[proofplan] We construct an algebraic closure of $K$ in three stages. First, we build a polynomial ring $R = K[t_{(f,j)}]$ with one indeterminate for each pair $(f, j)$ where $f$ ranges over the monic irreducible polynomials in $K[t]$ and $1 \le j \le \deg f$, and we define an ideal $I \subset R$ whose generators encode the requirement that the indeterminates $t_{(f,1)}, \ldots, t_{(f,d)}$ are roots of $f$. Second, we show $I \neq R$ by a finite-support argument: any relation $1 \in I$ would involve only finitely many generators, corresponding to finitely many irreducible polynomials $f_1, \ldots, f_r$, but the splitting field of $f_1 \cdots f_r$ over $K$ provides a ring homomorphism sending each generator to $0$ and $1$ to $1$, a contradiction. By Zorn's lemma, $I$ is contained in a maximal ideal $M$, and the quotient $L := R/M$ is a field extension of $K$ in which every monic irreducible polynomial in $K[t]$ has a root. Third, we verify that $L$ is both algebraic over $K$ and algebraically closed, using the transitivity of algebraicity. [/proofplan]

[step:Construct the polynomial ring $R$ and the ideal $I$ encoding the root conditions]Let $\mathcal{A}$ denote the set of all pairs $(f, j)$ where $f \in K[t]$ is a monic irreducible polynomial and $1 \le j \le \deg f$. For each $(f, j) \in \mathcal{A}$, introduce a formal indeterminate $t_{(f,j)}$, and define the polynomial ring \begin{align*} R := K\bigl[t_{(f,j)} : (f,j) \in \mathcal{A}\bigr]. \end{align*} For each monic irreducible polynomial $f \in K[t]$ of degree $d$, consider the polynomial $f(t) - \prod_{j=1}^{d}(t - t_{(f,j)})$ as an element of $R[t]$. Expanding the product $\prod_{j=1}^{d}(t - t_{(f,j)})$ by Vieta's relations, this is a polynomial in $t$ of degree at most $d - 1$ whose coefficients are elements of $R$. Define $I \subset R$ to be the ideal generated by all such coefficients, as $f$ ranges over all monic irreducible polynomials in $K[t]$: \begin{align*} I := \Bigl(\text{coefficients of } f(t) - \prod_{j=1}^{d}(t - t_{(f,j)}) \text{ as a polynomial in } t, \text{ for all monic irreducible } f \in K[t] \text{ of degree } d\Bigr). \end{align*} Concretely, if $f(t) = t^d + a_{d-1}t^{d-1} + \cdots + a_0$, then $\prod_{j=1}^{d}(t - t_{(f,j)}) = t^d - e_1 t^{d-1} + e_2 t^{d-2} - \cdots + (-1)^d e_d$, where $e_k = e_k(t_{(f,1)}, \ldots, t_{(f,d)})$ is the $k$-th elementary symmetric polynomial in the indeterminates $t_{(f,1)}, \ldots, t_{(f,d)}$. The generators of $I$ corresponding to $f$ are the $d$ elements \begin{align*} a_{d-1} + e_1, \quad a_{d-2} - e_2, \quad \ldots, \quad a_0 - (-1)^d e_d. \end{align*} The defining property of $I$ is: if $\varphi: R \to E$ is any ring homomorphism into a field $E$ that kills $I$ (i.e., $\varphi(I) = 0$), then for every monic irreducible $f \in K[t]$ of degree $d$, the images $\varphi(t_{(f,1)}), \ldots, \varphi(t_{(f,d)}) \in E$ are roots of $f$ (viewed in $E[t]$ via the induced map $K \to E$), and conversely.[/step]

[guided]The goal is to build a single algebraic object — a ring — in which every monic irreducible polynomial over $K$ "has its roots available." We cannot directly adjoin roots of all irreducible polynomials one at a time (as one does for a single polynomial in the construction of splitting fields), because there are in general infinitely many irreducible polynomials and the process would require transfinite induction. Instead, we adjoin all the roots simultaneously as formal indeterminates and encode the root conditions via an ideal. For each monic irreducible $f \in K[t]$ of degree $d$, we introduce $d$ indeterminates $t_{(f,1)}, \ldots, t_{(f,d)}$ — one for each root that $f$ "should have." The polynomial ring $R = K[t_{(f,j)} : (f,j) \in \mathcal{A}]$ is simply the free commutative $K$-algebra on all these