## April 13, 2021

### Algebraic Closure

#### Posted by Tom Leinster This semester I’ve been teaching an undergraduate course on Galois theory. It was all online, which meant a lot of work, but it was also a lot of fun: the students were great, and I got to know them individually better than I usually would.

For a category theorist, Galois theory is a constant provocation: very little is canonical or functorial, or at least, not in the obvious sense (for reasons closely related to the nontriviality of the Galois group). One important not-obviously-functorial construction is algebraic closure. We didn’t get to it in the course, but I spent a while absorbed in an expository note on it by Keith Conrad.

Proving that every field has an algebraic closure is not entirely trivial, but the proof in Conrad’s note seems easier and more obvious than the argument you’ll find in many algebra books. As he says, it’s a variant on a proof by Zorn, which he attributes to “B. Conrad” (presumably his brother Brian). It should be more widely known, and now I find myself asking: why would you prove it any other way?

What follows is a somewhat categorical take on the Conrad–Zorn proof.

The reason why many constructions in field theory fail to be functorial is that they involve arbitrary choices, often of roots or irreducible factors of polynomials. For instance, to construct the splitting field of a polynomial, we have to make a finite number of arbitrary choices.

The algebraic closure of a field is something like the splitting field of all polynomials, so you’d expect it to require an infinite number of arbitrary choices (at least if the field is infinite). It’s no surprise, then, that it cannot be done without some form of the axiom of choice. Actually, it doesn’t require the full-strength axiom, but it does need a small dose: there are models of ZF in which some fields have no algebraic closure.

That explains why Zorn’s name came up, and it also suggests why no proof of the existence of algebraic closures can be entirely trivial.

An algebraic closure of a field $K$ is, by definition, an extension $M$ of $K$ that is algebraically closed (every polynomial over $M$ splits in $M$) and is minimal as such. So to construct an algebraic closure of $K$, it looks as if we have to manufacture an extension $M$ in which it’s possible to split not only every polynomial over $K$, but also every polynomial over $M$ itself.

Fortunately, there’s a lemma that saves us from that: any algebraic extension $M$ of $K$ in which every polynomial over $K$ splits is, in fact, an algebraic closure of $K$. This is proved by a standard field-theoretic argument that I won’t dwell on here, except to say that the main ingredient is the “transitivity of algebraicity”: given extensions $K \subseteq L \subseteq M$ and $\alpha \in M$, if $\alpha$ is algebraic over $L$ and $L$ is algebraic over $K$ then $\alpha$ is algebraic over $K$.

So: an algebraic closure of $K$ is an algebraic extension of $K$ in which every polynomial over $K$ splits. That’s the formulation I’ll use for the rest of this post.

Colin McLarty’s paper The rising sea: Grothendieck on simplicity and generality begins with a joke I find very resonant:

In 1949, André Weil published striking conjectures linking number theory to topology and a striking strategy for a proof [Weil, 1949]. Around 1953, Jean-Pierre Serre took on the project and soon recruited Alexander Grothendieck. Serre created a series of concise elegant tools which Grothendieck and coworkers simplified into thousands of pages of category theory.

How do we construct an algebraic closure of a given field? If you just want a concise presentation, I can’t do better than point you to Conrad’s note. But here I want to come at it a bit differently.

In my mind, the key to the Conrad–Zorn approach is to work in the world of rings for as long as possible and only descend to fields at the last minute. Behind this is the fact that the theory of rings is algebraic and the theory of fields is not, which is ultimately why constructions with fields often fail to be functorial.

(Here I mean “algebraic theory” in the technical sense. One could argue that little is more algebraic than the theory of fields, just as one could argue that “algebraic group” is a tautology…)

So we’re going to use rings, by which I mean commutative rings with $1$. A field extension $M$ of our field $K$ is just a homomorphism of fields $K \to M$, but we’re going to be interested in “ring extensions” of $K$, by which I mean homomorphisms $K \to A$, where $A$ is a ring. Such a thing is just a (commutative) $K$-algebra, which is why you don’t hear anyone talking about “ring extensions”.

