Proof of existence of algebraic closure with Zorn's lemma

Question

I read the proof of existence of algebraic closure by Zorn's lemma, whose idea is to to consider the "set" $$ \mathcal{P} := \{ K \mid \text{the algebraic extension of } F \}. $$ Using Zorn's lemma, one can show that $(\mathcal{P}, \subset)$ has a maximal element, which is an algebraic closure of $F$. The only problem is that $\mathcal{P}$ is actually not a set but a class, hence we cannot apply Zorn's lemma on it.

However, I wonder if the following consideration works. Let us fix a Grothendieck universe $\mathcal{U}$, which is big enough for most cases. In the future, we only consider $\mathcal{U}$-sets, $\mathcal{U}$-groups, $\mathcal{U}$-fields and so on. Now $$ \mathcal{P} := \{ K \mid \text{the $\mathcal{U}$-algebraic extension of } F \} $$ is indeed a set, hence we can apply Zorn's lemma on it.

The only thing I worry about is whether something wrong will appear when we play Zorn's lemma over sets "as big as the fixed Grothendieck universe", such as $\mathcal{P}$.

Answer 1

This is a perfectly valid approach. But notice that the existence of a Grothendieck universe is independent from ZFC. This is why this approach is probably usually not taken. On the other hand, the existence of Grothendieck universes is useful and also used in so many contexts nowadays, that it is also not a big deal. And I will explain why your proof eventually does not need Grothendieck universes anyway.

Let me recall how Zorn's Lemma follows from the axiom of choice. Then we use this to construct the algebraic closure with your idea.

Let $(P,\leq)$ be a partially ordered set in which every chain has an upper bound, which however has no maximal element. By the axiom of choice, we have a function that chooses an upper bound for every chain, and for every element another element strictly above it. We then define a strictly increasing function $f : \mathbf{On} \to P$ by transfinite recursion as follows. If $\alpha$ is an ordinal number and all $f(\beta)$ for $\beta < \alpha$ have been defined, then let $p \in P$ be the chosen upper bound of the chain $\{f(\beta) : \beta < \alpha\}$, and finally let $f(\alpha) \in P$ be the chosen element $p < f(\alpha)$. Then $f(\beta) < f(\alpha)$ for all $\beta < \alpha$. So $f$ is an injective function from the class of ordinals to a set, which cannot exist, say by Hartog's Lemma. $\checkmark$

Informally, the construction constructs the maximal element of $P$ as follows. Start with any element of $P$. If it is maximal, done. If not, take an element which is larger. Take an element which is larger, etc. If you have infinitely many elements, which are strictly increasing already, take an upper bound. If that is maximal, done. Otherwise, increase that one even more. Then you have an element which is strictly larger than all the previous elements. Continue this ordinal many times. At some stage this process has to stop, though. We will find a maximal element.

In our case, $P$ is the set of algebraic extensions of $K$ in the universe $\mathcal{U}$. Notice that $P \notin \mathcal{U}$, but we still may work in the collection of all sets (or sets in $\mathcal{U}^+$ for an even larger universe $\mathcal{U}^+$, if you like). For algebraic extensions $L,L'$ of $K$ we write $L \leq L'$ when $L$ is a subfield of $L'$ in the usual sense. The empty chain has an upper bound, the extension $K$ of $K$. If $C \subseteq P$ is a non-empty chain, then one checks that $\bigcup C$ is an upper bound of $C$ in $P$.

Therefore, by Zorn's Lemma, there is a maximal element of $(P,\leq)$. It is an algebraic extension $L$ of $K$ in $\mathcal{U}$ such that for every other algebraic extension $L'$ in $\mathcal{U}$ with $L \subseteq L'$ we have $L = L'$. But then, $L$ must be an algebraic closure of $K$. This means that $L = \overline{K}$ is algebraic over $K$ and that every monic polynomial over $K$ splits into linear factors over $L$. This more elementary property makes it clear that $\overline{K}$ is also maximal with respect to all algebraic extensions of $K$, regardless whether they lie in $\mathcal{U}$ or not.

By the preceding discussion, we may construct $\overline{K}$ as follows. If $K$ is algebraically closed, done. If not, take a proper algebraic extension of $K$. Take a proper algebraic extension of that. Continue. If you have infinitely many strictly increasing algebraic extensions of $K$, take their union, and a strict algebraic extension of that (if it doesn't exist, we would be done). Continue ordinal many times. At some point we have found a maximal element, an algebraic closure.

The point is, however, that we don't need to repeat this process so many times. Here is a more refined construction of $\overline{K}$.

For every monic $f \in K[T]$, recall that $K[T]/\langle f \rangle$ is a $K$-algebra in which $f$ has a root. It follows immediately that the tensor product of $K$-algebras $\bigotimes_{f \in K[T] \text{ monic}} K[T] / \langle f \rangle$ has the property that every $f$ has a root inside it. By linear algebra, it is non-zero (in fact, an explicit $K$-basis is known). Hence, it has a maximal ideal (axiom of choice). Then, the quotient by that maximal ideal is an algebraic field extension $K'$ of $K$ such that every monic polynomial over $K$ has a root in $K'$. Now the union of $K \subseteq K' \subseteq K'' \subseteq \cdots$ is an algebraic closure of $K$. (Actually, but that it is non-trivial, we have $\overline{K} = K'$ already.)

This means that we just need to run our transfinite construction once for every polynomial, and then repeat that countably many times. We don't need all ordinal numbers. We know exactly when the process ends. And in particular, we see that the choice of the Grothendieck universe does not make any difference, and that we don't need it (just the axiom of choice).

Answer 2

I don't know about the Grothendieck universe, but a standard way to resolve such set-theoretical issues is as follows.* We fix a set $X$ with $|X| > |K| + \aleph_0$ and an injective map $i : K \to X$. Now we take $\mathcal{P}$ as the family of all fields $(L, +, \cdot, 0, 1)$ such that

• $i[K] \subseteq L \subseteq X$,
• $i : K \to L$ is a field embedding,
• the extension $i[K] \subseteq L$ is algebraic.

Such $\mathcal{P}$ is clearly a set. Then we order $\mathcal{P}$ by the relation of being a subfield.

In this setting we can use the Zorn's lemma and the usual proof goes through. There is maybe one bit that needs adjusting: when proving by contradiction that the maximal element $L$ of $\mathcal{P}$ is algebraically closed, we need to carefully move the structure from $L[X]/(w)$ to a set $M$ such that $L \subseteq M \subseteq X$ so that the constructed field is still in $\mathcal{P}$.

*Well ok, a standard way is to ignore these issues completely until pressed sufficiently hard. :P