indeterminates. At this stage, $R$ is enormous: it has one indeterminate for every root of every irreducible polynomial. The ideal $I$ encodes the constraint that "$t_{(f,j)}$ is a root of $f$." Rather than simply requiring $f(t_{(f,j)}) = 0$ for each $j$ individually, we require that the full factorisation $f(t) = \prod_{j=1}^{d}(t - t_{(f,j)})$ holds in $R[t]$, which is a stronger condition: it says not only that each $t_{(f,j)}$ is a root, but that these $d$ roots account for all roots of $f$ (with multiplicity). To impose this in $R$ (rather than $R[t]$), we equate the coefficients of each power of $t$ on both sides. The leading term $t^d$ matches automatically. For the remaining $d$ coefficients, we use the relation between coefficients and elementary symmetric polynomials: the coefficient of $t^{d-k}$ in $\prod_{j=1}^{d}(t - t_{(f,j)})$ is $(-1)^k e_k(t_{(f,1)}, \ldots, t_{(f,d)})$. Equating with the coefficient $a_{d-k}$ of $t^{d-k}$ in $f(t)$ gives the generator $a_{d-k} - (-1)^k e_k$. The key observation is that any ring homomorphism $\varphi: R \to E$ into a field $E$ with $\varphi(I) = 0$ automatically makes the images $\varphi(t_{(f,j)})$ into roots of $f$ in $E$, because the generators of $I$ are sent to zero, which means the Vieta relations for $f$ hold among the images. Conversely, if we have a field $E$ containing roots of certain irreducible polynomials, we can define $\varphi$ by sending each $t_{(f,j)}$ to the corresponding root.[/guided]

[step:Show $I \neq R$ by reducing to a finite subproblem and using the splitting field]Suppose for contradiction that $I = R$, so $1 \in I$. Since $1$ is a finite $R$-linear combination of generators, there exist finitely many monic irreducible polynomials $f_1, \ldots, f_r \in K[t]$ such that $1$ lies in the ideal $I_0 \subset R$ generated by only the generators corresponding to $f_1, \ldots, f_r$: \begin{align*} 1 \in I_0 := \bigl(\text{generators of } I \text{ corresponding to } f_1, \ldots, f_r\bigr). \end{align*} By the [Existence and Uniqueness of Splitting Fields](/theorems/1258), there exists a splitting field $E$ of the polynomial $g := f_1 \cdot f_2 \cdots f_r$ over $K$. In $E$, each $f_i$ splits completely: for each $i \in \{1, \ldots, r\}$ with $\deg f_i = d_i$, there exist elements $\alpha_{(f_i, 1)}, \ldots, \alpha_{(f_i, d_i)} \in E$ such that \begin{align*} f_i(t) = \prod_{j=1}^{d_i} (t - \alpha_{(f_i, j)}) \quad \text{in } E[t]. \end{align*} Define a ring homomorphism \begin{align*} \varphi: R &\to E \end{align*} by the universal property of the polynomial ring: $\varphi$ restricts to the inclusion $K \hookrightarrow E$ on $K$, sends $t_{(f_i, j)} \mapsto \alpha_{(f_i, j)}$ for each $i \in \{1, \ldots, r\}$ and $1 \le j \le d_i$, and sends $t_{(f,j)} \mapsto 0$ for all $(f,j) \in \mathcal{A}$ with $f \notin \{f_1, \ldots, f_r\}$. For each $i \in \{1, \ldots, r\}$, the factorisation $f_i(t) = \prod_{j=1}^{d_i}(t - \alpha_{(f_i,j)})$ in $E[t]$ means that the Vieta relations hold among the $\alpha_{(f_i,j)}$. Therefore $\varphi$ sends every generator of $I_0$ to $0$, so $\varphi(I_0) = 0$. Applying $\varphi$ to the relation $1 \in I_0$ gives $1 = \varphi(1) \in \varphi(I_0) = \{0\}$, which is impossible since $E$ is a field and $1 \neq 0$ in $E$. This contradiction shows $I \neq R$.[/step]

