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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.

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up vote 3 down vote accepted

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.

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You can also get each $I_k$ to be prime, generated by (say) the $(k+2)^{th}$ prime integer. – Graham Leuschke May 30 '10 at 18:54
And of course in the second example $J$ and the $I_k$'s are prime. – Jonas Meyer May 30 '10 at 21:11
Thanks to all! Audrey. – Audrey Kirilova May 31 '10 at 4:07

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