[proofplan] We use the commutativity of the finite $A$-algebra $T$ to localize $T$ at the maximal ideal determined by the residual eigenvalue system $\lambda$. The residual eigenspace condition implies that this localization acts nontrivially on the localized module, so some characteristic-zero component of the localized algebra must survive. Choosing a minimal prime on such a component gives a finite domain over $A$, whose fraction field is a finite extension of $K$ and whose integral closure is the required coefficient ring. Finally, the eigenspace over this field is nonzero because the chosen component survives after inverting the uniformizer, and reduction of the resulting character recovers the original residual system. [/proofplan] [step:Localize at the residual eigenvalue system] Let $\mathfrak m_A\subset A$ denote the maximal ideal of the complete discrete valuation ring $A$, and define \begin{align*} \mathfrak m_T:=\ker(\lambda). \end{align*} Since $T$ is commutative, localization at prime and maximal ideals is available. Since $\lambda:T\to k$ is an $A$-algebra homomorphism and $k$ is a field, $\mathfrak m_T$ is a maximal ideal of $T$ containing $\mathfrak m_A T$. Define \begin{align*} M_{\mathfrak m_T}:=M\otimes_T T_{\mathfrak m_T}, \end{align*} where $T$ acts on $M$ through the given homomorphism $T\to \operatorname{End}_A(M)$. The hypothesis gives a nonzero vector $\bar v\in M\otimes_A k$ satisfying $(t-\lambda(t))\bar v=0$ for every $t\in T$. Hence the one-dimensional $k$-subspace $k\bar v\subset M\otimes_A k$ is a nonzero $T$-submodule on which $T$ acts through $T/\mathfrak m_T\cong k$. Therefore \begin{align*} (M\otimes_A k)_{\mathfrak m_T}\neq 0. \end{align*} Since localization commutes with tensoring over $A$ in this finite setting, this is \begin{align*} M_{\mathfrak m_T}\otimes_A k\neq 0. \end{align*} By [Nakayama's lemma](/page/Nakayama%27s%20Lemma) for the local ring $T_{\mathfrak m_T}$, applied to the finite $T_{\mathfrak m_T}$-module $M_{\mathfrak m_T}$, we obtain \begin{align*} M_{\mathfrak m_T}\neq 0. \end{align*} [guided] The residual eigenvalue system determines the maximal ideal \begin{align*} \mathfrak m_T:=\ker(\lambda)\subset T. \end{align*} Here we use that $T$ is commutative, so kernels of maps to fields are maximal ideals and localization at $\mathfrak m_T$ is the usual commutative localization. The ideal $\mathfrak m_T$ is maximal because $T/\mathfrak m_T$ embeds as the image of $\lambda$ in the field $k$, and the image contains the image of $A/\mathfrak m_A$ coming from the $A$-algebra structure. We now isolate the part of $M$ on which this residual system can occur. Define the localized module \begin{align*} M_{\mathfrak m_T}:=M\otimes_T T_{\mathfrak m_T}. \end{align*} The vector $\bar v\in M\otimes_A k$ is killed by every element $t-\lambda(t)$, so the line $k\bar v$ is a nonzero $T$-submodule on which $T$ acts through $T/\mathfrak m_T$. In particular the localization of $M\otimes_A k$ at $\mathfrak m_T$ is nonzero: \begin{align*} (M\otimes_A k)_{\mathfrak m_T}\neq 0. \end{align*} Because $T$ is finite over $A$ and $M$ is finite over $A$, localization and reduction modulo $\mathfrak m_A$ are compatible here: \begin{align*} (M\otimes_A k)_{\mathfrak m_T}\cong M_{\mathfrak m_T}\otimes_A k. \end{align*} Thus \begin{align*} M_{\mathfrak m_T}\otimes_A k\neq 0. \end{align*} The ring $T_{\mathfrak m_T}$ is local, and its maximal ideal contains $\mathfrak m_A T_{\mathfrak m_T}$. [Nakayama's lemma](/page/Nakayama%27s%20Lemma) applied to the finite $T_{\mathfrak m_T}$-module $M_{\mathfrak m_T}$ gives \begin{align*} M_{\mathfrak m_T}\neq 