[proofplan] We use Zorn's Lemma. Consider the set of all pairs $(L, \tau)$ where $L$ is an intermediate field $k \subset L \subset K$ and $\tau: L \to \Omega$ is a field embedding extending $\sigma$. This set is non-empty (it contains $(k, \sigma)$) and is partially ordered by extension. We verify that every chain has an upper bound (by taking the union), so Zorn's Lemma yields a maximal element $(M, \tau_M)$. We then show $M = K$ by contradiction: if $M \neq K$, we extend $\tau_M$ to $M(\alpha)$ for some $\alpha \in K \setminus M$, contradicting maximality. [/proofplan] [step:Define the partially ordered set of extensions] Define the set \begin{align*} \mathcal{S} = \{(L, \tau) : k \subset L \subset K, \; L \text{ is a subfield of } K, \; \tau: L \to \Omega \text{ is a field embedding}, \; \tau|_k = \sigma\}. \end{align*} Equip $\mathcal{S}$ with the partial order $(L_1, \tau_1) \leq (L_2, \tau_2)$ if and only if $L_1 \subset L_2$ and $\tau_2|_{L_1} = \tau_1$. The set $\mathcal{S}$ is non-empty since $(k, \sigma) \in \mathcal{S}$. [/step] [step:Verify that every chain in $\mathcal{S}$ has an upper bound] Let $\{(L_i, \tau_i)\}_{i \in I}$ be a totally ordered chain in $\mathcal{S}$. Define $L = \bigcup_{i \in I} L_i$ and define $\tau: L \to \Omega$ by $\tau(x) = \tau_i(x)$ whenever $x \in L_i$. This is well-defined: if $x \in L_i \cap L_j$ and $(L_i, \tau_i) \leq (L_j, \tau_j)$, then $\tau_j(x) = \tau_i(x)$ since $\tau_j$ extends $\tau_i$. We verify that $L$ is a subfield of $K$: for any $x, y \in L$, there exist $i, j \in I$ with $x \in L_i$ and $y \in L_j$. Since the chain is totally ordered, either $L_i \subset L_j$ or $L_j \subset L_i$. In either case, $x$ and $y$ both lie in a single $L_m$, so $x - y$, $xy$, and $x^{-1}$ (for $x \neq 0$) all lie in $L_m \subset L$. The map $\tau$ is a field homomorphism: for any $x, y \in L$, choosing $m$ with $x, y \in L_m$, we have $\tau(x + y) = \tau_m(x + y) = \tau_m(x) + \tau_m(y) = \tau(x) + \tau(y)$, and similarly for multiplication. Since $\tau$ is a field homomorphism between fields, it is injective (its kernel is an ideal of $L$, hence $\{0\}$). Moreover, $\tau|_k = \sigma$ since each $\tau_i$ extends $\sigma$. Therefore $(L, \tau) \in \mathcal{S}$ and $(L_i, \tau_i) \leq (L, \tau)$ for all $i \in I$. [guided] We need to verify the hypothesis of Zorn's Lemma: every chain has an upper bound. The natural candidate is the union of all fields in the chain with the map defined piecewise. Let $\{(L_i, \tau_i)\}_{i \in I}$ be a totally ordered chain in $\mathcal{S}$. Set $L = \bigcup_{i \in I} L_i$. Why is this a field? Given any two elements $x, y \in L$, there exist indices $i, j$ with $x \in L_i$ and $y \in L_j$. Since the chain is totally ordered, WLOG $L_i \subset L_j$, so both $x, y \in L_j$. Since $L_j$ is a field, all field operations on $x, y$ yield elements of $L_j \subset L$. This argument works for addition, multiplication, subtraction, and division (when the denominator is non-zero). Define $\tau: L \to \Omega$ by $\tau(x) = \tau_i(x)$ for any $i$ with $x \in