Converse of Wiener's lemma Let $A$ be a commutative Banach algebra with unit. 
It is well known that if the Gelfand transform $\hat{x}$ of $x\in A$ is non-zero, then $x$ is invertible in $A$ (the so called Wiener Lemma in the case when $A$ is the Banach algebra of absolutely convergent Fourier series).
As a converse of the above, let $B$ be a Banach space contained in $A$ and suppose $B$ is closed under inversion - i.e.: If $x\in B$ and $x^{-1}\in A$ then $x^{-1}\in B$.
(1) Prove that $B$ is a Banach algebra.
(2) Must $A$ and $B$ have the same norm? If not are the norms similar?
(3) Do $A$ and $B$ have the same maximal ideal space?
 A: For now, I shall assume that $B$ is a closed subspace of $A$ (and is treated simply as a topological vector space for the moment).
(1) Let $x \in B$; we need to prove that $x^2 \in B$. From this elementary algebra will imply that $B$ is closed under multiplication. Now, when $t$ is close to zero, we have $1-tx$ invertible in $A$, hence in $B$; so $1+tx + t^2x^2 + \dots$ lies in $B$ for $t$ close to zero. Taking the second derivative at $0$ (recall that $B$ is a closed subspace) shows that $x^2 \in B$.
I don't fully understand what you are asking in (2) yet, so:
(3) No. Take $A$ to be any Banach algebra not equal to the complex numbers and $B$ the subalgebra spanned by the identity.
A: I think the OP needs to give a precise formulation of what he/she means by "a Banach space inside a Banach algebra", because this affects the answers to the questions in non-trivial ways.
Contrary to what the OP seems to claim in comments to Akhil Matthew's answer, if $B$ is not closed in $A$, then it can be an inverse-closed subspace without the given Banach norm on $B$ being submultiplicative. However, by Akhil's argument, $B$ is a subalgebra of $A$.
It is still not clear to me what precisely is meant by (2). So I shall just point out for the record that if we let $A({\mathbb T})$ be the space of all continuous functions ${\mathbb T}\to {\mathbb C}$ with absolutely convergent Fourier series, equipped with pointwise product, this is a Banach algebra and it is a unital subalgebra of $C({\mathbb T})$. Clearly the norm on $A({\mathbb T})$ is not equivalent to the sup-norm inherited from $C({\mathbb T})$; but $A({\mathbb T})$ is inverse-closed in $C({\mathbb T})$, by Gelfand's version of Wiener's $1/f$ lemma.
Note also in (3) that one can have commutative, semisimple, unital Banach algebras $A$ and $B$, and an injective, contractive, unital algebra homomorphism $B\to A$ with dense range, such that $A$ and $B$ have different maximal ideal spaces. The example I have in mind is due, I think, to Honary, but I am out of the office right now and can't look this up.
