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Let $\overline{\mathbb Q}$ be the algebraic closure of $\mathbb Q$. One usually viewes $\overline{\mathbb Q}$ as a subfield of $\mathbb C$. But $\mathbb R$, and hence $\mathbb C$, is defined analytically (e.g., as the completion of $\mathbb Q$). Is there a simple, purely algebraic way to construct $\overline{\mathbb Q}$, and prove that it is algebraically closed? Of course one can just follow the proof that all fields have an algebraic closure, but that seems quite complicated. Could the construction of $\overline{\mathbb Q}$ from $\mathbb Q$ be as easy as the construction of $\mathbb C$ from $\mathbb R$?

zxcv
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    It depends on how explicit you want this "construction" to be. Adjoining all roots of rational polynomials would run in the problem that in general you cannot write them "explicit" if the degree is $>4$, so this naive definition of explicit will not do it. – Matthias Sep 12 '22 at 11:50
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    The construction of $;\Bbb C;$ out of $;\Bbb R;$ is so simple among other things because $;\Bbb C/\Bbb R;$ is an extension of degree $;2;$...the lowest degree any non-trivial extension can have. Not the case for $;\overline{\Bbb Q}/\Bbb Q;$ – DonAntonio Sep 12 '22 at 11:54
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    I agree that before the 20th century it was common to view the algebraic closure of $\mathbf Q$ as a subfield of $\mathbf C$, and it is still standard to do that when you are first learning about fields, but in (algebraic) number theory today it is better to regard $\overline{\mathbf Q}$ as an abstract field with no a priori embedding into $\mathbf C$ because it can equally well be embedded in an algebraic closure of $\mathbf Q_p$ or in $\mathbf C_p$ for primes $p$ and none of these different embeddings should be considered as more important than the others. – KCd Sep 12 '22 at 11:59
  • There is always Artin's construction, which is purely algebraic and works for any field. See for instance https://wordpress.nmsu.edu/pamorand/files/2018/10/algebraicclosure.pdf – lhf Sep 12 '22 at 13:55

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I'm not aware of a "simple" or "easy" way to do this. It's worth knowing that, in general, both the existence and uniqueness of algebraic closures are independent of ZF and require some use of choice. This MO discussion is quite nice and shows that the ultrafilter lemma suffices, but it's not known whether the two are equivalent. There is also a discussion at this math.SE question.

To be clear, proving that the algebraic closure of $\mathbb{Q}$ exists can be done in ZF without any choice; you can just explicitly describe an order in which to adjoin the roots of all irreducible polynomials. However, it's independent of ZF whether the algebraic closure is unique, and it's consistent with ZF that fields exist which have no algebraic closure.

This is all to give some indication that some "real work" is required to prove that algebraic closures exist. For $\mathbb{Q}$ I don't know a way to bypass this work without constructing $\mathbb{C}$, which as you say you want to avoid.

The nicest proof I know that algebraic closures exist in general is explained in this note by Keith Conrad, who attributes it to Brian Conrad and indirectly to Zorn. It's discussed in this $n$-category cafe post by Tom Leinster. The idea is that for a field $K$ we can (canonically, with no choice) construct a "splitting ring" $R$ (not field) of all polynomials over $K$, by writing down generators and relations corresponding to formally adjoining all roots of all (nonzero) polynomials. Then we need either the axiom of choice to find a maximal ideal $M$ of $R$, which gives the algebraic closure as $R/M$, or the ultrafilter lemma to find a prime ideal $P$ of $R$, which gives the algebraic closure as $\text{Frac}(R/P)$.

This argument by Joel David Hamkins proving both existence and uniqueness from the compactness theorem for first-order logic (equivalent to the ultrafilter lemma) is also quite nice and conceptual but requires some familiarity with first-order logic.

Qiaochu Yuan
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  • I don't understand your answer, there is of course an algorithm generating a tower of fields $\Bbb{Q}=F_0,F_n\subset F_{n+1}$ such that $\overline{\Bbb{Q}} = \bigcup_{n\ge 1} F_n$. – reuns Sep 12 '22 at 19:36
  • @reuns: that doesn't contradict anything I've said. The OP asks for a construction "as easy as the construction of $\mathbb{C}$ from $\mathbb{R}$" and refers to the general construction of an algebraic closure as "quite complicated"; this is what I'm responding to. – Qiaochu Yuan Sep 12 '22 at 19:47
  • You wrote "I don't know a way without first constructing $\Bbb{C}$" – reuns Sep 12 '22 at 19:52
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    I said "I don't know a way to bypass this work without first constructing $\mathbb{C}$"; that is, the OP says they want to avoid the general proof that fields have algebraic closures, and I don't know a way to avoid basically running through this proof one way or another, except by constructing $\mathbb{C}$. – Qiaochu Yuan Sep 12 '22 at 19:55