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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
The algebraic closure of a field is its injective hull ( - perhaps this is what you are thinking of? – David Roberts Nov 9 '12 at 3:39
Though this is in the category of algebraic extensions, not the full category of fields. – David Roberts Nov 9 '12 at 3:40

1 Answer 1

There is at least one significant reason why we don't expect algebraic closure to be a fibrant replacement. Fibrant replacements are usually "unique up to contractible choice", so that there should be no indeterminacy about what the fibrant replacement is. On the order hand, an algebraic closure $K \to \bar K$ does not determine the target uniquely because there are typically nontrivial automorphisms that fix $K$.

However, this has a relatively straightforward fix: we can consider the relative case where we have a map $K \to L$ of fields, and consider the algebraic closure of $K$ inside $L$ (alias the integral closure). This is unique. Moreover, we might hope that there's a model structure somewhere so that perhaps the fibrant replacement of $K$ is boring, but the algebraic closure of $K$ inside $L$ is a factorization $K \to K' \to L$ into an acyclic cofibration followed by a fibration (so it's fibrant replacement in the "over $L$" category).

The following describes an attempt to do so. Here are some definitions.

  • For a ring homomorphism $R \to S$, let $R^c \subset S$ be the set of elements in $S$ which are integral over $R$: they satisfy a monic polynomial with coefficients in $R$. This is a subring of $S$ (the integral closure of $Im(R) \subset S$).

  • A map $R \to S$ is a weak equivalence if $R^c$ contains all the units of $S$.

  • A map $R \to S$ is a cofibration if $S$ is a localization of $R^c$.

  • A map $R \to S$ is a fibration if $R \to S$ is a monomorphism and $Im(R) = R^c$.


  • A map $R \to S$ is an acyclic cofibration if and only if $S = R^c$.

  • A map $R \to S$ is an acyclic fibration if and only if $R \to R^c$ is an isomorphism and $R^c$ contains the units of $S$.

We'd like to use this to define a (cofibrantly generated) model structure on the category of commutative rings with unit. If it does, then $R \to R^c \hookrightarrow S$ is always a factorization into an acyclic cofibration followed by a fibration, and factorization into a cofibration followed by an acyclic fibration also adds the units of $S$ to $R^c$. In particular, for fields this recovers the integral closure as desired.

To prove that this is a model structure, I'll write down some generating classes and apply the recognition theorem.

  • Let $J$ be the set containing the map $\Bbb Z[x] \to \Bbb Z$ sending $x$ to zero, together with the maps $\Bbb Z[a_1,\dots,a_n] \to \Bbb Z[a_1,\dots,a_n,y]/(y^n = \sum a_i y^{n-i})$.

  • Let $I$ be the union of $J$ with the map $\Bbb Z[x] \to \Bbb Z[x^{\pm 1}]$.

Pushouts along the maps in $J$ are either quotients or adjoining roots of monic polynomials, and so the closure of $J$ under pushouts, retracts, and transfinite compositions is the collection of maps $R \to S$ such that $S = R^c$. This is precisely the set of acyclic cofibrations.

Pushouts along the maps in $I$ are quotients, algebraic extensions, or localizations. The closure of $I$ under pushouts, retracts, and transfinite compositions is the collection of maps $R \to S$ such that $S$ is a localization of $R^c$. This is precisely the set of cofibrations.

Maps $R \to S$ with the right lifting property with respect to $J$ have the right lifting property with respect to $\Bbb Z[x] \to \Bbb Z$ (i.e., are monomorphisms) and with respect to adjoining solutions to monic polynomials (i.e. $R \to R^c$ is surjective). These are precisely the fibrations.

Maps $R \to S$ with the right lifting property with respect to $I$ also have the right lifting property with respect to $\Bbb Z[x] \to \Bbb Z[x^{\pm 1}]$, and hence $R$ must contain the units of $S$. These are precisely the acyclic fibrations.

As a result, it almost looks like we get a model structure on the category of commutative rings with the desired properties. However, a critical part of the recognition theorem fails. Our definition of "weak equivalence" does not satisfy 2-out-of-3 for a really stupid reason: any map $R \to 0$ is a weak equivalence.

If we restrict our attention to some class of rings and monomorphisms, we fix the 2-out-of-3 axiom but lose completeness and cocompleteness. If we restrict our attention to the lattice of subrings of a fixed large ring (like a field $L$), we still inherit the lifting and factorization axioms from the category of rings, and we do get a model structure (but it's somewhat less satisfying than if it applied to the category of rings). It seems most promising to instead ask that our weak equivalences also be monomorphisms, but my time is up on this problem for now.

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I don't think the $RLP(I)$ are the same as what you call the acyclic fibrations: rather they seem like $R\to S$ s.t. $R=R^c$ and $R^c\cap S^\times=(R^c)^\times$. – Charles Rezk Oct 24 at 17:10
So maybe w.e.s should be $R\to S$ such that $R^c\cap S^\times = (R^c)^\times$. Still not a model category ... – Charles Rezk Oct 24 at 17:11
@CharlesRezk Right, my bad. And still not a model category, but just a couple of factorization systems. – Tyler Lawson Oct 25 at 0:58
I forgot to mention that Clark pointed me towards a paper of Anel in a related earlier question of mine: – Tyler Lawson Oct 25 at 0:59

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