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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?

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@Paul: Are you aware that in fact the process ends at the first step? That is, $K_1$ is already algebraically closed (so $K_{i+1} = K_i$ for all $i > 0$); the proof of this requires rather more input from the theory of fields than is needed to handle the usual discussion with the infinite process. Not sure to what extent this affects your interest in this particular construction. –  user27056 Nov 9 '12 at 3:25
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The algebraic closure of a field is its injective hull (ncatlab.org/nlab/show/injective+hull) - perhaps this is what you are thinking of? –  David Roberts Nov 9 '12 at 3:39
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Though this is in the category of algebraic extensions, not the full category of fields. –  David Roberts Nov 9 '12 at 3:40

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