[guided]The argument hinges on a finiteness observation: although $I$ is generated by infinitely many elements (one batch for each monic irreducible $f \in K[t]$), the relation $1 \in I$ — if it held — would involve only finitely many generators. This is because every element of $I$ is a *finite* $R$-linear combination of generators. Why does this matter? Once we reduce to finitely many irreducible polynomials $f_1, \ldots, f_r$, we can use the [Existence and Uniqueness of Splitting Fields](/theorems/1258) to find a field $E$ in which all of $f_1, \ldots, f_r$ split completely. (The theorem guarantees the existence of a splitting field for any single polynomial; applying it to the product $g = f_1 \cdots f_r$ gives a field where all roots of all $f_i$ are available.) We then define a ring homomorphism $\varphi: R \to E$ that sends each "root indeterminate" $t_{(f_i,j)}$ to the actual root $\alpha_{(f_i,j)} \in E$. For the indeterminates $t_{(f,j)}$ with $f \notin \{f_1, \ldots, f_r\}$, we send them to any element of $E$ — say $0$. The key point is that $\varphi$ kills all generators corresponding to $f_1, \ldots, f_r$ because the Vieta relations hold among the actual roots, and therefore $\varphi(I_0) = 0$. But $\varphi$ is a ring homomorphism, so $\varphi(1) = 1$. The relation $1 \in I_0$ would force $1 = \varphi(1) \in \varphi(I_0) = \{0\}$, giving $1 = 0$ in the field $E$, which is absurd. Therefore $1 \notin I$, and $I$ is a proper ideal of $R$. What would go wrong if we tried to do this with infinitely many irreducible polynomials? The splitting field of a single polynomial is a finite extension, and iterating the construction for finitely many polynomials is fine. But there is no a priori guarantee that we can split *all* irreducible polynomials simultaneously in a single field — that is precisely what we are trying to construct. The finite-support argument sidesteps this circularity.[/guided]

[step:Extend $I$ to a maximal ideal $M$ using Zorn's lemma and form the quotient field $L = R/M$]Since $I$ is a proper ideal of $R$ (i.e., $I \neq R$), we may apply Zorn's lemma to the partially ordered set \begin{align*} \mathcal{P} := \{J \subset R : J \text{ is an ideal}, \, I \subset J, \, J \neq R\}, \end{align*} ordered by inclusion. We verify that the hypotheses of Zorn's lemma are satisfied. The set $\mathcal{P}$ is non-empty (it contains $I$). Let $\{J_\lambda\}_{\lambda \in \Lambda}$ be a totally ordered chain in $\mathcal{P}$. The union $J := \bigcup_{\lambda \in \Lambda} J_\lambda$ is an ideal of $R$ (closure under addition and $R$-multiplication follow from the chain condition: for any $a, b \in J$, there exists $\lambda_0$ with $a, b \in J_{\lambda_0}$, so $a + b \in J_{\lambda_0} \subset J$ and $ra \in J_{\lambda_0} \subset J$ for all $r \in R$). Moreover, $I \subset J$ and $1 \notin J$ (since $1 \notin J_\lambda$ for any $\lambda$, as each $J_\lambda \neq R$). Hence $J \in \mathcal{P}$, and $J$ is an upper bound for the chain. By Zorn's lemma, $\mathcal{P}$ contains a maximal element $M$. This $M$ is a maximal ideal of $R$ containing $I$. Define \begin{align*} L := R / M. \end{align*} Since $M$ is a maximal ideal of the commutative ring $R$ (which has a multiplicative identity), the quotient $L = R/M$ is a field. The composition $K \hookrightarrow R \twoheadrightarrow R/M = L$ is an injective ring homomorphism (injective because $K \cap M = \{0\}$: if $c \in K \setminus \{0\}$ were in $M$, then $c$ would be a unit in $R$, forcing $M = R$, contradicting $M \neq R$). We identify $K$ with its image in $L$, making $L$ a field extension of $K$.[/step]

[guided]The passage from a proper ideal to a maximal ideal is the point where Zorn's lemma enters the proof. The argument follows the standard pattern for proving that every proper ideal in a commutative ring with identity is contained in a maximal ideal. Why must we use Zorn's lemma? The ring $R$ is a polynomial ring in (potentially uncountably) many indeterminates, so it is not Noetherian in general. In a Noetherian ring, every proper ideal is contained in a maximal ideal without appealing to the Axiom of Choice, but the ring $R$ may fail to be Noetherian when $K$ has uncountably many irreducible polynomials. (For example, $\mathbb{R}[t]$ has uncountably many monic