# If a sub-C*-algebra does not contain the unit, is it contained in a proper ideal?

Suppose that $R$ and $S$ are unital rings and that $S$ is a subring of $R$ in the weak sense where the multiplicative identities $1_R$ and $1_S$ are not assumed to be the same. In fact, assume $1_R \neq 1_S$. Then $S$ cannot contain any invertible element of $R$ (if $s \in S$ and there is an $r \in R$ with $sr=1_R$ then $0 = (1_Rs - 1_Ss)r = 1_R - 1_S$ so $1_R = 1_S$). This is sort of reminiscent of the statement a proper ideal in $R$ cannot contain any invertible element of $R$. My question is whether there must actually be some proper ideal $I$ of $R$ with $S \subset I \subset R$. Actually, I would like to ask the question with the rings replaced by C*-algebras - but a positive answer to the corresponding question about rings would clearly suffice.

Let $A$ and $B$ be unital C*-algebras with $B$ a sub-C*-algebra of $A$, but suppose $1_A \neq 1_B$. Does it necessarily follow that there is a proper closed ideal $I$ of $A$ satisfying $B \subset I \subset A$?

I'm also curious what happens if the assumption that $B$ is unital is dropped. In this case I'm not even certain whether $B$ can contain an invertible element of $A$.

Let $B$ be a non-unital sub-C*-algebra of a unital C*-algebra $A$. Can $B$ contain an invertible element of $A$? If not, does there have to exist a proper closed ideal $I$ of $A$ with $B \subset I \subset A$?

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Hmm regarding my second question, it is clearly possible for a nonunital subring to contain an invertible from the larger ring as the example $2 \mathbb{Z} \subset \mathbb{Q}$ shows. – Mike F Dec 4 '11 at 0:01

For the first question, consider $2$-by-$2$ matrices, and the subalgebra of matrices of the form $\begin{bmatrix}a&0\\0&0\end{bmatrix}$.

For the first part of the second question, the answer is no. Suppose that $a\in B$ is invertible in $A$. Then by spectral permanence, $a^{-1}$ is in $B+\mathbb C\cdot 1_A$, so $a^{-1}=b+\lambda\cdot 1_A$ for some $b\in B$ and $\lambda\in \mathbb C$. This means that $1_A=a^{-1}a=ba+\lambda a\in B$.

(Note that this is special of C*-algebras, and is not true for general algebras or even general Banach algebras. Consider $X\in X\cdot\mathbb C[X]\subset \mathbb C[X,X^{-1}]$. For a similar Banach algebra example, consider $f(z)=z$ in the algebra $C(\mathbb T)$ of continuous functions on the unit circle in $\mathbb C$, which is invertible in $C(\mathbb T)$ but not in the closed subalgebra generated by $f$.)

For the second part of the second question, the answer is no. Let $C_0(\mathbb R)$ act on $L^2(\mathbb R)$ as multiplication operators. Then it forms a nonunital C*-subalgebra of $B(L^2(\mathbb R))$ contained in no proper ideal (its nonzero elements are noncompact).

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Hmm ok, so the algebra of 2-by-2 matrixes has no proper ideals whatsoever it would seem? – Mike F Dec 4 '11 at 6:43
Right: $M_n(F)$ is simple when $n\in\mathbb N$ and $F$ is a field. – Jonas Meyer Dec 4 '11 at 6:47
Also, if $A$ has unit $1$ and $B \subset A$ has $1 \notin A$ then vector space sum $B + \mathbb{C} \cdot 1 = \{b + \lambda \cdot 1 : b \in B, \lambda \in \mathbb{C}\}$ is actually automatically closed and a C*-algebra in its own right? Also, it contains $A$ as an ideal of co-dimension 1 so it is actually, essentially, the unitization of $A$! I don't think I realized this was true before... – Mike F Dec 4 '11 at 6:51
Mike: I think there are some typos in your last comment (a couple of places where "A" should be "B"), but I'm pretty sure I know what you mean, and yes, that's right. – Jonas Meyer Dec 4 '11 at 7:02
Yes, a couple As should really be Bs - oops. Thanks for your answer. I've read some notes which developed, as quickly and painlessly as possible, the basic results of C*-algebra theory. Unfortunately, this has made my grasp of the subject pretty superficial. I know very few examples and can't figure out 90% of the zillions of little questions I've been asking myself lately to try and improve my grasp of this stuff. – Mike F Dec 4 '11 at 7:17

The second question has a more general answer. Let $\mathcal{H}$ be any infinite-dimensional Hilbert space. Then the set $\mathcal{K}(\mathcal{H})$ of compact operators on $\mathcal{H}$ has the following three properties:

• It is a non-unital $C^{*}$-subalgebra of $\mathcal{B}(\mathcal{H})$.

• It does not contain any invertible element of $\mathcal{B}(\mathcal{H})$.

• It is a maximal two-sided ideal of $\mathcal{B}(\mathcal{H})$.

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Thanks for your response, but I don't think this is quite what I had in mind (although it has been a long time since I thought about this). Let me try to clarify what I meant. Here are two properties a subset $B$ of a unital ring $A$ may or may not have: (1) $B$ doesn't contain any invertible element of $A$, (2) $B$ there is a proper ideal $I$ of $A$ with $B \subset I$. Notice that (2) is a stronger than (1). In the case where $B$ is a nonunital sub-C*-algebra of a unital C*-algebra $A$, Jonas showed that (1) holds. He also showed by example that (2) does not have to hold. – Mike F Oct 25 '13 at 5:20
In your example, $B = K(H)$ contains no unit of $A = B(H)$, so (1) holds. Of course, as Jonas showed, (1) must always hold. But we also have $B \subset I = K(H)$ where $I$ is a proper (closed) ideal. So (2) holds as well in your example, and we do not learn that (2) need not hold in general. – Mike F Oct 25 '13 at 5:22
What is this supposed to generalize? I don't think it answers the question. – Jonas Meyer Nov 1 '13 at 2:51