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If $I_1, \dots, I_n \subset R$ are distinct ideals with the property that $I_i + I_j = R$ for any $i \ne j$, show that $I_i + \prod_{j \ne i} I_j = R$ for any $i$ where the product is take over all the integers $1, \dots, n$ except $i$.

My failed attempts : $\prod_{j \ne i} I_j \subset I_j$ for all $j \ne i$ because of definition of ideals, but that does not imply $I_i + \prod_{j \ne i} I_j = R$ because still may not $\prod_{j \ne i} I_j = I_j$. Also, $I_i + I_j = R$ implies that for any $r \in R$, $r=i_1+i_2=i_1+i_3=\dots$ for fixed $i_1 \in I_1$ and some $i_j \in I_j$, though $r-i_1 \in \cap_{j \ne i} I_j$ but that does not lead to any thing!

Source of the exercise : Introduction to Algebraic Geometry by Justin R. Smith

hm2020
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  • Is the ring commutative with unit? – Bernard Sep 11 '21 at 08:46
  • @Bernard the book doesn't say that but I think yes (bc the book is about algebraic geometry). Also the answer below is not helpful at all! –  Sep 11 '21 at 08:49
  • @TheMagicMountain- an example: If you have 3 coprime ideals $I_1,I_2,I_3$ you want $I_1+I_2I_3=(1)$. Choose $(a_2,b_2)\in I_1\times I_2, (a_3,b_3)\in I_1\times I_3$ with $a_i+b_i=1$. It follows $(a_2+b_2)(a_3+b_3)=1=a_2a_3+a_2b_3+a_3b_2+b_2b_3+\in I_1+I_2I_3$, the the claim follows. Do you see how this generalize? – hm2020 Sep 11 '21 at 10:06
  • More generally: Let $(a_i,b_i)\in I_1\times I_i$ for $i\neq 1$ be elements with $a_i+b_i=1$. Consider the element $\prod_i (a_i+b_i)=1$. Is this an element in $I_1 + \prod_{j\neq 1}I_j$? – hm2020 Sep 11 '21 at 10:13
  • @hm2020 First you suppose $a_2 \in I_1$ then you conclude $a_2b_3 \in I_2I_3$. How do you infer $a_2 \in I_2$ too? –  Sep 11 '21 at 10:18
  • Since $I_1$ is an ideal it follows $a_2a_3+a_2b_3+a_3b_2\in I_1$. Also $b_2b_3\in I_2I_3$ and hence $a_2a_3+a_2b_3+a_3b_2 +b_2b_3=1\in I_1+I_2I_3$. – hm2020 Sep 11 '21 at 10:19
  • @hm2020 Thanks! If you would you like you may post your comments as an answer. –  Sep 11 '21 at 10:22

3 Answers3

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$R=\prod_{j\ne i}(I_i+I_j)=L+\prod_{j\ne i}I_j$ where $L$ is a sum of products of ideals including $I_i$. Hence, $L\subseteq I_i$. Maybe it helps to write $1=\prod_{j\ne i}(x_{ij}+x_j)$ with $x_{ij}\in I_i$ and $x_j\in I_j$ for $j\ne i$.

Brauer Suzuki
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  • How do you conclude $L \subseteq I_i$ implies $L = I_i$ ? –  Sep 11 '21 at 09:39
  • Also I couldn't understand you last sentence. –  Sep 11 '21 at 09:52
  • Being an ideal means that $I_i$ is closed under addition and $RI_i\subseteq I_i$ and $I_iR\subseteq I_i$. When you expand the product over $(I_i+I_j)$, all summands except the "last" one contain at least one factor $I_i$. You collect those summands in $L$. My last sentence (which I just corrected) is only an alternative (perhaps more elementary) approach. – Brauer Suzuki Sep 11 '21 at 10:39
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Question: "@hm2020 Thanks! If you would you like you may post your comments as an answer. – TheMagicMountain"

Answer: If $I_i+I_j=(1)$ for $i\neq j$, it follows there are elements (let $i=1$) $(a_j,b_j)\in I_1 \times I_j$ ($j\neq 1$) with $a_j+b_j=1$. You may check that the element

$$\prod_j (a_j+b_j)=1 \in I_1 + \prod_{j\neq 1}I_j$$

hence

$$ I_1 + \prod_{j\neq 1}I_j=(1)$$

is the unit ideal.

hm2020
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Supposing that $R$ is a commutative ring with unit, it suffices to show that no prime ideal contains both $I_i$ and $\prod_{j\ne i}I_j$.

Now if a prime ideal contains $\prod_{j\ne i}I_j$, it contains one of the $I_j$ ($j\ne i$), so it cannot contain $I_i$ since $I_i+I_j=R$.

Bernard
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  • I don't understand how each term of the sum belongs to $\prod_{j\ne i}I_j$! Say $i=3$ and $n=5$, then $1=(u_1+v_1)(u_2+v_2)(u_3+v_3)(u_4+v_4)$ so how this product minus the term $u_1u_2u_4u_5$ belongs to your mentioned product? –  Sep 11 '21 at 09:36
  • You're right. I was somewhat too fast. I've replaced it with a proof based on prime ideals. – Bernard Sep 11 '21 at 10:01
  • Unf, I have no idea about the theorem you have used relating to the prime ideals. –  Sep 11 '21 at 10:11
  • It is a direct consequence of the definition of a prime ideal: if a prime ideal contains a product of ideals, it contains at least one of them (proved by contradiction). – Bernard Sep 11 '21 at 12:04
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    The anonymous troll is here again! – Bernard Sep 13 '21 at 22:03