Given a polynomial $f \neq 0$ over $K$, we could consider the splitting field $SF_K(f)$ of $f$ over $K$. But instead, let’s consider what I’ll call its splitting ring $SR_K(f)$. Informally, $SR_K(f)$ is the $K$-algebra you get by freely adjoining to $K$ the right number of roots, say

$\alpha_{f, 1}, \ldots, \alpha_{f, \deg(f)},$

and imposing relations to say that they are the roots of $f$. Write

$\hat{f}(x) = a \prod_{i = 1}^{\deg(f)} (x - \alpha_{f, i}),$

where $a$ is the leading coefficient of $f$. The relations should make $\hat{f}(x)$ and $f(x)$ be equal as polynomials over $SR_K(f)$.

Formally, then, we define $SR_K(f)$ to be the $K$-algebra

$SR_K(f) = K[\alpha_{f, 1}, \ldots, \alpha_{f, \deg(f)}]/\sim$

where $\sim$ identifies the $r$th coefficient of $f$ with the $r$th coefficient of $\hat{f}$ for each $r \geq 0$. So $SR_K(f)$ is the free $K$-algebra in which $f$ splits.

For example, take $f(x) = x(x - 1)$. The splitting field of $f$ over $K$ is just $K$ itself. But the splitting ring $SR_K(f)$ is another story. It’s the polynomial ring $K[\alpha, \beta]$ quotiented out by the relations that identify the coefficients of $f(x) = x(x - 1)$ with those of

$\hat{f}(x) = (x - \alpha)(x - \beta).$

Comparing constant terms gives $\alpha\beta = 0$, and comparing coefficients of degree $1$ gives $\alpha + \beta = 1$. So

$SR_K(f) = K[\alpha, \beta]/(\alpha\beta = 0, \ \alpha + \beta = 1),$

or equivalently,

$SR_K(f) = K[\alpha]/(\alpha^2 = \alpha).$

This is bigger than the splitting field! A field can’t have nontrivial idempotents: $\alpha^2 = \alpha$ forces $\alpha \in \{0, 1\}$. But a ring can, and $SR_K(f)$ does.

More generally, we can form the splitting ring of not just one polynomial but any set of them (providing they’re nonzero). For $\mathcal{P} \subseteq K[x] \setminus \{0\}$, put

$SR_K(\mathcal{P}) = K[\{ \alpha_{f, i}: f \in \mathcal{P}, \ 1 \leq i \leq \deg(f)\}]/\sim,$

where $\sim$ identifies the coefficients of $f$ with the corresponding coefficients of $\hat{f}$ for each $f \in \mathcal{P}$. Equivalently, $SR_K(\mathcal{P})$ is the coproduct $\coprod_{f \in \mathcal{P}} SR_K(f)$ in the category of $K$-algebras. It’s the free $K$-algebra in which all the polynomials in $\mathcal{P}$ split.

The Conrad proof goes like this: put $R = SR_K(K[x] \setminus \{0\})$, then (i) show that $R$ is nontrivial, (ii) quotient $R$ out by a maximal ideal to get a field $M$, and (iii) show that $M$ is an algebraic closure of $K$. Step (i) is where the substance of the argument lies. Step (ii) calls on the fact that every nontrivial ring has a maximal ideal (a routine application of Zorn’s lemma). And step (iii) is pretty much immediate by the lemma I started with, since by construction every polynomial over $K$ splits in $R$, hence in $M$.

But you can come at it a bit differently, as follows. First let me list some apparent trivialities.

• A ring $A$ is trivial if and only if $0_A = 1_A$. (Triviality is an equational property.)

• The forgetful functor from rings to sets preserves filtered colimits. This is equivalent to the statement that the theory of rings is finitary: the ring operations (such as addition) take only finitely many inputs.

• A filtered colimit of nontrivial rings is nontrivial. This isn’t as obvious as it might seem: for instance, the analogous statement for abelian groups is false. It follows from the previous two bullet points and the way filtered colimits work in $Set$.