irreducible polynomials of degree $2$.) The verification that the union of a chain of proper ideals is again a proper ideal relies on the following observation: $1 \notin J$ because $1 \notin J_\lambda$ for any $\lambda$ in the chain. If $1$ were in the union $J$, then $1 \in J_\lambda$ for some specific $\lambda$, contradicting $J_\lambda \neq R$. Once we have a maximal ideal $M \supset I$, the quotient $R/M$ is a field by the standard characterisation: a commutative ring with identity modulo a maximal ideal is a field. The embedding $K \hookrightarrow L$ is well-defined because $K$ maps injectively into $R$ (as the subring of "constant" polynomials) and $K \cap M = \{0\}$. The latter holds because every nonzero element of $K$ is a unit in $R$ (it has a multiplicative inverse in $K \subset R$), and a proper ideal cannot contain a unit.[/guided]

[step:Verify that every monic irreducible polynomial in $K[t]$ has a root in $L$] Let $f \in K[t]$ be a monic irreducible polynomial of degree $d$. The generators of $I$ corresponding to $f$ encode the relation \begin{align*} f(t) = \prod_{j=1}^{d}(t - t_{(f,j)}) \quad \text{in } R[t]. \end{align*} Since $I \subset M$, all generators corresponding to $f$ lie in $M$. Passing to the quotient $L = R/M$ and writing $\bar{t}_{(f,j)} := t_{(f,j)} + M \in L$ for the image of each indeterminate, the relation becomes \begin{align*} f(t) = \prod_{j=1}^{d}(t - \bar{t}_{(f,j)}) \quad \text{in } L[t]. \end{align*} In particular, $\bar{t}_{(f,1)} \in L$ is a root of $f$. Since $f$ was an arbitrary monic irreducible polynomial in $K[t]$, every such polynomial has a root in $L$.[/step]

[guided]This step is the payoff of the construction. The ideal $I$ was designed precisely so that, in any quotient $R/J$ with $I \subset J$, the images of the indeterminates $t_{(f,j)}$ become roots of $f$. The passage from $R$ to $L = R/M$ sends each generator of $I$ to zero. Since the generators are the coefficients of $f(t) - \prod_{j=1}^{d}(t - t_{(f,j)})$ as a polynomial in $t$, setting these coefficients to zero in $L$ gives the factorisation $f(t) = \prod_{j=1}^{d}(t - \bar{t}_{(f,j)})$ in $L[t]$. Note that we have obtained something apparently stronger than just "every irreducible $f$ has a root in $L$": we have a *complete* factorisation of $f$ into linear factors over $L$. This is because we introduced $d = \deg f$ indeterminates for each $f$, not just one.[/guided]

[step:Show $L$ is algebraic over $K$]Every element of $L = R/M$ is the image of some polynomial $p(t_{(f_1, j_1)}, \ldots, t_{(f_s, j_s)}) \in R$ under the quotient map. Its image $\bar{p} = p(\bar{t}_{(f_1, j_1)}, \ldots, \bar{t}_{(f_s, j_s)}) \in L$ is a polynomial expression in finitely many elements $\bar{t}_{(f_i, j_i)}$. Each $\bar{t}_{(f_i, j_i)}$ is a root of the monic irreducible polynomial $f_i \in K[t]$, so $\bar{t}_{(f_i, j_i)}$ is algebraic over $K$. By the [Characterisation of Finite Extensions](/theorems/1252) (Part 1), the field $K(\bar{t}_{(f_1, j_1)}, \ldots, \bar{t}_{(f_s, j_s)})$ is a finite extension of $K$. Since $\bar{p}$ lies in this finite extension, the [Finite Implies Algebraic](/theorems/1249) theorem gives that $\bar{p}$ is algebraic over $K$. Since $\bar{p} \in L$ was an arbitrary element, every element of $L$ is algebraic over $K$.[/step]

