In abstract algebra lecture, the lecturer wants to construct an algebraic closure of $\mathbb{Q}$. The construction is as follow:

Suppose $\mathbb{Q_1}=\mathbb{Q}$. Let $\mathbb{Q}_2$ be the set which contains $\mathbb{Q}_1$ and all roots of polynomials which have degree less than or equal to $1$. Some for $\mathbb{Q}_3$ which contains $\mathbb{Q}_2$ and all roots of polynomial which have degree less than or equal to $2$. In general , $\mathbb{Q}_n$ is the set which contains $\mathbb{Q}_{n-1}$ and all roots of polynomials which have degree less than oe equal to $n-1$. Take $E=\cup_{n \geq 1} \;\mathbb{Q}_n$

Claim: $E$ is an algebraic extension of $\mathbb{Q}$

Let $\alpha \in E$. Then $\alpha \in \mathbb{Q}_n$ for some $n$. By the contruction, we know that $\mathbb{Q}_n$ is an algebraic extension of $\mathbb{Q}_{n-1}$ and $\mathbb{Q}_{n-1}$ is an algebraic extension of $\mathbb{Q}_{n-2}$. By using transitivity of algebraic extension, we have $\mathbb{Q}_n$ is an algebraic extension of $\mathbb{Q}$. Hence, $\alpha$ is algebraic over $\mathbb{Q}$.

Claim: $E$ is algebraically closed

Let $f \in E[X]$. Then $f(X)=a_0+a_1X+...+a_nX^n$ where $a_i \in E$ for all $1 \leq i \leq n$. Then $a_i \in \mathbb{Q}_m$ for some $m$. Choose $t$ large enough such that all $a_i \in \mathbb{Q}_t$. Then by the construction , $f$ has root in $\mathbb{Q}_{t+1} \subset E$

Hence, $E$ is an algebraic closure of $\mathbb{Q}$.

Can anyone help me to check whether I got the proof correct or not. Because I not sure whether I leave out some details.

  • $\begingroup$ When you say "$f$ has root in $\Bbb Q_{t+1}\subset E$", this is not quite true, since you haven't taken the degree of the polynomial into account. The correct place to look for the roots would be in $\Bbb Q_{\max(n+1, t+1)}$. Otherwise this looks fine to me. $\endgroup$ – Arthur Oct 18 '13 at 11:44
  • $\begingroup$ From this construction, do we know that $E$ is a field? Because I sort of can't see why #E# is a field. $\endgroup$ – Idonknow Oct 18 '13 at 12:00
  • $\begingroup$ You might have to take for $\mathbb{Q}_n$ the field generated by the set $\mathbb{Q}_n$ for having $E$ as the union of fields (equals the process of adjoining all the roots of polynomials of increasing degree) $\endgroup$ – BIS HD Oct 18 '13 at 12:25
  • $\begingroup$ @BISHD can you elaborate more? $\endgroup$ – Idonknow Oct 18 '13 at 13:05
  • $\begingroup$ @Idonknow Sure! Since you construct a increasing chain of sets, it is not clear, wheather the union is a field. But if you construct an increasing chain of fields, the (even infinite) union is a field. So instead of taking the set $\mathbb{Q_2}$ of all roots of polynomials of degree $1$ and the $\mathbb{Q}$ you have to take the field generated by this set, say $L_2$. This is the smallest field that contains the set $\mathbb{Q_2}$. (this construction can for example be found in this thread. So now you get an... $\endgroup$ – BIS HD Oct 18 '13 at 13:32

The proof is incorrect [for a general field. Remarks added at the end, about special cases where it can work.]. A minor problem is that the description says to add all the roots of polynomials of some degree in one stage but the way this argument is usually written out is to add one root per (irreducible) polynomial at each step. This is not the real difficulty, since at each step numbered $\geq k$ every irreducible polynomial of degree $\leq k$ will acquire at least one new root, and it will be split after a number of steps bounded by twice its degree (plus the number of the step at which it came into existence, if not originally in the field). And one can canonically add at least one root for every polynomial in any given set.