• A ring $A$ is nontrivial if and only if there is a homomorphism from $A$ to some field, if and only if there is a surjective homomorphism from $A$ to some field. Indeed, the first bullet point together with the nontriviality of fields shows that if $A$ admits a homomorphism to a field then $A$ is nontrivial. On the other hand, if $A$ is nontrivial then by Zorn’s lemma, we can find a maximal ideal $J$ of $A$, and the natural homomorphism $A \to A/J$ is then a surjection to a field.

• Given a homomorphism $A \to B$ of $K$-algebras and a polynomial $f$ over $K$, if $f$ splits in $A$ then $f$ splits in $B$.

Now, here’s a proof that every field $K$ has an algebraic closure.

• Put $S = colim_{\mathcal{F}} SR_K(\mathcal{F})$, where the colimit is over finite subsets $\mathcal{F}$ of $K[x]\setminus\{0\}$. Here I’m implicitly using the fact that $SR_K(\mathcal{F})$ is functorial in $\mathcal{F}$ with respect to inclusion. This is a filtered colimit.

• For each $\mathcal{F}$, the polynomial $\prod_{f \in \mathcal{F}} f$ has a splitting field $L$. Then every polynomial in $\mathcal{F}$ splits in $L$, so by the universal property of $SR_K(\mathcal{F})$, there is a homomorphism $SR_K(\mathcal{F}) \to L$. It follows that $SR_K(\mathcal{F})$ is nontrivial.

• So the $K$-algebra $S$ is a filtered colimit of nontrivial rings, and is therefore nontrivial.

• Hence there is a surjective homomorphism from $S$ to some field $M$. Composing the natural maps $K \to S \to M$ makes $M$ into an extension of $K$.

• For each nonzero $f \in K[x]$, we have homomorphisms of $K$-algebras $SR_K(f) \to S \to M$, and $f$ splits in $SR_K(f)$, so $f$ splits in $M$. Hence the field $M$ is an extension of $K$ in which every polynomial over $K$ splits.

• Each of the $K$-algebras $SR_K(\mathcal{F})$ is generated by elements algebraic over $K$, so the same is true of $M$. Hence $M$ is algebraic over $K$, and is, therefore, an algebraic closure of $K$.

In any construction of algebraic closures, the challenge is to find some extension $M$ of $K$ in which every polynomial over $K$ splits. It may be that $M$ is wastefully big, but you can cut it down to size by taking the subfield $M'$ of $M$ consisting of the elements algebraic over $K$, and $M'$ is then an algebraic closure of $K$. This provides an alternative to the last step above. But I like it less, because the fact is that the $M$ we constructed is already the algebraic closure.

Whether one uses the approach in Conrad’s note or the slight variant I’ve just given, I prefer this to the Artin (?) proof that appears in many books and is also summarized on page 2 of Conrad’s note. It seems much more direct and appealing.

There’s another proof I also like: the existence of algebraic closures follows very quickly from the compactness theorem of model theory. Like the proof above, this argument relies on the knowledge that for any finite collection of polynomials, there’s some field in which they all split. In other words, it uses the existence of splitting fields.

And there’s one more thing I can’t resist saying. Although algebraic closure isn’t functorial in the obvious sense, I believe it’s functorial in a non-obvious sense — what I’d think of as a “Hakim-type” sense.

Let me explain what I mean, taking a bit of a run-up.

The category of rings has a subcategory consisting of the local rings. The inclusion has no left adjoint: there’s no way to take the free local ring on a ring. But there is if you allow the ambient topos to vary! In other words, consider all pairs $(\mathcal{E}, R)$ where $\mathcal{E}$ is a topos and $R$ is a ring in $\mathcal{E}$. Given such a pair, one can look for the universal map from $(\mathcal{E}, R)$ to some pair $(\mathcal{E}', R')$ where $R'$ is local.

Monique Hakim proved, among other things, that if we start with a ring $R$ in the topos $Set$, then this construction gives the topos $Sh(Spec(R))$ of sheaves on the prime spectrum of $R$, together with the structure sheaf $\mathcal{O}_R$, which is a sheaf of local rings on $Spec(R)$, or equivalently a local ring in $Sh(Spec(R))$. This wonderful result appeared in her 1972 book Topos Annelés et Schémas Relatifs.

