A note I'm studying says its primitive ideals space is $\lbrace \lbrace 0 \rbrace, \mathcal{B}_0(\mathcal{H}) \rbrace $. I think it might be just $\lbrace \lbrace 0 \rbrace \rbrace $ so I'm somewhat confused now. Anyone please help clarify?

Adjoining a unit correspond to one point compactifiction of the primitive ideal space.
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Marc PalmJul 22 '11 at 7:48

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Please read the FAQ on how to ask: why do you think the answer should be $\{\{0\}\}$. Is the confusion over the terms "ideal" or "primitive"??
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Matthew DawsJul 22 '11 at 16:38

Say, the algebra is $A$. Then $\pi(I) \in \pi(A)'$ and by irreducibility, $\pi(I) = cI$. Since $\pi(I) = \pi(I)^2$, $c$ must be either 0,1. But $c=0$ contradicts the irreducibility of $\pi$. So, $\pi(I) = I$ and hence any irrep with $\mathcal{B}_0(\mathcal{H})$ as its kernel, would be zero on $\mathcal{B}_0(\mathcal{H})$ and map $I$ to $I$. This again contradicts the irreducibility. Thus, there is no irrep with kernel $\mathcal{B}_0(\mathcal{H})$. I might have misunderstood some big point.
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Chao K.Jul 22 '11 at 22:46

To fix notation, let $A^\sim = \{ (a,t) : a\in A, t\in\mathbb C\}$. Then the irrep you need is $A^\sim\rightarrow\mathbb C; (a,t) \mapsto t$. This is a homomorphism, and has kernel $A$. It's irreducible, as, well, it's non-zero, and $\mathbb C$ is one-dimensional! So your argument is wrong when you claim "This again contradicts the irreducibility".
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Matthew DawsJul 25 '11 at 13:12

Oh, yeah! Thank you very much. I somehow only thought of irreps on the same Hilbert space $\mathcal{H}$ and later came to a conclusion that $\mathcal{H}$ must be one-dimensional. So, let's say $x = \{0\}$, $y = A$ and we now have the primitive ideal space $\{x,y\}$. The space is not even $T_1$ since $\clos(\{x\}) = \{x, y\}$. However, the primitive ideal space of $A$ is $\{x\}$. According to pm′s comment, the primitive ideal space of $A^\sim$ is supposed to be the one−point compactification of $\{x\}$ which is $\{x, \infty\}$ with the discrete topology, which is $T_1$. What's wrong here?
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Chao K.Jul 25 '11 at 20:37

whydo you think the answer should be $\{\{0\}\}$. Is the confusion over the terms "ideal" or "primitive"?? – Matthew Daws Jul 22 '11 at 16:38`$A^\sim = \{ (a,t) : a\in A, t\in\mathbb C\}$`

. Then the irrep you need is $A^\sim\rightarrow\mathbb C; (a,t) \mapsto t$. This is a homomorphism, and has kernel $A$. It's irreducible, as, well, it's non-zero, and $\mathbb C$ is one-dimensional! So your argument is wrong when you claim "This again contradicts the irreducibility". – Matthew Daws Jul 25 '11 at 13:12