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Consider a rng $R$ and ideals $I_1,\dots , I_n$ such that $I_i+I_j=R$ for $i\neq j$. Show that the inclusion $\prod I_i\subseteq \bigcap I_i$ is an equality (Exercise 1.1.4, Bosch's Algebraic Geometry and Commutative Algebra)

I already proved Chinese remainder theorem (in the previous exercise), but I don't know how to approach this one, and I've been thinking for a while. Even if I show $R/\prod (I_i)\cong R/ \bigcap (I_i)$, the thesis is not proven, and I can't think of anything else. Can you only give a hint to start? Thank you

Bill Dubuque
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Dr. Scotti
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    I would do this by induction. Of course $IJ\subset I\cap J$. Also, there are $a\in I,b\in J$ such that $a+b=1$, so, if $z\in I\cap J$, then $z=z(a+b)=za+zb\in IJ$. Hence, $IJ=I\cap J$. Now, for the step, use the fact that there exist $a_i\in I_i$ and $b_i\in J_{n+1}$ such that $a_i+b_i=1$ for all $i\leq n$. – defacto Jan 14 '22 at 17:36
  • @XuguiManuel Why $za + zb \in IJ,$ could you please explain this? – weird Nov 26 '22 at 19:33
  • @MathIgnorance Note that elements of $IJ$ are of the form $\sum^n_{i=1}a_i b_i$, with $a_i\in I$ and $b_i\in J$. – defacto Nov 27 '22 at 21:01

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First assume $n=2$. If $I+J=R$ then we can write $1=x+y$ where $x\in I, y\in J$. So if $z\in I\cap J$ then we have:

$z=z(x+y)=zx+zy\in IJ$

Now try to apply induction in order to prove the general case.

Mark
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  • I do not understand why the line before last implies that $z \in IJ,$ could you please explain? – weird Nov 26 '22 at 19:30
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    @MathIgnorance We have $x\in I$ and $z\in J$, and thus $zx\in IJ$. Similarly, $z\in I, y\in J$ and so $zy\in IJ$. So the sum $zx+zy$ is in $IJ$ as well. – Mark Nov 26 '22 at 19:55
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Below is a sketch of a simple inductive proof. Note that since $\,\cap = {\rm lcm}\,$ for principal ideals, this an ideal-theoretic generalization of lcm = product for pair coprime integers.

Theorem $\ \bigcap I_i = \prod I_i\, $ if $\ I_j + I_k = 1\, $ for $\,j\neq k$.

Proof $ $ (Sketch) $\ $ We proceed by (complete) induction on $n$.

Base case: $\,n = 2,\,$ i.e. $\, I\cap J = I J\ $ if $\ \color{#c00}{I+J = 1}.\,$ Direction $\, I\cap J \supseteq IJ$ is clear. Conversely $\, I\cap J = (\color{#c00}{I+J})(I\cap J) = I(I\cap J) + J(I\cap J) \subseteq IJ.\,$

Induction step: $ $ assume it is true for all $\,k< n.\,$ Then

$$\qquad\qquad\begin{align} &I_n \cap (I_{n-1}\cap\:\! \cdots\:\!\cap I_1)\\[.3em] =\ &I_n \cap (I_{n-1}\times \cdots\times I_1)\ \ {\rm by\ induction}\\[.3em] =\ &I_n \!\times (I_{n-1}\times \cdots\times I_1)\ \ \text{by $\,k\!=\!2\,$ case & Lemma below } \end{align}$$

Lemma $\ \forall i\!: \color{#0a0}{I + I_i = 1}\Rightarrow I + I_1\cdots I_{n-1} = 1\,\ $ [Euclid's Lemma in ideal form]

Proof $ $ Induct using $\,I + I_i J = I + I_i J + I J = I + (\color{#0a0}{I_i+I})J = I+J$

Or $\bmod I\!:\ I_k\equiv 1\,\Rightarrow\, \prod I_k \equiv 1^n\! = 1\ $ [ideal form of coprimes are closed under product]

Bill Dubuque
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