Now if I’m not mistaken, there’s a similar description of algebraic closure. Consider all pairs $(\mathcal{E}, K)$ where $\mathcal{E}$ is a topos and $K$ is a field in $\mathcal{E}$. Given such a pair, one can look for the universal map from it to a pair $(\mathcal{E}', K')$ where $K'$ is algebraically closed. And I believe that if we start with a field $K$ in $Set$, the resulting pair is the topos of sets equipped with a continuous action by the absolute Galois group $G$ of $K$, together with the algebraic closure $\overline{K}$ with its natural $G$-action. There are lots of details I’ve skipped here, but perhaps someone can tell me whether that’s more or less right and, if so, where this is all written up.

Posted at April 13, 2021 12:13 PM UTC

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### Re: Algebraic Closure

Back in the day, we’d sometimes hear around these parts of the groupoid of algebraic closures of a field (and again). Is that not a way to get around making arbitrary choices?

Posted by: David Corfield on April 13, 2021 1:49 PM | Permalink | Reply to this

### Re: Algebraic Closure

I’m sure that’s at least indirectly relevant. But at some point you have to do the work of showing that algebraic closures actually exist!

Posted by: Tom Leinster on April 13, 2021 2:59 PM | Permalink | Reply to this

### Re: Algebraic Closure

A really nice post, thanks!

I think someone’s preference for a proof of this theorem will depend largely on what they are familiar with. And if you are teaching it, with what techniques you want to teach! (The model-theoretic proof can be found in my book, An Invitation to Model Theory!)

For me, the key point of the proof is that the finitary operation of finding splitting fields commutes with filtered colimits. Then you need some version of the axiom of choice to show the existence of the filtered colimits. The model-theoretic perspective is to do as much work as possible in the finitary setting (finite extensions of fields), whereas your proof does the work “up top”. One advantage of the finitary approach is that it is more-or-less clear that the absolute Galois group of K is the (projective) limit of the Galois groups of the finite extensions. I don’t think that comes out of your proof very directly.

The algebraic closure of K is unique up to non-unique isomorphism, and this absolute Galois group exactly measures the non-uniqueness of that isomorphism, and the lack of functorality. For a Galois theory course one might argue that this is really the object of interest!

Posted by: Jonathan Kirby on April 14, 2021 10:00 AM | Permalink | Reply to this

### Re: Algebraic Closure

Thanks, Jonathan. I was hoping you might read this and have something to say.

From a pedagogical point of view, the Conrad–Zorn proof is the one I would have taught if we’d got that far in the course. The students don’t know about Zorn’s lemma, or maximal ideals for that matter, but I could have asked them to take on trust that every nontrivial ring has some quotient that’s a field. The scope of our course is fairly limited compared to some other Galois theory courses (e.g. those that happen at graduate level), because the students begin with not very much background in ring theory. So there’s a lot we didn’t do. But I couldn’t help thinking about it anyway.

One advantage of the finitary approach is that it is more-or-less clear that the absolute Galois group of K is the (projective) limit of the Galois groups of the finite extensions. I don’t think that comes out of your proof very directly.

Right, that seems like an important point. I should think about that.

Posted by: Tom Leinster on April 14, 2021 1:28 PM | Permalink | Reply to this

### Re: Algebraic Closure

By the way, anyone who isn’t aware of Jonathan’s model theory book should be! Among other virtues, it’s the shortest introduction to model theory I know — much shorter, for instance, than a well-known one with “shorter” in the title :-)

Posted by: Tom Leinster on April 14, 2021 1:34 PM | Permalink | Reply to this

### Re: Algebraic Closure

Thanks for this, Tom. By a coincidence, on Monday Thierry Coquand on constructive algebra which mentions algebraic closure, see

http://www.cse.chalmers.se/~coquand/verona.pdf

Posted by: Nicola Gambino on April 14, 2021 3:20 PM | Permalink | Reply to this

### Re: Algebraic Closure A filtered colimit of nontrivial rings is nontrivial. This isn’t as obvious as it might seem: for instance, the analogous statement for abelian groups is false. It follows from the previous two bullet points and the way filtered colimits work in $\mathrm {Set}$.