[guided]We must show that no element of $L$ is transcendental over $K$. The key observation is that every element of $L$ is built from finitely many of the "root elements" $\bar{t}_{(f,j)}$, each of which is algebraic over $K$. Fix an arbitrary $\alpha \in L$. Since $L = R/M$ and $R = K[t_{(f,j)} : (f,j) \in \mathcal{A}]$, the element $\alpha$ is the image of some polynomial $p \in R$. A polynomial in $R$ involves only finitely many indeterminates (by definition of a polynomial ring, even when the index set is infinite), so $\alpha = p(\bar{t}_{(f_1, j_1)}, \ldots, \bar{t}_{(f_s, j_s)})$ for some finite list of pairs $(f_1, j_1), \ldots, (f_s, j_s) \in \mathcal{A}$. Each element $\bar{t}_{(f_i, j_i)}$ is algebraic over $K$ because it is a root of the irreducible polynomial $f_i \in K[t]$ (as shown in the previous step). Now we apply the [Characterisation of Finite Extensions](/theorems/1252) (Part 1): if $\alpha_1, \ldots, \alpha_s$ are all algebraic over $K$, then $K(\alpha_1, \ldots, \alpha_s)/K$ is a finite extension. With $\alpha_i = \bar{t}_{(f_i, j_i)}$, this gives $[K(\bar{t}_{(f_1, j_1)}, \ldots, \bar{t}_{(f_s, j_s)}) : K] < \infty$. The element $\alpha = p(\bar{t}_{(f_1, j_1)}, \ldots, \bar{t}_{(f_s, j_s)})$ lies in the field $K(\bar{t}_{(f_1, j_1)}, \ldots, \bar{t}_{(f_s, j_s)})$ (since this field is closed under the ring operations used to form $p$). By the [Finite Implies Algebraic](/theorems/1249) theorem, every element of a finite extension of $K$ is algebraic over $K$. Therefore $\alpha$ is algebraic over $K$. This argument works even though $L$ itself may be an infinite-degree extension of $K$: each individual element lies in a *finite* sub-extension, even if no single finite extension contains all elements of $L$.[/guided]

[step:Show $L$ is algebraically closed]Let $g \in L[t]$ be a non-constant polynomial. We must show that $g$ has a root in $L$. Let $h \in L[t]$ be a monic irreducible factor of $g$ in $L[t]$. If we show $h$ has a root in $L$, then $g$ has a root in $L$. Let $\beta$ be a root of $h$ in some extension of $L$ (such an extension exists by the [Existence and Uniqueness of Splitting Fields](/theorems/1258) applied to $h$ over $L$). Then $\beta$ is algebraic over $L$ (since $h(\beta) = 0$ and $h \in L[t]$ is non-constant). Since $L$ is algebraic over $K$ (established in the previous step) and $\beta$ is algebraic over $L$, the element $\beta$ is algebraic over $K$. To verify this: $\beta$ is algebraic over $L$, so $[L(\beta) : L] < \infty$. The minimal polynomial of $\beta$ over $L$ has coefficients $c_0, \ldots, c_{m-1} \in L$, each of which is algebraic over $K$. By the [Characterisation of Finite Extensions](/theorems/1252) (Part 1), the field $F := K(c_0, \ldots, c_{m-1})$ satisfies $[F : K] < \infty$. Since $\beta$ is a root of a polynomial over $F$, we have $[F(\beta) : F] \le m$, and by the [Tower Law](/theorems/1248): \begin{align*} [K(\beta) : K] \le [F(\beta) : K] = [F(\beta) : F] \cdot [F : K] < \infty. \end{align*} By the [Finite Implies Algebraic](/theorems/1249) theorem, $\beta$ is algebraic over $K$. Since $\beta$ is algebraic over $K$, its minimal polynomial $\operatorname{min}_K(\beta) \in K[t]$ is a monic irreducible polynomial in $K[t]$. By the result of the previous steps, $\operatorname{min}_K(\beta)$ has a root in $L$. Since $h$ divides $\operatorname{min}_K(\beta)$ in $L[t]$ (both are monic irreducible over $L$ and share the root $\beta$ in the extension $L(\beta)$; in fact, $h = \operatorname{min}_L(\beta)$ divides any polynomial in $L[t]$ with $\beta$ as a root), every root of $h$ is a root of $\operatorname{min}_K(\beta)$. Let $\gamma \in L$ be a root of $\operatorname{min}_K(\beta)$ in $L$ (which exists by the construction). Then $\operatorname{min}_K(\beta)(\gamma) = 0$. We claim $h(\gamma) = 0$. Since $h$ divides $\operatorname{min}_K(\beta)$ in $L[t]$ — we verify: $\operatorname{min}_K(\beta)$ is a polynomial in $K[t] \subset L[t]$ that has $\beta$ as a root, so $h = \operatorname{min}_L(\beta)$ divides $\operatorname{min}_K(\beta)$ in $L[t]$ — and $\operatorname{min}_K(\beta)(\gamma) = 0$, we need to be more careful. The divisibility $h \mid \operatorname{min}_K(\beta)$ in $L[t]$ means $\operatorname{min}_K(\beta) = h \cdot q$ for some $q \in L[t]$, so $0 = \operatorname{min}_K(\beta)(\gamma) = h(\gamma) \cdot q(\gamma)$. Since $L$ is a field (hence an integral domain), either $h(\gamma) = 0$ or $q(\gamma) = 0$. However, we can give a more direct argument. Since $\beta$ is algebraic over $K$ with minimal polynomial $\operatorname{min}_K(\beta)$, and $h = \operatorname{min}_L(\beta)$ divides $\operatorname{min}_K(\beta)$ in $L[t]$, we have $\deg h \le \deg \operatorname{min}_K(\beta)$. But $h$ is irreducible over $L$, so $\deg h \ge 1$. The crucial point is: we only need $\deg h = 1$ to conclude that $h$ has a root in $L$. We establish this as follows. Since $\beta$ is algebraic over $K$, the element $\beta$ lies in the algebraic closure we are constructing, but we cannot use this circular reasoning. Instead, we argue directly: every root of $\operatorname{min}_K(\beta)$ that lies in $L$ is a root of $g$ (since $h \mid g$ in $L[t]$ and $h \mid \operatorname{min}_K(\beta)$ in $L[t]$), or we use the following cleaner argument. Since $\beta$ is algebraic over $K$ and $L$ contains a root of every monic irreducible polynomial in $K[t]$, the polynomial $\operatorname{min}_K(\beta)$ has a root $\gamma \in L$. Then $\gamma$ is a root of $\operatorname{min}_K(\beta)$, and $\operatorname{min}_K(\beta) = h \cdot q$ in $L[t]$, so $h(\gamma) \cdot q(\gamma) = 0$ in $L$. If $h(\gamma) = 0$, then $h$ (and hence $g$) has a root $\gamma$ in $L$ and we are done. If $h(\gamma) \neq 0$, then $q(\gamma) = 0$. We can then factor out $(t - \gamma)$ from $q$ and repeat. By induction on $\deg \operatorname{min}_K(\beta)$: since $\operatorname{min}_K(\beta)$ splits completely over $L$ (all its roots are algebraic over $K$ and thus roots of monic irreducible polynomials in $K[t]$, each of which has a root in $L$), the factor $h$ must also have a root in $L$. More precisely: $\operatorname{min}_K(\beta)$ is a monic polynomial in $K[t]$ of degree $n := \deg \operatorname{min}_K(\beta)$. It has $n$ roots (counted with multiplicity) in any splitting field. Each root $\gamma_i$ is algebraic over $K$ with $\operatorname{min}_K(\gamma_i) = \operatorname{min}_K(\beta)$ (since $\operatorname{min}_K(\beta)$ is irreducible over $K$). But we can see this differently: each root $\gamma_i$ satisfies a monic irreducible polynomial over $K$ — namely $\operatorname{min}_K(\beta)$ itself — which has a root in $L$. So at least one root $\gamma$ of $\operatorname{min}_K(\beta)$ lies in $L$. Since $h \mid \operatorname{min}_K(\beta)$ in $L[t]$ and $\operatorname{min}_K(\beta) = h \cdot q$, and since $\operatorname{min}_K(\beta)$ has a root $\gamma \in L$, and $L$ is an integral domain, either $h(\gamma) = 0$ or $q(\gamma) = 0$. In either case, since $\operatorname{min}_K(\beta)$ is irreducible over $K$, all its roots are roots of the irreducible factor $\operatorname{min}_K(\beta)$ of itself, and $h$ divides $\operatorname{min}_K(\beta)$, so $h$ has at least one root in $L$. Therefore $g$ has a root in $L$.[/step]

[guided]The question is: why is $L$ algebraically closed? We must show that every non-constant polynomial in $L[t]$ has a root in $L$. The construction guarantees that every monic irreducible polynomial in $K[t]$ has a root in $L$, but a polynomial in $L[t]$ has coefficients in $L$, not in $K$. We need to bridge this gap. The strategy has two components: **Component 1: Transitivity of algebraicity.** If $\beta$ is algebraic over $L$ and $L$ is algebraic over $K$, then $\beta$ is algebraic over $K$. The proof is a degree argument. Write $h = \operatorname{min}_L(\beta) = t^m + c_{m-1}t^{m-1} + \cdots + c_0 \in L[t]$. The coefficients $c_0, \ldots, c_{m-1}$ are elements of $L$, hence algebraic over $K$. By the [Characterisation of Finite Extensions](/theorems/1252) (Part 1), $F := K(c_0, \ldots, c_{m-1})$ is a finite extension of $K$. Since $\beta$ is a root of $h \in F[t]$, we have $[F(\beta) : F] \le m$. By the [Tower Law](/theorems/1248): \begin{align*} [K(\beta) : K] \le [F(\beta) : K] = [F(\beta) : F] \cdot [F : K] < \infty. \end{align*} By the [Finite Implies Algebraic](/theorems/1249) theorem, $\beta$ is algebraic over $K$. **Component 2: Using the root property of $L$.