0. \end{align*} This is the point of the localization: it discards all components of $T$ irrelevant to the residual eigenvalue system while preserving a nonzero part of the module. [/guided] [/step] [step:Choose a characteristic-zero component that acts nontrivially] Let $\pi\in A$ be a uniformizer, so $\mathfrak m_A=(\pi)$. Since $M$ is finite free over the discrete valuation ring $A$, multiplication by $\pi$ on $M$ is injective. [Localization is exact](/theorems/2847), so multiplication by $\pi$ is also injective on $M_{\mathfrak m_T}$; equivalently, $M_{\mathfrak m_T}$ is $A$-torsion-free. The residual nonzero vector from the preceding step lies in the localization at $\mathfrak m_T$, so some prime of the finite commutative $A$-algebra $T_{\mathfrak m_T}$ lying inside the support of $M_{\mathfrak m_T}$ contains $\mathfrak m_A T_{\mathfrak m_T}$. Choose a minimal prime $\mathfrak p$ of this support contained in such a prime. The ring $T_{\mathfrak m_T}$ is Noetherian because it is finite over the Noetherian ring $A$. For a finite module over a Noetherian commutative ring, the minimal primes in its support are among its [associated primes](/page/Associated%20Prime): equivalently, each such prime is the annihilator of some nonzero element after passing to a suitable primary component. Since multiplication by $\pi$ is injective on $M_{\mathfrak m_T}$, $\pi$ is not a zero divisor on $M_{\mathfrak m_T}$ and therefore is not contained in any associated prime of $M_{\mathfrak m_T}$. Hence $\pi\notin\mathfrak p$, and because the only prime ideals of the discrete valuation ring $A$ are $(0)$ and $(\pi)$, we have $\mathfrak p\cap A=(0)$. Therefore the generic fibre survives: \begin{align*} M_K:=M_{\mathfrak m_T}\otimes_A K\neq 0. \end{align*} Define the finite-dimensional commutative $K$-algebra \begin{align*} T_K:=T_{\mathfrak m_T}\otimes_A K. \end{align*} The algebra $T_K$ acts on the nonzero finite-dimensional $K$-[vector space](/page/Vector%20Space) $M_K$. Since $T_K$ is a finite-dimensional commutative algebra over the field $K$, it is an [Artinian ring](/page/Artinian%20Ring), and the support of the finite $T_K$-module $M_K$ contains a maximal ideal. Choose a maximal ideal $\mathfrak n\subset T_K$ such that \begin{align*} (M_K)_{\mathfrak n}\neq 0. \end{align*} Set \begin{align*} K':=T_K/\mathfrak n. \end{align*} Then $K'/K$ is a finite [field extension](/page/Field%20Extension). [guided] Let $\pi\in A$ be a uniformizer, so $\mathfrak m_A=(\pi)$. Because $A$ is a discrete valuation ring, $A$ is a domain, and because $M$ is finite free over $A$, multiplication by $\pi$ on $M$ is injective. [Localization is exact](/theorems/2845), so multiplication by $\pi$ remains injective on $M_{\mathfrak m_T}$. Thus $M_{\mathfrak m_T}$ is $A$-torsion-free. The preceding step showed that the localization at the residual maximal ideal is nonzero. To pass to characteristic zero, we choose a component of the support of $M_{\mathfrak m_T}$ that is not killed by inverting $\pi$. More precisely, choose a prime in the support of the finite $T_{\mathfrak m_T}$-module $M_{\mathfrak m_T}$ [lying over](/theorems/2944) the closed point, and then choose a minimal prime $\mathfrak p$ of the support contained in it. The ring $T_{\mathfrak m_T}$ is Noetherian because it is finite over the Noetherian ring $A$. For a finite module over a Noetherian commutative ring, the minimal primes in its support are [associated primes](/page/Associated%20Prime). Since $\pi$ acts injectively on $M_{\mathfrak m_T}$, it is not a zero divisor on that module and therefore is not contained in any associated