[proofplan] The strategy combines completion with the spectrum theorem. A division algebra is, by definition, a unital algebra in which every non-zero element is invertible. After completing $A$ to a Banach algebra $B$ in which $A$ embeds isometrically, the [Spectrum is Non-Empty and Compact](/theorems/2669) theorem provides, for each $x \in A$, a scalar $\lambda \in \mathbb{C}$ with $\lambda 1 - x \notin G(B)$. Since $A$ is dense in $B$ and $G(B)$ is open, this non-invertibility transfers from $B$ down to $A$ via a perturbation argument, and the division-algebra hypothesis then forces $\lambda 1 - x = 0$, so $x = \lambda 1$. Hence the canonical map $\mathbb{C} \to A$, $\lambda \mapsto \lambda 1$, is surjective; it is automatically isometric because $\|\lambda 1\| = |\lambda| \|1\| = |\lambda|$. [/proofplan] [step:Recall what a division algebra is and pass to the completion] A **complex unital normed division algebra** $A$ is a complex algebra equipped with a submultiplicative norm $\|\cdot\|$ and a multiplicative identity $1 \neq 0$ such that the group of invertibles is $G(A) = A \setminus \{0\}$, i.e.\ every non-zero element of $A$ has a two-sided multiplicative inverse in $A$. We assume the standard normalisation $\|1\| = 1$ (which holds for any Banach algebra after rescaling). The completion of $A$ as a normed space is a Banach space $B$ in which $A$ is dense. The algebra operations extend by continuity: for $a, b \in B$ with $a_n \to a$ and $b_n \to b$ along sequences in $A$, the product $a_n b_n$ is Cauchy in $A$ (since $\|a_n b_n - a_m b_m\| \leq \|a_n\| \|b_n - b_m\| + \|a_n - a_m\| \|b_m\|$ and $(a_n), (b_n)$ are bounded Cauchy), so we define $ab := \lim a_n b_n$. The submultiplicative inequality, associativity, and distributivity all extend by continuity. Hence $B$ is a unital Banach algebra in which $A$ embeds as an isometric subalgebra with the same unit; we identify $A \subseteq B$. [/step] [step:Apply the spectrum theorem in the completion to find a non-invertible scalar shift] Fix $x \in A$. Apply [Spectrum is Non-Empty and Compact](/theorems/2669) to $x$ regarded as an element of the unital Banach algebra $B$: the spectrum $\sigma_B(x) = \{\lambda \in \mathbb{C} : \lambda 1 - x \notin G(B)\}$ is non-empty. Choose $\lambda \in \sigma_B(x)$, so that \begin{align*} \lambda 1 - x \notin G(B). \end{align*} [/step] [step:Transfer non-invertibility from $B$ down to $A$] We claim $\lambda 1 - x \notin G(A)$ as well. Since $\lambda 1 - x \in A$ (as $x \in A$ and $1 \in A$), suppose for contradiction $\lambda 1 - x \in G(A)$, with inverse $y \in A$ satisfying $(\lambda 1 - x) y = y(\lambda 1 - x) = 1$ in $A$. Because the inclusion $A \hookrightarrow B$ is a unital algebra embedding (Step 1), the same identities $(\lambda 1 - x) y = y(\lambda 1 - x) = 1$ hold in $B$, so $y$ is a two-sided inverse of $\lambda 1 - x$ in $B$, i.e.\ $\lambda 1 - x \in G(B)$. This contradicts $\lambda \in \sigma_B(x)$. Hence $\lambda 1 - x \notin G(A)$. [guided] This step is the bridge between the spectrum theorem (which lives in the complete algebra $B$) and the division-algebra hypothesis (which lives in $A$). The point is that non-invertibility *in the larger algebra $B$* automatically implies non-invertibility *in the smaller algebra $A$*. Why? An inverse, if it exists in $A$, is also an inverse in $B$ (because $A$ embeds in $B$ as a unital subalgebra with the same multiplication). So \begin{align*} \lambda 1 - x \in G(A) \implies \lambda 1 - x \in G(B). \end{align*} Contrapositively, \begin{align*} \lambda 1 - x \notin G(B) \implies \lambda 1 - x \notin G(A). \end{align*} This is a one-way implication. The reverse — non-invertibility in $A$ implies non-invertibility in $B$ — can fail in general (an element may become invertible upon enlarging the algebra). But we only need the direction we have. It is critical that we picked $\lambda \in \sigma_B(x)$ and not $\sigma_A(x)$. The spectrum theorem [Spectrum is Non-Empty and Compact](/theorems/2669) requires a Banach algebra, so we must work in the complete algebra $B$. Then we descend to $A$ using the implication above. [/guided] [/step] [step:Use the division-algebra hypothesis to conclude $x$ is a scalar multiple of $1$] We have $\lambda 1 - x \in A$ and $\lambda 1 - x \notin G(A)$ (Step 3). By the division-algebra hypothesis, the only non-invertible element of $A$ is $0$: \begin{align*} A \setminus G(A) = \{0\}. \end{align*} Hence $\lambda 1 - x = 0$, i.e.\ $x = \lambda 1$. [/step] [step:Define the canonical map and verify it is an isometric isomorphism] Define \begin{align*} \theta : \mathbb{C} &\to A \\ \lambda &\mapsto \lambda 1. \end{align*} *Linearity.* $\theta(\alpha \lambda + \beta \mu) = (\alpha \lambda + \beta \mu) 1 = \alpha (\lambda 1) + \beta (\mu 1) = \alpha \theta(\lambda) + \beta \theta(\mu)$. *Multiplicativity.* $\theta(\lambda \mu) = (\lambda \mu) 1 = (\lambda 1)(\mu 1) = \theta(\lambda) \theta(\mu)$, where the second equality uses $1 \cdot 1 = 1$ and bilinearity of multiplication. *Unital.* $\theta(1) = 1 \cdot 1 = 1$. *Surjectivity.* By Step 4, every $x \in A$ has the form $\lambda 1$ for some $\lambda \in \mathbb{C}$ (uniquely determined since $1 \neq 0$ implies $\lambda 1 = \mu 1 \implies (\lambda - \mu) 1 = 0 \implies \lambda = \mu$). So $\theta$ is surjective. *Injectivity.* From the previous paragraph, $\theta(\lambda) = \theta(\mu) \implies (\lambda - \mu) 1 = 0 \implies \lambda = \mu$. *Isometric.* For $\lambda \in \mathbb{C}$, \begin{align*} \|\theta(\lambda)\|_A = \|\lambda 1\|_A = |\lambda| \cdot \|1\|_A = |\lambda|, \end{align*} using homogeneity of the norm and $\|1\|_A = 1$ (Step 1). Hence $\theta : \mathbb{C} \to A$ is a unital algebra isomorphism that is isometric, i.e.\ an isometric isomorphism. This completes the proof. [/step]