L_i$. Is this well-defined? If $x \in L_i \cap L_j$ with $(L_i, \tau_i) \leq (L_j, \tau_j)$, then $\tau_j|_{L_i} = \tau_i$, so $\tau_j(x) = \tau_i(x)$. The compatibility condition in the partial order is precisely what makes $\tau$ well-defined. To see $\tau$ is a field homomorphism: for any $x, y \in L$, pick $m$ with $x, y \in L_m$. Then $\tau(x + y) = \tau_m(x + y) = \tau_m(x) + \tau_m(y) = \tau(x) + \tau(y)$, and $\tau(xy) = \tau_m(xy) = \tau_m(x)\tau_m(y) = \tau(x)\tau(y)$. A non-zero field homomorphism is automatically injective (its kernel is an ideal of a field, hence $\{0\}$ or the whole field; since $\tau(1) = 1 \neq 0$, the kernel is $\{0\}$). Finally, $\tau|_k = \sigma$ because every $\tau_i$ extends $\sigma$. Therefore $(L, \tau) \in \mathcal{S}$ and it is an upper bound for the chain. [/guided] [/step] [step:Apply Zorn's Lemma to obtain a maximal extension] By Zorn's Lemma, $\mathcal{S}$ has a maximal element $(M, \tau_M)$. That is, $M$ is an intermediate field $k \subset M \subset K$, $\tau_M: M \to \Omega$ is a field embedding extending $\sigma$, and no proper extension of $(M, \tau_M)$ exists in $\mathcal{S}$. [/step] [step:Show the maximal element satisfies $M = K$ by extending over a simple algebraic extension] Suppose for contradiction that $M \neq K$. Choose $\alpha \in K \setminus M$. Since $K/k$ is algebraic, $\alpha$ is algebraic over $k$ and hence algebraic over $M$. Let $m_{\alpha, M} \in M[x]$ denote the minimal polynomial of $\alpha$ over $M$, which is irreducible in $M[x]$ with $\deg m_{\alpha, M} \geq 1$. Write $m_{\alpha, M}(x) = \sum_{j=0}^{d} c_j x^j$ with $c_j \in M$ and $d = \deg m_{\alpha, M}$. Define the image polynomial \begin{align*} \tau_M(m_{\alpha, M})(x) = \sum_{j=0}^{d} \tau_M(c_j) x^j \in \tau_M(M)[x] \subset \Omega[x]. \end{align*} This is a non-constant polynomial in $\Omega[x]$. Since $\Omega$ is algebraically closed, $\tau_M(m_{\alpha, M})$ has a root $\beta \in \Omega$. We now extend $\tau_M$ to $M(\alpha)$. Since $\alpha$ is algebraic over $M$ with minimal polynomial $m_{\alpha, M}$, the simple extension $M(\alpha)$ is isomorphic to $M[x] / (m_{\alpha, M})$ via the evaluation map. Define \begin{align*} \tau': M(\alpha) &\to \Omega \end{align*} by $\tau'(g(\alpha)) = \tau_M(g)(\beta)$ for any $g \in M[x]$, where $\tau_M(g)$ denotes the polynomial obtained by applying $\tau_M$ to each coefficient of $g$. This map is well-defined: if $g(\alpha) = h(\alpha)$, then $m_{\alpha, M}$ divides $g - h$ in $M[x]$, so $\tau_M(m_{\alpha, M})$ divides $\tau_M(g - h) = \tau_M(g) - \tau_M(h)$ in $\Omega[x]$ (since $\tau_M$ is a ring homomorphism on coefficients). Since $\beta$ is a root of $\tau_M(m_{\alpha, M})$, evaluating at $\beta$ gives $\tau_M(g)(\beta) = \tau_M(h)(\beta)$. The map $\tau'$ is a field homomorphism (it is the composition of the isomorphism $M(\alpha) \cong M[x]/(m_{\alpha, M})$ with the map induced by $\tau_M$ on coefficients and evaluation at $\beta$). Since it is a non-zero field homomorphism, it is injective. Moreover, $\tau'|_M = \tau_M$ (for $c \in