prime. Hence $\pi\notin\mathfrak p$. The only prime ideals of the discrete valuation ring $A$ are $(0)$ and $(\pi)$, so $\mathfrak p\cap A=(0)$. Therefore this component remains after tensoring with the fraction field $K$: \begin{align*} M_K:=M_{\mathfrak m_T}\otimes_A K\neq 0. \end{align*} Now define the finite-dimensional commutative $K$-algebra \begin{align*} T_K:=T_{\mathfrak m_T}\otimes_A K. \end{align*} It acts on the nonzero finite-dimensional $K$-vector space $M_K$. Since $T_K$ is a finite-dimensional commutative algebra over the field $K$, it is an [Artinian ring](/page/Artinian%20Ring). Therefore the support of the finite $T_K$-module $M_K$ contains at least one maximal ideal. Choose a maximal ideal $\mathfrak n\subset T_K$ such that \begin{align*} (M_K)_{\mathfrak n}\neq 0. \end{align*} The quotient \begin{align*} K':=T_K/\mathfrak n \end{align*} is a field, and it is finite over $K$ because $T_K$ is finite-dimensional over $K$. This field is the coefficient field over which the residual eigenvalue system will lift. [/guided] [/step] [step:Construct the lifted character] Let \begin{align*} \theta:T\longrightarrow T_{\mathfrak m_T}\longrightarrow T_{\mathfrak m_T}\otimes_A K\longrightarrow K' \end{align*} be the composite $A$-algebra homomorphism. Let $A'$ be the ring of integers of the finite extension $K'/K$, equivalently the [integral closure](/page/Integral%20Closure) of $A$ in $K'$, and let $\mathfrak m_{A'}\subset A'$ denote its maximal ideal. Since $T$ is finite over $A$, every element $\theta(t)\in K'$ is integral over $A$. Because $A'$ is the integral closure of $A$ in $K'$, we have $\theta(t)\in A'$ for every $t\in T$. Thus $\theta$ factors through an $A$-algebra homomorphism \begin{align*} \widetilde\lambda:T\longrightarrow A'. \end{align*} We prove that the reduction of $\widetilde\lambda$ is $\lambda$. Since $A$ is complete local and $T$ is finite over $A$, the localized algebra $T_{\mathfrak m_T}$ is a finite local $A$-algebra. If $s\in T\setminus\mathfrak m_T$, then $s/1$ is a unit in $T_{\mathfrak m_T}$. Its inverse belongs to the finite $A$-algebra $T_{\mathfrak m_T}$, so its image under $\theta$ is integral over $A$ and hence lies in $A'$. Therefore $\widetilde\lambda(s)$ has an inverse in $A'$, and $\widetilde\lambda(s)\notin\mathfrak m_{A'}$. Let $\varphi:T_{\mathfrak m_T}\to A'$ be the induced local map. The preimage \begin{align*} \mathfrak P:=\varphi^{-1}(\mathfrak m_{A'}) \end{align*} is a prime ideal of $T_{\mathfrak m_T}$ containing $\mathfrak m_A T_{\mathfrak m_T}$. The quotient $T_{\mathfrak m_T}/\mathfrak m_A T_{\mathfrak m_T}$ is an Artinian local $k$-algebra with unique prime ideal $\mathfrak m_TT_{\mathfrak m_T}/\mathfrak m_A T_{\mathfrak m_T}$. Hence $\mathfrak P=\mathfrak m_TT_{\mathfrak m_T}$. It follows that exactly the elements of $\mathfrak m_T$ reduce to zero modulo $\mathfrak m_{A'}$, so the reduction of $\widetilde\lambda$ modulo $\mathfrak m_{A'}$ has kernel $\mathfrak m_T$ and factors as \begin{align*} T\longrightarrow T/\mathfrak m_T\cong k. \end{align*} By construction this reduced character is precisely $\lambda$. [guided] We define the candidate lifted eigenvalue system by following $T$ into the characteristic-zero quotient chosen in the previous step: \begin{align*} \theta:T\longrightarrow T_{\mathfrak m_T}\longrightarrow T_{\mathfrak m_T}\otimes_A K\longrightarrow K'. \end{align*} This is an $A$-algebra homomorphism into the field $K'$. We need the values to lie not just in $K'$ but in its ring of integers. Let $A'$ denote the [integral closure](/page/Integral%20Closure) of $A$ in $K'$, equivalently the ring