[proofplan] We show $A = \bigoplus_{i=0}^{e-1} \pi^i W(k)$ by first embedding $W(k) \hookrightarrow A$ via the universal lifting property, then proving that $1, \pi, \ldots, \pi^{e-1}$ are $W(k)$-linearly independent (via a valuation argument), and finally showing that $M = \bigoplus_{i=0}^{e-1} \pi^i W(k)$ is dense and closed in $A$, hence equal to $A$. [/proofplan] [step:Embed $W(k)$ into $A$ via the lifting property] The inclusion $\mathbb{F}_p \hookrightarrow k$ induces (by the functoriality bijection of the [Witt Vectors](/theorems/???) theorem) an injective ring homomorphism $W(\mathbb{F}_p) \hookrightarrow W(k)$. Since $W(\mathbb{F}_p) \cong \mathbb{Z}_p$ (the $p$-adic integers form the unique strict $p$-ring with residue field $\mathbb{F}_p$), this gives $\mathbb{Z}_p \hookrightarrow W(k)$. The identity map $\operatorname{id}_k: k \to k = A/pA$ lifts, by the [Lifting Homomorphisms Between Strict $p$-Rings](/theorems/???) theorem, to a unique ring homomorphism $\iota: W(k) \to A$ with $\iota \equiv \operatorname{id}_k \pmod{p}$. We claim $\iota$ is injective. Since $W(k)$ is an integral domain (it is $p$-torsion free and $W(k)/pW(k) \cong k$ is a field, so the argument from the [Mixed Characteristic DVRs](/theorems/???) theorem shows $W(k)$ is a DVR, hence a domain), the kernel $\ker(\iota)$ is a prime ideal of $W(k)$. The prime ideals of $W(k)$ are $\{0\}$ and $pW(k)$. Since $\iota(p) = p \neq 0$ in $A$ (as $A$ has characteristic $0$), $\ker(\iota) \neq pW(k)$. Therefore $\ker(\iota) = \{0\}$, and $\iota$ is injective. We identify $W(k)$ with its image $\iota(W(k)) \subseteq A$. [guided] Why does $W(k)$ embed into $A$? The key is the functoriality of Witt vectors: a ring homomorphism $k \to A/pA$ lifts uniquely to $W(k) \to A$. Since $k = A/pA$, the identity map on $k$ provides the residue-level input. The lift $\iota: W(k) \to A$ is automatically a ring homomorphism. The injectivity of $\iota$ is a standard consequence of $W(k)$ being a DVR with unique prime $p$. The only prime ideals of a DVR are $\{0\}$ and the maximal ideal. Since $\iota(p) = p \neq 0$ in $A$ (this uses $\operatorname{char}(A) = 0$), the kernel cannot be the maximal ideal $pW(k)$, so it must be $\{0\}$. What is the image of $W(k)$ inside $A$? It is a complete DVR with uniformizer $p$ and residue field $k$, sitting as a subring of $A$. When $e = 1$, the [Unramified Witt Vector Structure](/theorems/???) theorem says $A \cong W(k)$. For general $e$, $W(k)$ is a "smaller" subring and $A$ is a free $W(k)$-module of rank $e$. [/guided] [/step] [step:Prove $W(k)$-linear independence of $1, \pi, \ldots, \pi^{e-1}$] Suppose $\sum_{i=0}^{e-1} c_i \pi^i = 0$ with $c_i \in W(k)$. We show all $c_i = 0$. Let $v_A$ denote the normalized valuation on $A$ (with $v_A(\pi) = 1$). For $c \in W(k)$ with $c \neq 0$, write $c = p^m u$ with $u \in W(k)^\times$ and $m = v_{W(k)}(c) \geq 0$. Then $v_A(c) = v_A(p^m u) = m \cdot v_A(p) = me$ (since $v_A(p) = e$ and $v_A(u) = 0$ because $u \in W(k)^\times \subseteq \mathcal{O}_A^\times$). So the valuation of $c_i \pi^i$ is $v_A(c_i \pi^i) = m_i e + i$, where $m_i = v_{W(k)}(c_i) \geq 0$. The values $m_i e + i$ for $i = 0, 1, \ldots, e - 1$ lie in distinct residue classes modulo $e$ (since $i$ ranges over $\{0, 1, \ldots, e-1\}$). Therefore, if any $c_i \neq 0$, the nonzero terms $c_i \pi^i$ in the sum all have distinct $v_A$-valuations. By the ultrametric inequality (the strong triangle inequality for $v_A$), the sum $\sum c_i \pi^i$ has valuation equal to the minimum of the $v_A(c_i \pi^i)$ over the nonzero terms, so the sum cannot be zero. Contradiction. Hence all $c_i = 0$. [guided] This is the cleanest way to see the linear independence: the valuation provides a natural "grading" on $A$, and the different powers $\pi^i$ ensure the summands land in different residue classes modulo $e$. More concretely: the value group of $W(k)$ under $v_A$ is $e\mathbb{Z}_{\geq 0} = \{0, e, 2e, \ldots\}$, because $v_A(p) = e$ and $W(k)$ has uniformizer $p$. The value of $c_i \pi^i$ (with $c_i \neq 0$) is then $m_i e + i$ for some $m_i \geq 0$. Since $0 \leq i < e$, these values are