Constructing Algebraic Closures in ETCS

Question

It seems to be folklore that "ETCS suffices to develop most of algebraic geometry", formally backed by the fact that "ETCS is equivalent to Bounded ZFC". But I have some doubts about that former fact:

If I am not mistaken the axioms of ETCS amount to requiring a wellpointed topos $\mathsf{Set}$ with natural numbers object and choice. As such there is nothing telling us that this topos is complete or cocomplete. In fact, if it were assumed to be cocomplete the object $$\coprod\limits_{n\geq0}\mathcal{P}^{(n)}(\mathbb{N})$$ should exist. But it is remarked here that it is consistent with ETCS that this object does not exist (at least this is my takeaway).

Now consider the construction of an algebraic closure $\overline{K}$ of a field $K$. One standard way is to iteratively construct algebraic field extensions $L^{(n)} \mid L^{(n)}$, such that $L^{(0)}=K$ and in $L^{(n+1)}$ every polynomial with coefficients in $L^{(n)}$ splits. The union $$L = \bigcup\limits_{n\geq0} L^{(n)}$$ turns out to be an algebraic closure of $K$. But this unit is an infinitely large filtered colimit, so a priori there is no reason for this to exist in our topos $\mathsf{Set}$. Other variants of this construction (as explained here) also inherently use transfinite colimits.

Up until now I was under the impression that reduction to the algebraically closed case was one of the more important tools of algebraic geometry. So with the slogan from the beginning of this question in mind this rises the question:

Can the algebraic closure of a field be constructed in ETCS without further assumptions on the topos $\mathsf{Set}$?

Thank you for your time.

Answer 1

Any elementary topos is complete and cocomplete in an internal sense. If you have an NNO then you will have certain inductive constructions, though not all. As you say, the iterated powerset is not available, because it cannot be bounded a priori. On the other hand, the algebraic closure can be bounded a priori – in fact, it can be constructed in non-iteratively:

1. Let $A$ be the polynomial algebra over $K$ generated by the symbols $T_f$, one for each polynomial $f$ in one variable over $K$.
2. Let $I$ be the ideal of $A$ generated by $f (T_f)$.
3. Let $\mathfrak{m}$ be any maximal ideal of $A$ containing $I$.
4. Then $\bar{K} = A / \mathfrak{m}$ is an algebraic field extension of $K$, and by construction every polynomial in one variable over $K$ splits in $\bar{K}$, so $\bar{K}$ is an algebraic closure of $K$.

That said, there are hidden uses of induction (even transfinite induction) in the above. For example, in step 2, we use induction to construct $I$ as a subset of $A$, and in step 3, we use transfinite induction to construct $\mathfrak{m}$ as a subset of $A$. These are legitimate essentially because the objective is a subset of a set we already have.

Answer 2

As such there is nothing telling us that this topos is complete or cocomplete.

It is definitely complete and cocomplete in the internal sense, i.e., every internal diagram has a limit and a colimit. In particular, the existence of infinite products follows from cartesian closure.

On the other hand, an elementary topos need not be complete or cocomplete with respect to some external set theory.

In fact, if it were assumed to be cocomplete the object $∐_{n≥0}P^{(n)}(N)$ should exist.

In order to talk about the coproduct of a family internally, the family itself must be constructed, e.g., as a morphism in the topos. It is not possible to construct such a family in ETCS because counterexamples exist.

If you are considering an elementary topos in some external set theory that (say) has the axiom of replacement, then it is possible to construct such a family externally. However, as already mentioned above, there is no reason for an elementary topos to be externally cocomplete.

Can the algebraic closure of a field be constructed in ETCS without further assumptions on the topos Set?

Yes, of course. The standard construction of Steinitz from 1910 starts by adjoining the roots of all monic polynomials, and then quotients out by a certain maximal ideal constructed using Zorn's lemma. This requires the axiom of choice, but makes no use of replacement.

Answer 3

Several answers have addressed an approach to taking the algebraic closure. I write to assert that one can indeed carry out the construction linked to by OP and presented by Leinster here.

If one starts with a field $K$ and some set $F \subseteq K[X]$, we can define $SR_K(F) = K[\{\alpha_{f, i} \mid f \in F, 0 \leq i < deg(f)\}] / ((f(\alpha_{f, i})_{f \in F, 0 \leq i < deg(f)})$.