M$, $\tau'(c) = \tau_M(c)$). Therefore $(M(\alpha), \tau') \in \mathcal{S}$ with $(M, \tau_M) < (M(\alpha), \tau')$ (strict inequality since $\alpha \in M(\alpha) \setminus M$). This contradicts the maximality of $(M, \tau_M)$. [guided] The heart of the proof is showing that the maximal element must satisfy $M = K$. The argument proceeds by contradiction: if $M \subsetneq K$, we show we can extend $\tau_M$ to a strictly larger intermediate field, contradicting maximality. Pick $\alpha \in K \setminus M$. Since $K/k$ is algebraic, $\alpha$ is algebraic over $k$, and since $k \subset M$, $\alpha$ is also algebraic over $M$. Let $m_{\alpha, M} \in M[x]$ be the minimal polynomial of $\alpha$ over $M$. This is an irreducible polynomial of degree $d \geq 1$ (it has degree $\geq 1$ because $\alpha \notin M$). The key idea is to use $\tau_M$ to "transport" $m_{\alpha, M}$ into $\Omega[x]$, then find a root of the transported polynomial in $\Omega$ (which exists because $\Omega$ is algebraically closed), and use that root to define the extension. Write $m_{\alpha, M}(x) = \sum_{j=0}^{d} c_j x^j$ with $c_j \in M$. Apply $\tau_M$ coefficient-by-coefficient to get $\tau_M(m_{\alpha, M})(x) = \sum_{j=0}^{d} \tau_M(c_j) x^j \in \Omega[x]$. Since $\Omega$ is algebraically closed and this polynomial has degree $d \geq 1$, it has a root $\beta \in \Omega$. Now define $\tau': M(\alpha) \to \Omega$ as follows. Every element of $M(\alpha)$ can be written as $g(\alpha)$ for some $g \in M[x]$ with $\deg g < d$ (using the isomorphism $M(\alpha) \cong M[x]/(m_{\alpha, M})$). Set $\tau'(g(\alpha)) = \tau_M(g)(\beta)$, where $\tau_M(g)$ is obtained by applying $\tau_M$ to the coefficients of $g$. Why is $\tau'$ well-defined? If $g(\alpha) = h(\alpha)$ in $M(\alpha)$, then $(g - h)(\alpha) = 0$, so $m_{\alpha, M}$ divides $g - h$ in $M[x]$. Write $g(x) - h(x) = m_{\alpha, M}(x) \cdot q(x)$ for some $q \in M[x]$. Applying $\tau_M$ to coefficients: $\tau_M(g)(x) - \tau_M(h)(x) = \tau_M(m_{\alpha, M})(x) \cdot \tau_M(q)(x)$. Evaluating at $\beta$: $\tau_M(g)(\beta) - \tau_M(h)(\beta) = \tau_M(m_{\alpha, M})(\beta) \cdot \tau_M(q)(\beta) = 0 \cdot \tau_M(q)(\beta) = 0$. One can verify directly that $\tau'$ preserves addition and multiplication (it is defined via the ring homomorphism $M[x] \to \Omega$ given by applying $\tau_M$ to coefficients and evaluating at $\beta$, which factors through $M[x]/(m_{\alpha, M})$ because $\beta$ is a root of $\tau_M(m_{\alpha, M})$). As a non-zero homomorphism between fields, $\tau'$ is injective. For any $c \in M$, $\tau'(c) = \tau_M(c)$, so $\tau'$ extends $\tau_M$. This gives $(M(\alpha), \tau') \in \mathcal{S}$ with $(M, \tau_M) < (M(\alpha), \tau')$, since $M \subsetneq M(\alpha)$ (as $\alpha \notin M$). This contradicts the maximality of $(M, \tau_M)$. [/guided] [/step] [step:Conclude that the maximal embedding extends $\sigma$ to all of $K$] The contradiction shows $M = K$. Therefore $\tau_M: K \to \Omega$ is a field embedding with $\tau_M|_k = \sigma$. Setting $\tau = \tau_M$ completes the proof. [/step]