Just to make sure I understand this I verified that this is false for abelian groups: The point is that the filtered colimit is zero if for any $G$ in the colimit diagram and $g \in G$ there’s a map $G \to H$ in the diagram which maps $g$ to zero, but this does not imply that any particular group in the diagram is trivial. The first counterexample I thought of is to take the doubling function $f : \mathbb {Z} / 4 \mathbb {Z}$, $f (x) = 2 x$ and repeat it infinitely in a diagram: $\mathbb {Z} / 4 \mathbb {Z} \xrightarrow {f} \mathbb {Z} / 4 \mathbb {Z} \xrightarrow {f} \mathbb {Z} / 4 \mathbb {Z} \xrightarrow {f} \cdots$ But then I realized there is an even simpler example, just using the zero function $0 (x) = 0$ instead: $\mathbb {Z} \xrightarrow {0} \mathbb {Z} \xrightarrow {0} \mathbb {Z} \xrightarrow {0} \cdots$ The colimit of either diagram is zero.

Posted by: Itai Bar-Natan on April 14, 2021 4:22 PM | Permalink | Reply to this

### Re: Algebraic Closure

Thanks for that — I had a much more complicated argument in mind, so I’m glad you came up with that simpler one.

Here’s a variant on your example: splitting an idempotent is an example of a filtered colimit, and if you split the idempotent $0: \mathbb{Z} \to \mathbb{Z}$ in $Ab$, you also get the trivial group.

The argument I had in mind (to show that a filtered colimit of abelian groups needn’t be abelian) is interesting for other reasons. It’s to do with the fact that not every abelian group has a maximal subgroup (meaning, of course, a maximal proper subgroup).

At some point I fell into the trap of believing that every nontrivial module has a maximal submodule, lazily thinking that it must be a consequence of Zorn’s lemma. But it’s not, because the union of a chain of proper submodules needn’t be proper.

For example, the $\mathbb{Z}$-module $\mathbb{Q}$ has no maximal submodule (subgroup). For suppose that $N \subset \mathbb{Q}$ is maximal. Then $\mathbb{Q}/N$ has no nontrivial proper subgroups, so it must be either $\mathbb{Z}$ or $\mathbb{Z}/p\mathbb{Z}$ for some prime $p$. In either case, you can easily derive a contradiction from the existence of the natural homomorphism $\mathbb{Q} \to \mathbb{Q}/N$ (for what would map to the generator?).

From this it follows that a cofiltered limit of nontrivial abelian groups isn’t always nontrivial. For suppose that it is. Given a nonempty chain $\mathcal{C}$ of proper subgroups of an abelian group $A$, the corresponding system of quotient groups forms a cofiltered diagram, and its colimit is $A/\bigcup\mathcal{C}$. Since this colimit is nontrivial, $\bigcup\mathcal{C}$ is a proper subset of $A$. Hence the union of any nonempty chain of proper subgroups of an abelian group is proper. Now Zorn’s lemma implies that every nontrivial abelian group has a maximal subgroup. And we’ve just seen that that’s false.

Posted by: Tom Leinster on April 14, 2021 7:17 PM | Permalink | Reply to this

### Re: Algebraic Closure

I remember that Conrad write-up! I used it to write the nLab article on splitting fields, which has a certain amount in common with your write-up here.

One thing I was doing in those days was seeing how much mathematical juice could be gotten out of the ultrafilter principle, as opposed to the stronger axiom of choice. Thus, the existence and uniqueness up to isomorphism of splitting fields can be proven just using the ultrafilter principle as the principle of choice. The general flavor of the arguments is similar to arguments in model theory that invoke the compactness theorem.

Posted by: Todd Trimble on April 16, 2021 6:33 PM | Permalink | Reply to this

### Re: Algebraic Closure

Aha, great. Thanks. I looked at the nLab article on algebraic closure before I wrote this post, but I didn’t think of looking at the article on splitting fields.

Probably not coincidentally, I also tried to do everything with just the ultrafilter principle rather than the full axiom of choice. But I didn’t manage.

Am I right in thinking that the compactness theorem only requires some version of the ultrafilter principle, but its proof via ultraproducts uses the full axiom of choice? This question isn’t very deeply considered; I only ask because when I tried proving compactness using ultraproducts, I found myself wanting something like full choice.

