And, in fact, $Z(\mathfrak{a}) = \emptyset$ if and only if $\mathfrak{a} = \langle 1 \rangle$.

So, in the language of algebraic geometry, the proof would read: the hypothesis implies $\bigcap_{i=1}^n Z(I_i) = \emptyset$, so $Z(\sum_{i=1}^n I_i^{c_i}) = \bigcap_{i=1}^n Z(I_i^{c_i}) = \bigcap_{i=1}^n Z(I_i) = \emptyset$, which implies $\sum_{i=1}^n I_i^{c_i} = \langle 1 \rangle$.

Answer 3

Here is another perspective. Let $J=I_1^{c_1}+I_2^{c_2}+\dots + I_n^{c_n}$ and consider the quotient ring $R/J$. We write $\pi:R\to R/J$ for the quotient map. Note that if $x\in I_j$ then $x^{c_j}\in J$ and so $\pi(x)$ is nilpotent. So, $\pi(I_j)$ is contained in the nilradical of $R/J$ for each $j$. Since the nilradical is an ideal, $\pi(I_1+I_2+\dots+I_n)$ is contained in the nilradical. But since $1\in I_1+I_2+\dots+I_n$, this means $\pi(1)=1$ is nilpotent in $R/J$. Thus $R/J$ is the zero ring and $J=R$, as desired.

(This is really Lord Shark the Unknown's argument in disguise. Indeed, if you look at the proof that the nilradical is closed under addition, which was the key step, it is very similar to Lord Shark the Unknown's argument. But this method of shifting the perspective to the quotient ring is a really handy general tool that often makes these questions easier to think about.)