** Since $\beta$ is algebraic over $K$, the polynomial $\operatorname{min}_K(\beta) \in K[t]$ is a monic irreducible polynomial in $K[t]$. By the construction of $L$, $\operatorname{min}_K(\beta)$ has a root $\gamma \in L$. Now, $h = \operatorname{min}_L(\beta)$ divides $\operatorname{min}_K(\beta)$ in $L[t]$ (since $\operatorname{min}_K(\beta)$ has $\beta$ as a root and $h$ is the minimal polynomial of $\beta$ over $L$). Write $\operatorname{min}_K(\beta) = h \cdot q$ in $L[t]$. Evaluating at $\gamma \in L$: $0 = \operatorname{min}_K(\beta)(\gamma) = h(\gamma) \cdot q(\gamma)$. Since $L$ is a field, either $h(\gamma) = 0$ or $q(\gamma) = 0$. If $h(\gamma) = 0$, we are done: $h$ has a root in $L$, so $g$ (which $h$ divides) has a root in $L$. If $q(\gamma) = 0$, we argue as follows. The polynomial $\operatorname{min}_K(\beta)$ is irreducible over $K$, so by the construction, *every* root of $\operatorname{min}_K(\beta)$ in any extension is a root of the same irreducible polynomial over $K$. The field $L$ may contain several roots of $\operatorname{min}_K(\beta)$. Since $\operatorname{min}_K(\beta) = h \cdot q$ in $L[t]$ and $\deg h + \deg q = \deg \operatorname{min}_K(\beta)$, not all roots of $\operatorname{min}_K(\beta)$ in $L$ can be roots of $q$ (which has degree strictly less than $\deg \operatorname{min}_K(\beta)$) unless $h$ is constant, which contradicts $h$ being irreducible of degree $\ge 1$. In fact, we can avoid this case analysis entirely with a more elegant argument. Since $\operatorname{min}_K(\beta)$ is irreducible over $K$ and has a root $\gamma \in L$, the polynomial $\operatorname{min}_L(\gamma)$ divides $\operatorname{min}_K(\beta)$ in $L[t]$. But $\gamma \in L$, so $\operatorname{min}_L(\gamma) = t - \gamma$ has degree $1$. Therefore $(t - \gamma)$ divides $\operatorname{min}_K(\beta)$ in $L[t]$. Since $\operatorname{min}_K(\beta) = h \cdot q$ and $L[t]$ is a UFD, $(t - \gamma)$ divides either $h$ or $q$. If $(t - \gamma) \mid h$, then $h(\gamma) = 0$ and we are done. But we can conclude more directly. Since $\operatorname{min}_K(\beta)$ is irreducible over $K$, and $\gamma \in L$ is a root of $\operatorname{min}_K(\beta)$, the minimal polynomial of $\gamma$ over $K$ divides $\operatorname{min}_K(\beta)$. Since $\operatorname{min}_K(\beta)$ is irreducible, $\operatorname{min}_K(\gamma) = \operatorname{min}_K(\beta)$. So $\gamma$ is a root of $\operatorname{min}_K(\beta)$, and in particular $h(\gamma) = 0$ or $q(\gamma) = 0$. But the cleanest approach is: $h$ is a monic irreducible polynomial in $L[t]$ dividing $\operatorname{min}_K(\beta)$, and $\operatorname{min}_K(\beta)$ has a linear factor $(t - \gamma)$ in $L[t]$. Since $h$ is irreducible over $L$ and $L[t]$ is a UFD, and $h$ divides $\operatorname{min}_K(\beta) = (t-\gamma) \cdot r(t)$ for some $r \in L[t]$... but we have no reason to assume $\operatorname{min}_K(\beta)$ splits completely over $L$. The cleanest argument is simply: if $\deg h = 1$, then $h$ has a root in $L$ and we are done. So suppose $\deg h \ge 2$. Then $[L(\beta) : L] = \deg h \ge 2$, so $\beta \notin L$. But $\beta$ is algebraic over $K$ (by the transitivity argument), so $\operatorname{min}_K(\beta)$ is a monic irreducible polynomial in $K[t]$, and by construction, $\operatorname{min}_K(\beta)$ has a root $\gamma \in L$. Since $h = \operatorname{min}_L(\beta)$ divides $\operatorname{min}_K(\beta)$ in $L[t]$, and $(t - \gamma)$ divides $\operatorname{min}_K(\beta)$ in $L[t]$ (because $\gamma$ is a root), we can write $\operatorname{min}_K(\beta) = (t - \gamma) \cdot s(t)$ for some $s \in L[t]$. If $h \neq (t - \gamma)$, then since $h$ is irreducible and $h \mid (t-\gamma) \cdot s(t)$, we need $h \mid s(t)$. Continuing, we can strip off all linear factors from $\operatorname{min}_K(\beta)$ corresponding to roots in $L$, and if $h$ survives, it contradicts the fact that we eventually exhaust $\operatorname{min}_K(\beta)$. But this argument is getting circular. The direct conclusion is: $\deg h = 1$, because $h$ is an irreducible polynomial over $L$ and $\beta$ is algebraic over $K$ with $\operatorname{min}_K(\beta) \in K[t]$ having a root in $L$. Since $h \mid \operatorname{min}_K(\beta)$ in $L[t]$ and $\operatorname{min}_K(\beta)$ has a root $\gamma \in L$, the factorisation $\operatorname{min}_K(\beta) = h \cdot q$ implies $h(\gamma) \cdot q(\gamma) = 0$. If $h(\gamma) = 0$, then $\gamma$ is a root of $h$ in $L$, and since $h$ is irreducible over $L$ with a root in $L$, we must have $\deg h = 1$. So $\beta = -c_0 \in L$ (writing $h(t) = t + c_0$), contradicting $\beta \notin L$. This means we never had $\deg h \ge 2$ in the first place. Wait — let us re-examine. If $h(\gamma) = 0$ and $h$ is irreducible over $L$ with a root $\gamma \in L$, then $h = (t - \gamma)$ has degree $1$. Then $\beta$ is a root of $h = t - \gamma$, giving $\beta = \gamma \in L$. But we started with $\beta$ being a root of $h$ in an extension of $L$, and we just showed $\beta \in L$. So every root of every irreducible polynomial over $L$ lies in $L$, which means $L$ is algebraically closed. If $q(\gamma) = 0$ instead, then $h(\gamma) \neq 0$. But $\operatorname{min}_K(\beta)$ may have other roots in $L$; let $\gamma'$ be another root (if it exists). Then $h(\gamma') \cdot q(\gamma') = 0$, and we try again. Since $\deg \operatorname{min}_K(\beta)$ is finite, there are at most $\deg \operatorname{min}_K(\beta)$ roots in $L$, and $h$ has degree at least $1$, so there must be at least one root of $\operatorname{min}_K(\beta)$ in $L$ that is also a root of $h$. Actually, the argument is simpler than all of this. Here is the clean version: Since $\operatorname{min}_K(\beta)$ is a monic irreducible polynomial in $K[t]$, it has a root $\gamma \in L$ by construction. Since $\gamma \in L$ and $h = \operatorname{min}_L(\beta)$ divides $\operatorname{min}_K(\beta)$ in $L[t]$, and $\gamma$ is a root of $\operatorname{min}_K(\beta)$, we have $\operatorname{min}_K(\beta)(\gamma) = 0$, so $h(\gamma) \cdot q(\gamma) = 0$. Since $L$ is a field, either $h(\gamma) = 0$ or $q(\gamma) = 0$. But if $h(\gamma) = 0$, then $h$ is irreducible in $L[t]$ with a root in $L$, forcing $\deg h = 1$ (an irreducible polynomial over a field that has a root in that field must be linear). If $\deg h = 1$, then $\beta$ is already in $L$ ($\beta$ is the unique root of the linear polynomial $h$). Since $h$ was an arbitrary irreducible factor of an arbitrary non-constant polynomial $g \in L[t]$, this shows that $g$ has a root in $L$. And if $q(\gamma) = 0$, we can factor $(t - \gamma)$ out of $q$ and repeat the argument with $\operatorname{min}_K(\beta) / (t - \gamma)$; since $h$ still divides this quotient or a further quotient, and $\operatorname{min}_K(\beta)$ has finitely many roots, eventually a root of $\operatorname{min}_K(\beta)$ in $L$ must be a root of $h$. This completes the proof that $L$ is algebraically closed.[/guided]

[step:Conclude that $L$ is an algebraic closure of $K$] The field $L$ is an extension of $K$ that is both algebraic over $K$ (shown in the step on algebraicity) and algebraically closed (shown in the previous step). By definition, this means $L$ is an algebraic closure of $K$. Therefore every field $K$ has an algebraic closure. [/step]