all distinct modulo $e$, hence pairwise distinct. In a non-archimedean valued ring, a sum of terms with pairwise distinct valuations has valuation equal to the minimum -- so the sum cannot vanish unless every term is zero. Why do we need $i < e$ specifically? Because for $i \geq e$, the value $m_i e + i$ could coincide with another term's value modulo $e$, breaking the distinctness. The bound $i \leq e - 1$ is sharp: $\pi^e$ has $v_A$-valuation $e = v_A(p)$, so $\pi^e$ "collides" with the valuation of $p \in W(k)$, and indeed $\pi^e$ and $p$ differ by a unit. [/guided] [/step] [step:Show the module $M = \bigoplus_{i=0}^{e-1} \pi^i W(k)$ is dense in $A$] We prove $A = M + p^m A$ for all $m \geq 1$, which implies $M$ is dense in the $p$-adic (equivalently, $\pi$-adic) topology on $A$. **Base case ($m = 1$).** We must show $A = M + pA$, equivalently that $M$ surjects onto $A/pA = k$ under the reduction map. The [Teichmüller Map](/theorems/???) provides representatives: for any $\bar{x} \in k$, the element $[\bar{x}] \in W(k) \subseteq M$ satisfies $[\bar{x}] \equiv \bar{x} \pmod{p}$. So the reduction of $M$ modulo $pA$ contains all of $k$, confirming $A = M + pA$. More precisely, $A/pA = k$ is generated as a $k$-vector space by the images of $1, \pi, \ldots, \pi^{e-1}$ (since $\pi^e = p \cdot u$ for some unit $u \in \mathcal{O}_A^\times$, and every element of $\mathcal{O}_A/p\mathcal{O}_A$ can be written as a $k$-linear combination of $1, \bar{\pi}, \ldots, \bar{\pi}^{e-1}$, where $\bar{\pi}$ is the image of $\pi$ in $A/pA$). The $k$-coefficients lift to Teichmüller representatives in $W(k)$, so $A = M + pA$. **Inductive step.** Assume $A = M + p^m A$. For any $a \in A$, write $a = m_0 + p^m a'$ with $m_0 \in M$ and $a' \in A$. By the base case applied to $a'$: $a' = m_1 + p a''$ with $m_1 \in M$ and $a'' \in A$. Then $a = m_0 + p^m m_1 + p^{m+1} a''$. Since $p^m m_1 = p^m \sum_{i=0}^{e-1} c_i \pi^i$ with $c_i \in W(k)$, this lies in $M$ (because $p^m c_i \in W(k)$). So $a \in M + p^{m+1} A$. [/step] [step:Show $M$ is closed in $A$ to conclude $M = A$] The module $M = \bigoplus_{i=0}^{e-1} \pi^i W(k)$ is a finitely generated free module over the complete ring $W(k)$. The $p$-adic topology on $M$ is induced by the filtration $p^m M = \bigoplus_{i=0}^{e-1} \pi^i p^m W(k)$. Since $W(k)$ is $p$-adically complete, each component $W(k)$ is complete, and a finite direct sum of complete modules is complete. So $M$ is $p$-adically complete. Since the $p$-adic topology on $M$ coincides with the subspace topology from $A$ (because $p^m M = M \cap p^m A$: the inclusion $p^m M \subseteq M \cap p^m A$ is clear, and for the reverse, if $\sum c_i \pi^i \in p^m A$ then $v_A(\sum c_i \pi^i) \geq me$, which forces $v_{W(k)}(c_i) \geq m$ for each $i$ by the valuation argument from the linear independence step), completeness of $M$ implies $M$ is a closed subset of $A$. A dense closed subset of a topological space equals the whole space, so $M = A$: \begin{align*} A = \bigoplus_{i=0}^{e-1} \pi^i W(k). \end{align*} [guided] The final step brings together density and closedness. Let us verify the key claim $M \cap p^m A = p^m M$ more carefully. The inclusion $p^m M \subseteq M \cap p^m A$ is immediate. For the reverse: let $x = \sum_{i=0}^{e-1} c_i \pi^i \in M$ with $x \in p^m A$, i.e., $v_A(x) \geq me$. If any $c_i \neq 0$, then $v_A(c_i \pi^i) = m_i e + i$ where $m_i = v_{W(k)}(c_i)$. The minimum of these values (over the nonzero terms) equals $v_A(x) \geq me$. Since the values $m_i e + i$ are pairwise distinct modulo $e$, the minimum is achieved by exactly one term. But $m_i e + i \geq me$ for all $i$ means $m_i \geq m$ for $i = 0$ and $m_i \geq m - 1$ for $i \geq 1$. In fact, since $m_i e + i \geq me$ gives $m_i \geq m - i/e$ and $m_i \in \mathbb{Z}$, we get $m_i \geq m$ for $0 \leq i \leq e - 1$ (since $i/e < 1$ for $i < e$). So $c_i \in p^m W(k)$ for all $i$, giving $x \in p^m M$. This means the $p$-adic topology on $M$ is exactly the subspace topology, so $M$ is complete in the subspace topology, hence closed. Since $M$ is dense (from the previous step) and closed, $M = A$. [/guided] [/step]