### 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.

## 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?