Dear all,
here is a question that has been bothering me. It goes without saying that I would appreciate any help in answering it. 


Let {I\_k} be a countable sequence of two sided closed ideals in a C\*-algebra (ring) and J be a two sided closed ideal in the same C\*-algebra (ring). 
Then "intersection of {I\_k + J} = (intersection of {I\_k}) + J" 


If needed we can relax the above hypothesis by assuming that {I\_k}'s and J are prime ideals. 


Thanks in advance,
Audrey Kirilova. 



Intersection of ideals in C*-algebra or even rings in general 
In the most general form, for arbitrary ideals over rings, this
is false. In the ring $\mathbb{Z}$ let $I\_k$ be generated by $2^k$
and let $J$ be generated by $3$. Then $I\_k+J=\mathbb{Z}$
for all $k$ and so $\cap\_{k=1}^\infty(I\_k+J)=\mathbb{Z}$.
But $\bigcap\_{k=1}^\infty I\_k=\lbrace0\rbrace$ and so
$J+\bigcap\_{k=1}^\infty I\_k=J\ne\mathbb{Z}$.


For an example in a $C^\*$-algebra let $R=C[0,1]$ the continuous
functions on $[0,1]$. Let $(a\_k)$ be an enumeration of
the rationals in $[0,1]$, and $I\_k$ be the ideal of functions
vanishing at $a\_k$. Let $J$ be the ideal of functions vanishing at
$1/\sqrt2$ say. Then $\bigcap\_{k=1}^\infty I\_k=\lbrace0\rbrace$
and $I\_k+J=R$ for all $k$, and the rest proceeds as above.


