Emil Artin's construction of the algebraic closure of a field $K$ is as follows. Let $K_0 = K$, and inductively let $\{x_f\}$ be a set of indeterminates indexed by the irreducible $f$ in one variable over $K_i$. Let $\mathfrak{m}_i$ be a maximal ideal of $K_i[x_f]$ containing the ideal generated by all $f(x_f)$ for $f$ irreducible (one can prove this ideal is proper). Define $K_{i+1} = K_i[x_f]/\mathfrak{m}_i$. Clearly, each $K_{i+1}$ is algebraic over $K_i$ (and thus over $K_0$), and every polynomial with coefficients in $K_i$ splits in $K_{i+1}$. In particular, the union $\bar{K}$ of the $K_i$ is algebraic over $K$ and algebraically closed.

To me, a fan of homotopy theory, this seems an awful lot like a small object argument. We can express a solution of the polynomial $f$ over the field $K$ as a solution to a lifting problem: we want the map $0:K[x] \to K$ sending $x$ to $0$ to factor through the map $f:K[x] \to K[x]$ sending $x$ to $f(x)$, so we want a lift to a square which has $K \to 0$ on the right, $f:K[x] \to K[x]$ on the left, and $0$ on the top. (I'd draw this here, but I don't know how.) To construct $K_1$, we are doing something like taking the pushout of the diagram $$\bigotimes_f K[x_f] \stackrel{\otimes f}{\leftarrow} \bigotimes_f K[x_f] \stackrel{0}{\rightarrow} K$$ whose left arrow is a coproduct of the left sides of the aforementioned lifting problems. The algebraic closure itself is then the colimit of these pushouts, which is doing something like giving a fibrant replacement for $K$.

As you've probably noticed, this naïve outline has quite a few problems. The pushouts constructed above are not actually the fields $K_i$, but rather become them after a quotient by a maximal ideal. Also, the lifting problems we are solving differ between the $K_i$: the left sides of the squares that appear are of the form $f:K_i[x] \to K_i[x]$, where $f$ is an irreducible polynomial over $K_i$. On the other hand, the small object argument is done via a single set of generating cofibrations which appear on the left sides of squares at every stage.

I'd like to think that, after passing from the category of $K$-algebras to something presumably more complicated, one can construct a model structure in which the algebraic closure of a field appears as its fibrant replacement. Does anyone know if this is true?

Emil Artin's construction of the algebraic closure of a field $K$ is as follows. Let $K_{0} = K$, and inductively let $\{x_f\}$ be a set of indeterminates indexed by the irreducible $f$ in one variable over $K_i$. Let $\mathfrak{m}_i$ be a maximal ideal of $K_{i}[x_f]$ containing the ideal generated by all $f(x_f)$ for $f$ irreducible (one can prove this ideal is proper). Define $K_{i+1} = K_i[x_f]/\mathfrak{m}_{i}$. Clearly, each $K_{i+1}$ is algebraic over $K_i$ (and thus over $K_0$), and every polynomial with coefficients in $K_i$ splits in $K_{i+1}$. In particular, the union $\bar{K}$ of the $K_i$ is algebraic over $K$ and algebraically closed.

To me, a fan of homotopy theory, this seems an awful lot like a small object argument. We can express a solution of the polynomial $f$ over the field $K$ as a solution to a lifting problem: we want the map $0:K[x] \to K$ sending $x$ to $0$ to factor through the map $f:K[x] \to K[x]$ sending $x$ to $f(x)$, so we want a lift to a square which has $K \to 0$ on the right, $f:K[x] \to K[x]$ on the left, and $0$ on the top. (I'd draw this here, but I don't know how.) To construct $K_1$, we are doing something like taking the pushout of the diagram $$\bigotimes_f K[x_f] \stackrel{\otimes f}{\leftarrow} \bigotimes_f K[x_f] \stackrel{0}{\rightarrow} K$$ whose left arrow is a coproduct of the left sides of the aforementioned lifting problems. The algebraic closure itself is then the colimit of these pushouts, which is doing something like giving a fibrant replacement for $K$.

As you've probably noticed, this naïve outline has quite a few problems. The pushouts constructed above are not actually the fields $K_i$, but rather become them after a quotient by a maximal ideal. Also, the lifting problems we are solving differ between the $K_i$: the left sides of the squares that appear are of the form $f:K_i[x] \to K_i[x]$, where $f$ is an irreducible polynomial over $K_i$. On the other hand, the small object argument is done via a single set of generating cofibrations which appear on the left sides of squares at every stage.

I'd like to think that, after passing from the category of $K$-algebras to something presumably more complicated, one can construct a model structure in which the algebraic closure of a field appears as its fibrant replacement. Does anyone know if this is true?