Posted by: Tom Leinster on April 17, 2021 1:05 AM | Permalink | Reply to this

### Re: Algebraic Closure

I believe Łoś’s theorem in full generality does need the axiom of choice, but for the applications of compactness needed here, one doesn’t need to work as hard. Roughly speaking, the “level of logic” needed for these applications is mere propositional logic.

I’d be be very happy to discuss this in greater detail. Actually, it would give me an opportunity to improve my own understanding of these matters.

Some time back I wrote up an article prime ideal theorem. But that’s harder than necessary for proving the prime ideal theorem for commutative rings, which is used in the article on splitting fields.

Posted by: Todd Trimble on April 17, 2021 2:27 AM | Permalink | Reply to this

### Re: Algebraic Closure

Consider all pairs (ℰ,K) where ℰ is a topos and K is a field in ℰ. Given such a pair, one can look for the universal map from it to a pair (ℰ’,K’) where K’ is algebraically closed. And I believe that if we start with a field K in Set, the resulting pair is the topos of sets equipped with a continuous action by the absolute Galois group G of K, together with the algebraic closure ¯K with its natural G-action.

How would the axiom of choice enter such a construction? Proving that an arbitrary field has an algebraic closure cannot be done in ZF alone.

Yet formulating algebraic closures as a left 2-adjoint would seem to imply that your construction is choice-free.

Or perhaps to show that the resulting topos is spatial one needs the axiom of choice?

Posted by: Dmitri Pavlov on April 16, 2021 7:16 PM | Permalink | Reply to this

### Re: Algebraic Closure

This a remark a out the end of your post. The theory of fields is not geometric (in the toposic sense) which is why one cannot hope to define a spectrum of fields in the way one recovers the Zariski spectrum of local rings à la Hakim (in particular, there no canonical way to define what is a field in a topos: what will you say of elements which are not non- zero). Furthermore, if there were a universal property of the algebraic closure, there wouldn’t be any Galois theory (all Galois groups would be trivial by lack of existence of automorphisms of the algebraic closure).

The algebraic closure of a field is the choice of a point of a suitable topos associated to the field. Given a topos there is canonically given category of points of the topos, but there is no reason why a specific point would be better than any other, even when the category of points is a connected non empty groupoid. It is like in topology: one can define the fundamental group of a space for each choice of base point, or the fundamental groupoid. Absolute Galois groups are exactly like fundamental groups (this is what Grothendieck taught us). In the case of the algebraic closure of a field $k$, one considers the topos of sheaves on the site defined as the opposite of finite reduced $k$-algebras (those which are finite products of finite field extensions of $k$). Coverings corresponds to finite families of flat maps of $k$-algebras $f_i\colon A\to B_i$ such that the induced map $A\to \prod_i B_i$ is injective. An algebraic closure of $k$ is exactly the same thing as a point of this topos: given a point, one may consider the system of its neighborhoods which will be a filtered diagram of finite $k$-algebras, the colimit of which will be an algebraic closure of $k$. We may put this topos in a spectral perspective à la Hakim, but not formulated in the language of fields. We should consider “algebraically closed henselian local rings” instead. If we only consider separable extensions instead, this is a way to introduce the étale topology (from a perspective à la Hakim, the étale topos of a ring is the spectrum obtained if we replace the notion of local rings by the one of strictly local henselian rings).

Posted by: Denis- Charles Cisinski on April 16, 2021 10:47 PM | Permalink | Reply to this

### Re: Algebraic Closure

Thanks for this comment, Denis-Charles. I was hoping some expert would come and say accurate things about that vague last part of my post. In particular, I was guiltily aware that separability should probably be a part of the story, because I’d been looking at an old paper of Kennison’s:

John F. Kennison. Separable algebraic closure in a topos. Journal of Pure and Applied Algebra 24 (1982), 7–24.

