Distributive law of ideals in $\mathbf Z$ relating $\cap$ and $+$.

Question

Let $\mathfrak a, \mathfrak b$ and $\mathfrak c$ be ideals in $\mathbf Z$.
Then show that
$$ \mathfrak a \cap (\mathfrak b + \mathfrak c) = \mathfrak a \cap \mathfrak b + \mathfrak a \cap \mathfrak c .$$

It is easy to prove $\supset$ side, but $\subset$ side is difficult for me. My questions are following.

1. How to prove $\subset$ side.
2. Is it true for all of principal ideal domains?

Answer 1

A ring which satisfies this property is called arithmetical. This paper has some nice results related to this: https://www.math.purdue.edu/~heinzer/preprints/irr15.pdf

For domains this is the same as being Prufer. I would direct you here, there is a very informative answer here: When does the distributive law apply to ideals in a commutative ring?

Edit: I got so excited about the second question, I suppose I forgot to comment on the first one. I think it is easiest for the integers to think about $$(a) + (b)= (\gcd(a,b))$$ and $$(a)\cap (b)=(\operatorname{lcm}(a,b)).$$ Then you have reduced the question to $$\operatorname{lcm}(a, \gcd(b,c))=\gcd(\operatorname{lcm}(a,b),\operatorname{lcm}(a,c)).$$

This is done here: https://proofwiki.org/wiki/GCD_and_LCM_Distribute_Over_Each_Other Since you have unique factorization in the integers, it is pretty clear to see by just looking at the prime factorizations why this is true.

Answer 2

@CPM Thank you for your link.
I just have obtained an elementary proof.

In a principal ideal domain, the relation $(a)\cap ((b)+(c)) = (a)\cap (b) + (a)\cap (c) $ holds.

I'll show the $\subset$ side.
Let $(b)+(c) = (d)$ and $(a)\cap (d) = (e)$
and we can write $b=b'd, c=c'd$ with $sb'+tc'=1$ and $ e=fa = gd$
Then
$$ e = e \cdot 1 = e(sb'+tc') = esb' + etc' $$
Since $ e=fa=gd , esb'\in (a)\cap (b)$ and $ etc' \in (a)\cap (c)$ and the proof is completed.