For 3 pairwise coprime ideals, is the intersection of two coprime to the third? Context
Hello, what I'm currently trying to prove is the generalization to this equality $JK = J \cap K$ which is:
$$\prod_{i=1}^{n}I_i=\bigcap_{i=1}^{n}I_i$$
I was told this could be done using mathematical induction on the smaller case's proof. So I used the small proof as my induction base, I asummed it worked for $n=k$ and now I havet to check it works for $n=k+1$.
The first contention $\prod_{i=1}^{n}I_i \subset \bigcap_{i=1}^{n}I_i$ is easy enough but on the second one I feel like I'm supposed to use that $ ( \bigcap_{i=1}^kI_i)$ and $ I_{k+1}$ are coprime but I don't immediately see it.
Question
Say I have three ideals $I, K,J$ and they're corpimne with eachother, this means there exists elemnts on said ideals such that:
$$i_1+k_1=1$$
$$j_1+k_2=1$$
$$i_2+j_2=1$$
Now, is it true then that $I\bigcap J$ is coprime with K?
The way I rationalized it was:
Given that $J$ and $I$ are coprimes we have that $J\bigcap I=JI$ then we take
$$i_1+k_1=1$$
$$j_1+k_2=1$$
And multiply them so we have $i_1 j_1 + i_1 k_2 + j_1 k_1 + k_1 k_2 =1 $ and since $K$ is and ideal we can name $i_1 k_2 + j_1 k_1 + k_1 k_2 =k_0 \in K$ and so we have that:
$$i_1 j_1 + k=1$$
which is the definition of coprime.
Is there a simpler way? Am I off on something?
Thankyou
 A: That's the classical proof - just like the analogous gcd proof - by multiplying the Bezout equations. As for a simpler way, it is more conceptual to employ the (ideal) distributive law as follows
$\begin{align}
\text{like gcds:}\, \ \ \ &(a,\ \ bc) \,=\,  (a,\,\ \ \ (a,\,\ b)\,c)\qquad \textbf{[GCD Euclid's Lemma]}\\[.4em]
{\bf Lemma}\ \ \ \ \ &\smash[b]{\underbrace{A\!+\!BC\, =\, A+(A\!+\!B)C}_{\textstyle\!\! =\, A(1\!+\!C)+BC }}\ \ \ \ \quad\textbf{[Ideal Euclid's Lemma]}\\[.4em]
\bf{Proof}\qquad &
\end{align}$
${\bf Corollary}\ \  \color{#c00}A+\color{#c00}BC\, =\, A+C\ \ {\rm if}\ \ \color{#c00}{A+B} = 1.\ $  So, if $\:\color{#c00}{I+I_k = 1},\,  \forall k,\, $ then induction yields
$\qquad\qquad\quad\ \color{#c00}{I}+ \color{#c00}{I_1} I_2\cdots I_n = \color{#c00}{I}+ \color{#c00}{I_2} I_3 \cdots I_n =\, \cdots\, =  \color{#c00}{I}+\color{#c00}{I_n} = 1\qquad\qquad\tag{E}$
i.e. just like gcds: $ $  if $I$ is coprime to all $I_k$ then $I$ remains coprime to their product.
Now let  $H_k$ denote our claim that $\, \bigcap_{i=1}^k I_i = \prod_{i=1}^k I_i\,$ for pair-coprime $I_i$. The base case $H_2$ is easy to prove. Suppose for induction that $H_k$  is true for all $\,k\le n.\,$ Then
$$\begin{align}
\smash[b]{{\Large \cap}_{i=1}^{n+1}I_i} = &\ \ I_{n+1}\cap (I_n\cap \cdots\cap I_1)\\[.3em]
= &\ \ I_{n+1}\cap(I_{n}  \cdots\  I_1) \ \ \,\text{by induction case}\ H_n\\[.3em]
= &\ \ I_{n+1}\, \cdot\,(  I_{n} \cdots\ I_1) \ \ \ \text{by induction case $\,H_2\,$ and $\rm(E)$ with $I = I_{n+1}$}\\[.3em]
= &\ \ \textstyle{{\prod}_{i=1}^{n+1}I_n}
\end{align}$$
i.e. $\,H_{n+1}\,$ is true, which completes the proof of the claim by (complete) induction.