It’ll take me a while to digest the main part of your comment, but a couple of quick things for now:

The theory of fields is not geometric

As you say, there’s more than one reasonable definition of field in a topos, and they’re not equivalent. But one of them is geometric: the one that says “every element is $0$ or invertible”. When you write that the theory of fields is not geometric, I guess what you mean is that the geometric theory just mentioned isn’t the “obvious” theory of fields, in which case it becomes a subjective matter of terminology. But in terms of the mathematical substance, when I was imagining a Hakim-type theorem on fields in a topos, I guess I was thinking of using the geometric theory of fields (or the theory of “geometric fields”, in Johnstone’s terminology).

if there were a universal property of the algebraic closure, there wouldn’t be any Galois theory (all Galois groups would be trivial by lack of existence of automorphisms of the algebraic closure).

That’s clear if we’re talking about universal properties in the obvious sense — staying in Set, without changing topos. As you say, that’s exactly because of the nontriviality of Galois theory. But it’s not so clear to me that this is an obstacle to a Hakim-type universal property.

Perhaps the answer to that is in the rest of your comment.

Posted by: Tom Leinster on April 17, 2021 1:32 AM | Permalink | Reply to this

### Re: Algebraic Closure

It is worth looking at the problem of constructing/characterizing the splitting field of a given polynomial. In some sense, everything is build on that (at least in Galois theory), and there are even examples where the algebraic closure is just a splitting field (e.g. the splitting field of $X^2+1$ in the case of real numbers). The whole point of Galois theory is that there is simply no reason to declare one root of a polynomial better than the other. I understand that we can do classical Galois theory internally in a topos, like in the work of Kennison and Wraith, but I do not see how this affects the question. In particular, spectral constructions à la Hakim involve the possibility of changing the underlying topos, while the whole point of Wraith and Kennison is to stay in a given topos.

Coming back to universal properties, the whole story about Zariski spectra is that, although there is no universal local ring associated to a (commutative) ring, there is a unique factorization system on the category of commutative rings: you may consider conservative maps (i.e. homomorphisms of rings $u:A\to B$ such that an element $a\in A$ is invertible if and only if $f(a)\in B$ is invertible as well as local maps (i.e. homomorphisms of rings $u:A\to B$ which exhibit $B$ as a localization of $A$ with respect to some subset $S\subset A$). In particular, given a morphism $f:A\to k$ with $k$ a field, there is a prime ideal $\mathfrak{p}=f^{0}\subset A$, and the localization $A_\mathfrak{p}=(A-\mathfrak{p})^{-1}A$ is a factorization of $f$ into a local map followed by a conservative one (and such factorization is indeed unique). To my knowledge, all spectral constructions à la Hakim are obtained from such factorization systems. You may have a look at this paper of Mathieu Anel to see a general framework as well as nice examples: https://arxiv.org/abs/0902.1130v2 There is also this paper of Gabber and Kelly, in which there is a variety of Grothendieck topologies which are characterized by their local objects (and thus belong to Anel’s framework): https://arxiv.org/abs/1407.5782

A way to make precise that fields are not obtained through a spectral construction is that they do not appear through a factorization system à la Anel, at least if we work on the category of commutative rings.

Posted by: Denis-Charles Cisinski on April 17, 2021 3:56 PM | Permalink | Reply to this

### Re: Algebraic Closure

Here’s one way to see that there’s no functorial way to construct algebraic closures. The category $Field_p$ of fields of characteristic $p$ (where $p$ is a prime or $0$) contains the category $ACL_p$ of algebraically closed fields of characteristic $p$, via a fully faithful inclusion functor $i: ACL_p \to Field_p$. A functorial algebraic closure wuould give us a functor $F: Field_p \to ACL_p$ and a natural transformation $1 \Rightarrow iF$. The inclusion of a field into its algebraic closure would also give us a natural transformation $1 \Rightarrow Fi$. Passing to classifying spaces, we would have a homotopy equivalence $|Field_p| \simeq |ACL_p|$. But $|Field_p|$ is contractible because $Field_p$ has an initial object, whereas $|ACL_p| = BGal(\overline{\mathbb{F}}_p/\mathbb{F}_p)$ is not contractible, a contradiction.