In other words, to get $SR_K(F)$, we begin by adjoining, for each $f \in F$, $deg(f)$ new elements $\alpha_{f, i}$. We then quotient by an appropriate ideal to force each $\alpha_i$ to be a root. Note that $SR_K$ is functorial: given $F \subseteq G$, there is a canonical map $SR_K(F \subseteq G) : SR_K(F) \to SR_K(G)$ which is functorial. $\DeclareMathOperator{colim}{colim}$

We can then take $R = \colim\limits_{F \subseteq K[X] \setminus \{0\}, F \text{ finite}} SR_K(F)$. It turns out that if we quotient this ring by a maximal ideal, this gives us an algebraic closure of $K$.

The only issue is in making sure that the colimit in question actually exists. To ensure that it does, we simply note that there is an upper bound on the cardinality of $SR_K(F)$. Note that $|SR_K(F)| \leq |K[\{\alpha_{f, i} \mid f \in F, 0 \leq i < deg(f)\}]| \leq |K[\{\alpha_{f, i} \mid f \in K[X] \mid 0 \leq i < deg(f)\}]|$. This cardinality upper bound allows us to ensure that the colimit goes through.

Similarly, the construction of the tower of $L$s also works if one is clever and constructs the $L$s with a cardinality bound. Formally, we first prove that given any field $J$, we can extend it to $J'$ where all polynomials with coefficients in $J$ split in $J'$ (this proof should go through in ETCS just fine). Then, take the subfield $Q \subseteq J'$ consisting only of elements algebraic over $J$. So we actually have a field extension $J \subseteq Q$, where $|J| \leq \max(\aleph_0, |K|)$, and where all polynomials in $J$ split in $Q$.

Now, we can choose the sequence of $L^{(i)}$s such that for all $i$, $|L^{(i)}| \leq \max(\aleph_0, |K|)$. Using this cardinal bound, we can then take $\colim\limits_{n \in \mathbb{N}} L^{(i)}$. This will be an algebraic closure of $K$.

Finally, a third approach also works which uses the compactness theorem. We can inductively define a set $S$ of constant symbols as follows:

• For each $k \in K$, we have a constant symbol $c_k \in S$
• For each $n \geq 1$, for all $s_0, s_1 , \ldots, s_n \in S$, and for each $i$ such that $1 \leq i \leq n$, we have a constant symbol $root_i(s_0, s_1, \ldots, s_n)$
• For all $s_1, s_2 \in S$, we have constant symbols $minus(s_1), plus(s_1, s_2), times(s_1, s_2) \in S$

More formally, $S$ is constructed as a $W$-type, which is permissible since all toposes have $W$-types.

We then consider the theory of a field. Add in the axioms that $c_0 = 0$, $c_1 = 1$, and, for each $x, y \in K$, the axioms that $c_{x + y} = c_x + c_y$ and that $c_{xy} = c_x c_y$. Also include, for all $s_1, s_2 \in S$, axioms requiring $-s_1 = minus(s_1)$, $s_1 + s_2 = plus(s_1, s_2)$, and $s_1 s_2 = times(s_1, s_2)$. Finally, for all $s_0, s_1, \ldots, s_n \in S$, formally expand the polynomial $\prod\limits_{i = 1}^n (X - root_i(s_0, \ldots, s_n))$ into $T_0 + T_1 X + \ldots + T_n X^n$, and add, for all $0 \leq j \leq n$, the axiom $s_n \neq 0 \to s_j = T_j$. This last group of axioms states that $s_0 + s_1 X + \cdots + s_n X^n = \prod\limits_{i = 1}^n X - root_i(s_0, \ldots, s_n)$ whenever $s_n$ is nonzero - we call this the splitting axiom for $s_0 + s_1 X + \cdots + s_n X^n$.

Let $M$ be a model of this theory with constant symbol inclusion $i : S \to M$. Then $i(S)$ is a subfield of $M$ containing $K$ and, moreover, is algebraically closed. So to get our result, it suffices to show that our theory is satisfiable.

To do this, we need only show that the theory is finitely satisfiable by the compactness theorem. We note that it's easy to get models of all the axioms except the splitting axioms. It suffices to show that given any finite set $A$ of splitting axioms, we can find some field $L$, together with functions $K \to S \to L$, where the composite is a field extension and all the splitting axioms in $A$ hold in $L$. In fact, let $Q = \prod\limits_{P \text{ has its splitting axiom in } A} P$, and let $L$ be a splitting field of $Q$.

This is another illustration of that very common technique of getting cardinality bounds on the things you want to construct to make sure the construction goes through.