# What does $(B+I)/I\sim B/(B\cap I)$ tell us?

Let $A$ be a $C^*$-algebra in which $B$ is a $C^*$-subalgebra and $I$ is a closed ideal. In several books on $C^*$-algebras I have encountered the following:

$(B+I)/I$ is $*$-isomorphic to $B/(B\cap I)$.

It seems important, but none of the books I read gives a hint why this is important.

So what does this isomorphism say actually?

Thanks!

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That is the second isomorphism theorem. Perhaps you have used it before on groups. –  azarel Jul 16 '12 at 19:51

Here's a general algebraic interpretation that I've always found compelling.

Within a fixed algebra $A$, we're given a subalgebra $B$ and an ideal $I$ (something you can quotient by, I'm skipping the specifics). We are interestered in the quotient $B/I$, but this does not make sense in general, since we don't necessarily have $I \subset B$.

1. either extend the subalgebra $B$ so that $I$ lies in this extension (the smallest such subalgebra is $B+I$)

2. or restrict $I$ so that this restriction lies in $B$ (the largest such ideal is $B \cap I$).

The isomorphism $(B+I)/I \cong B/(B\cap I)$ tells us that both approaches yield the same result.

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(+1) The best!${}$ –  Norbert Jul 17 '12 at 0:57
Very good explanation. Thanks! –  Hui Yu Jul 17 '12 at 1:00
If $I,J$ are closed ideals of the $C^*$-algebra $A$, then $I+J$ is a closed ideal as well.
(proof: the natural 'second isomorphism' map $I/(I\cap J)\to A/J$ is a $*$-(iso)morphism onto $(I+J)/J$, hence the latter is closed in $A/J$. Now $(I+J)/J$ and $J$ are Banach, hence $I+J$ as well.)