On the other hand, sometimes functorial “injective hulls” can be constructed. For instance, random graphs are characterized among all graphs by an injectivity property. One can functorially (with respect to graph embeddings) embed a nonempty graph $G$ into a random graph by iteratively adding, for each pair of tuples $\vec x$ in the graph, a point $y$ such that $\vec x$ is connected to all the $x_i$’s and no other points. This is okay because the classifying space of the category of nonempty graphs and the classifying space of the category of random graphs are both contractible.

I don’t think the classifying space obstruction is a complete obstruction, but I think it’s an interesting one nonetheless.

Posted by: Tim Campion on April 18, 2021 3:56 PM | Permalink | Reply to this

### Re: Algebraic Closure

Er– in the graph example, it should say that for every tuple $\vec x$ in the graph, one adds a point $y$ which is connected to $\vec x$ and to nothing else. Doing this “iteratively” means you do this for each tuple in $G$ to obtain $G'$, then do this for every tuple in $G'$ to obtain $G''$, etc, and repeat $\omega$-many times.

Posted by: Tim Campion on April 18, 2021 3:59 PM | Permalink | Reply to this

### Re: Algebraic Closure

Now if I’m not mistaken, there’s a similar description of algebraic closure.

Thanks for sharing this nice idea! I’d be surprised if this is written down anywhere, so if it’s true you should write it up!

I may be wrong, but it seems to me that whether this is true (not thinking about a construction à la Hakim, just about whether the universal property could hold) may rest upon the following: given a ‘fielded topos’ (pair of a topos and a field $k$ in it), can we construct the ‘Galois group’ of an extension $k'$ of that field as a group object in the topos? This is asking for some kind of ‘Galois descent’: if we take an ordinary field viewed as a (locally) constant sheaf on a site for the topos, then this is exactly asking that the Galois group in the ordinary sense be able to be viewed as a sheaf on that site, which it can be (a (locally) constant one).

Whether this ‘Galois descent’ property holds probably boils down to whether the definition of the Galois group can be expressed in geometric logic, which I haven’t thought carefully about, but which I’d guess is probably fine.

As an aside, one can of course describe the topos of $Gal(k^{sep}/k)$-sets as the étale topos of $Spec(k)$. If your construction is true, one will I think be able to recover from it things like the action of the Galois group on étale cohomology.

Posted by: Richard Williamson on April 20, 2021 1:06 AM | Permalink | Reply to this

### Re: Algebraic Closure

Hi Richard.

Thanks for sharing this nice idea! I’d be surprised if this is written down anywhere, so if it’s true you should write it up!

I should be clear that this idea — to the extent that it’s correct — is not mine. (Though I’m happy to take credit for the incorrect aspects.) It was based on my understanding of some comments by Gavin Wraith back in 2009. It may well be that I’ve misunderstood. In principle I’d like to take time to work through some of the literature on this and Denis-Charles’s comments here, but in reality there’s a heap of other stuff I need to do, so this will have to wait for another day.

Posted by: Tom Leinster on April 20, 2021 10:52 AM | Permalink | Reply to this

### Re: Algebraic Closure

I feel that the way you have formulated your understanding of it is sufficiently interesting to be worthy of more credit than you give yourself :-).

I am reasonably familiar with the things Denis-Charles mentions, e.g. the work of Anel that he mentioned, and I am not convinced (yet at least) that they exclude a universal property of the kind you formulate (which I would disassociate somewhat from the question of whether it arises as some kind of Spec construction).

Indeed, if one can construct in the topos the Galois group of a field extension as I mentioned, then I think it should be fairly formal that the universal property does hold. For then a geometric morphism of fielded topoi from $Gal(k'/k)$-sets to a $k$-fielded topos $T$ will amount to a field extension of $k$ to $k'$ ‘in $T$’, and asking for a universal property of $Gal(k^{alg}/k)$-sets of the kind you suggest will amount I think to being able to construct $k^{alg}$ in $T$, i.e. reduce to an internalisation in geometric logic of ‘ordinary’ Galois theory.

This is much the same as the way in which the topos-theoretic universal property of the Spec construction amounts (from one perspective) to expressing the ‘set-level’ construction of Spec of a ring in the internal logic of a ringed topos.

Posted by: Richard Williamson on April 20, 2021 12:15 PM | Permalink | Reply to this

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