relatively prime ideals $p_1\cap p_2+p_3=(1)$ when $p_1,p_2$ and $p_3$ are relatively prime

Question

Let $R$ be a ring. If $p_1,p_2,p_3$ are three pairwise relatively prime ideals, then $p_1\cap p_2+p_3=(1)$.

I just want to confirm my method is correct. Since $p_1+p_3=(1)$ and $p_2+p_3=(1)$, $(p_1+p_3)(p_2+p_3)=(1)$. $LHS=p_1p_2+p_3^2+p_3(p_1+p_2)=p_1p_2+p_3^2+p_3=p_1p_2+p_3$ as $p_3^2\subset p_3$. So we obtain equality by $p_1,p_2$ relatively prime.

Answer 1

Your proof is correct, but it can be shortened.

Let $1=a+b=c+d$, where $a\in p_1$, $c\in p_2$, $b,d\in p_3$; then $$ 1=(a+b)(c+d)=ac+(bc+ad+bd)\in p_1p_2+p_3 $$ Since $p_1p_2\subseteq p_1\cap p_2$, we are done.

Note that the hypothesis that $p_1$ and $p_2$ are coprime is not needed.