of integers of the finite extension $K'/K$, and let $\mathfrak m_{A'}\subset A'$ denote its maximal ideal. Since $T$ is finite as an $A$-algebra, every $t\in T$ is integral over $A$. The image of an integral element under a ring homomorphism is integral over the image of the base ring, so $\theta(t)$ is integral over $A$. By the defining property of $A'$, this gives \begin{align*} \theta(t)\in A' \end{align*} for all $t\in T$. Hence $\theta$ factors uniquely as \begin{align*} \widetilde\lambda:T\longrightarrow A'. \end{align*} It remains to check that this character lifts the prescribed residual system. The construction began by localizing at \begin{align*} \mathfrak m_T=\ker(\lambda). \end{align*} First take $s\in T\setminus\mathfrak m_T$. Then $s/1$ is a unit of $T_{\mathfrak m_T}$. Since $A$ is complete local and $T$ is finite over $A$, the local algebra $T_{\mathfrak m_T}$ is finite over $A$; hence the inverse $(s/1)^{-1}$ is integral over $A$. Its image under $\theta$ therefore lies in $A'$. Thus $\widetilde\lambda(s)$ has an inverse in $A'$, so it does not reduce to zero modulo $\mathfrak m_{A'}$. Now let $\varphi:T_{\mathfrak m_T}\to A'$ be the induced map. The preimage \begin{align*} \mathfrak P:=\varphi^{-1}(\mathfrak m_{A'}) \end{align*} is a prime ideal containing $\mathfrak m_A T_{\mathfrak m_T}$. After quotienting by $\mathfrak m_A$, we are looking at a prime ideal of \begin{align*} T_{\mathfrak m_T}/\mathfrak m_A T_{\mathfrak m_T}. \end{align*} This quotient is an Artinian local $k$-algebra, so it has exactly one prime ideal, namely its maximal ideal \begin{align*} \mathfrak m_TT_{\mathfrak m_T}/\mathfrak m_A T_{\mathfrak m_T}. \end{align*} Therefore \begin{align*} \mathfrak P=\mathfrak m_TT_{\mathfrak m_T}. \end{align*} This proves that an element of $T$ reduces to zero under $\widetilde\lambda$ modulo $\mathfrak m_{A'}$ exactly when it lies in $\mathfrak m_T$. Since $T/\mathfrak m_T\cong k$ through $\lambda$, the reduced character is exactly \begin{align*} \lambda:T\to k. \end{align*} [/guided] [/step] [step:Produce a nonzero eigenvector over the extended coefficient ring] Let \begin{align*} B':=T_K\otimes_K K' \end{align*} and let $\mathfrak q:=\ker(B'\to K')$ be the maximal ideal induced by the quotient $T_K\to K'$. Define the finite-dimensional $K'$-vector space \begin{align*} W:=M_K\otimes_K K'. \end{align*} The $T_K$-action on $M_K$ induces a $B'$-module structure on $W$. By the choice of $\mathfrak n$, the localization $W_{\mathfrak q}$ is nonzero as a module over the Artinian local ring $B'_{\mathfrak q}$. Since $B'_{\mathfrak q}$ is Artinian local and $W_{\mathfrak q}$ is a nonzero finite module, there exists a nonzero element $w\in W_{\mathfrak q}$ annihilated by $\mathfrak q B'_{\mathfrak q}$: choose $r\geq 0$ minimal with $(\mathfrak q B'_{\mathfrak q})^{r+1}W_{\mathfrak q}=0$ and take $0\neq w\in (\mathfrak q B'_{\mathfrak q})^rW_{\mathfrak q}$. By the product decomposition for commutative [Artinian rings](/page/Artinian%20Ring), there are pairwise orthogonal idempotents whose sum is $1$ and which decompose $B'$ as the product of its localizations at maximal ideals. Applying this decomposition to the finite $B'$-module $W$ identifies $W_{\mathfrak q}$ with the corresponding idempotent direct summand of $W$. We therefore regard $w$ as a nonzero element of $W\subset M\otimes_A K'$. Since $\mathfrak q$ is the kernel of the character $B'\to K'$, for every $t\in T$ the element $t-\widetilde\lambda(t)$ maps into $\mathfrak q$. Therefore \begin{align*} (t-\widetilde\lambda(t))w=0, \end{align*} so \begin{align*} tw=\widetilde\lambda(t)w \end{align*} for every $t\in T$. The natural map \begin{align*} M\otimes_A A'\longrightarrow M\otimes_A K' \end{align*} is injective because $M$ is finite free over $A$ and $A'\subset K'$ is a domain. Since $M\otimes_A A'$ is an $A'$-lattice in $M\otimes_A K'$, there exists a nonzero element $a\in A'$ such that \begin{align*} v:=aw\in M\otimes_A A'. \end{align*} Multiplying the eigenvector relation by $a$ gives \begin{align*} tv=\widetilde\lambda(t)v \end{align*} for every $t\in T$. Thus $v$ is the required characteristic-zero eigenvector with lifted eigenvalues. [guided] The quotient $M_K\otimes_{T_K}K'$ is not enough by itself, because an eigenrelation holding in a quotient need not hold for a representative in the original module. We therefore produce an actual vector in the module killed by the maximal ideal defining the character. Let \begin{align*} B':=T_K\otimes_K K' \end{align*} and let $\mathfrak q:=\ker(B'\to K')$ be the maximal ideal attached to the chosen character. Also define \begin{align*} W:=M_K\otimes_K K'. \end{align*} The $T_K$-action on $M_K$ induces a $B'$-module structure on $W$, so localization at $\mathfrak q$ is defined. The choice of $\mathfrak n$ means that the localization $W_{\mathfrak q}$ is nonzero. The ring $B'_{\mathfrak q}$ is Artinian local because $B'$ is a finite-dimensional commutative $K'$-algebra. A nonzero finite module over an [Artinian local ring](/page/Artinian%20Ring) has nonzero socle: choose $r\geq 0$ minimal such that \begin{align*} (\mathfrak q B'_{\mathfrak q})^{r+1}W_{\mathfrak q}=0. \end{align*} Then \begin{align*} (\mathfrak q B'_{\mathfrak q})^rW_{\mathfrak q}\neq 0, \end{align*} and any nonzero element \begin{align*} w\in (\mathfrak q B'_{\mathfrak q})^rW_{\mathfrak q} \end{align*} is annihilated by $\mathfrak q B'_{\mathfrak q}$. Because $B'$ is a finite-dimensional commutative $K'$-algebra, it is an [Artinian ring](/page/Artinian%20Ring). The product decomposition theorem for commutative Artinian rings gives pairwise orthogonal idempotents $e_1,\dots,e_r\in B'$ with $1=e_1+\cdots+e_r$ and \begin{align*} B'\cong \prod_{i=1}^r B'_{\mathfrak q_i}, \end{align*} where $\mathfrak q_1,\dots,\mathfrak q_r$ are the maximal ideals of $B'$. If $\mathfrak q=\mathfrak q_j$, then localization at $\mathfrak q$ is projection to the factor $B'_{\mathfrak q}$, and the module decomposes as \begin{align*} W\cong \prod_{i=1}^r (W_{\mathfrak q_i}). \end{align*} Thus $W_{\mathfrak q}$ is a direct summand of $W$, and we may regard $w$ as a nonzero element of $W\subset M\otimes_A K'$. For each $t\in T$, the scalar $\widetilde\lambda(t)$ is the image of $t$ under $B'\to K'$, so \begin{align*} t-\widetilde\lambda(t)\in \mathfrak q. \end{align*} Since $\mathfrak q$ annihilates $w$, we get \begin{align*} (t-\widetilde\lambda(t))w=0, \end{align*} which is exactly \begin{align*} tw=\widetilde\lambda(t)w \end{align*} for every $t\in T$. Finally we move from the field $K'$ back to the coefficient ring $A'$. The map \begin{align*} M\otimes_A A'\longrightarrow M\otimes_A K' \end{align*} is injective because $M$ is finite free over $A$ and $A'$ is a domain contained in $K'$. The module $M\otimes_A A'$ is an $A'$-lattice in $M\otimes_A K'$, so multiplying $w$ by a nonzero denominator $a\in A'$ gives \begin{align*} v:=aw\in M\otimes_A A'. \end{align*} The scalar multiplication does not change the eigenvalues: \begin{align*} tv=t(aw)=a(tw)=a\widetilde\lambda(t)w=\widetilde\lambda(t)v. \end{align*} Thus $v$ is a nonzero characteristic-zero eigenvector in $M\otimes_A A'$ whose eigenvalues are the lifted character $\widetilde\lambda$. [/guided] [/step]