The big problem is that the universal construction, adjunction of roots as "free variables" modulo known relations, will add infinitely many roots of every polynomial. If you formally endow every translated copy of an irreducible polynomial, $f(X+n)$, with at least one new root $X_n$, then these new roots are linearly independent in the enlarged ring, but $f(X_n+n)=0$ gives $f$ too many roots. Infinitely many of these $X_n+n$ will be equal in an algebraically closed field, where the number of roots of a polynomial is limited by the degree, but some extremely non-canonical choice has to be made of which formal roots correspond to which roots in the AC field (algebraically closed, or Axiom of Choice, for general fields it comes to the same thing), keeping things consistent for all polynomials at once, so as to not get unintended new relations like $7 = 0$, or contradictions.

So you get a ring, not a field, at every stage of the construction. The magical part of any formally correct proof using the Axiom of Choice, which is the place in the proof that contains all the work (given that Galois theory exists, a noncanonical choice of extension is the entire difficulty), is in the statement that a maximal ideal exists in this ring of formal solutions to polynomials.

This is where the zoo of ambiguously related roots is sorted out consistently. One gets a field extension by taking a quotient, and the process that includes this reduction at every stage is repeated $\omega$ long to crack more polynomials (reducing the degree of previously irreducible ones), finally splitting all of the old and new ones.

Or, Zorn's lemma is used to assert that a transfinite chain of one-polynomial root additions, which stay within the world of fields, does not run out of steam (solving only a subset of the polynomials), nor require a transfinite number of steps so long that it escapes the world of sets before finishing.

Not a very satisfying situation. For most purposes, finite or explicit profinite extensions solving known sets of polynomials, are enough.

For a field like $\mathbb{Q}$ that sits inside a much larger but known field such as $\mathbb{C}$ one can take the $\mathbb{Q_i}$ to be its field extensions by sets of roots that live in the larger field. Then having a field at each stage is not a problem, but the proof relies on the very special situation of somehow knowing that the larger field is algebraically closed itself, or big enough relative to the small field for this to be possible. However, it would be unusual to use the Artin type of proof to build algebraic numbers internal to the larger field, because closure has a simpler description as the set of all roots of polynomials with coefficients in the small field (so the whole construction can be done in one large step instead).

Another case with a canonical algebraic closure is the power series field in one variable, $\mathbb{C} ((X))$, where adding fractional powers of $X$ is enough. In both of these cases with explicit closures, the extra power to solve equations, without a transfinite process of adjoining roots, comes from analysis. The fields are complete in a topology, and roots can be found as limits of approximations.

  • $\begingroup$ So the construction of algebraic closure of rational numbers given by artin is correct? $\endgroup$ – Idonknow Oct 31 '13 at 3:31
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    $\begingroup$ Artin's proof that an arbitrary field has an algebraic closure is an example of a "formally correct proof using the Axiom of Choice" to hide Galois theoretic difficulties. The version of this proof for $Q$ stated in the question is incorrect (as it does not use AC or maximal ideals) if you don't assume the roots are complex numbers or in some other pre-existing field; correct if roots are taken in $C$; but unnecessary in the latter case because you could have done it more simply as "numbers in $C$ satisfying a polynomial equation over $Q$" without the infinitely layered construction. $\endgroup$ – zyx Oct 31 '13 at 5:01

The basic idea is correct, modulo the small fixes which user BIS HD has suggested in the comments, and the fact that one cannot add all of the roots of one polynomial at once (see the discussion in the comments). Perhaps the best way to proceed down this road is the method of Emil Artin, which is very elegant, and essentially what you have in mind (but it makes more explicit some of the steps that you left out).

Some remarks: this method is quite unsatisfying, in a way. While it works for an arbitrary field $F$, one is not so sure what the final product looks like. For instance, if you had started with the real numbers $\mathbf R$, the process would have stabilized at the first step without you ever knowing about it. The final result could be very simple, or very complicated. This seems to reflects the fact that an arbitrary field does not have a "canonical" algebraic closure.

A surprising thing about the algebraic closure of $\mathbf Q$ is that we can essentially obtain it in a single step, by using the fact that $\mathbf Q \subseteq \mathbf R$. The algebraic closure of $\mathbf R$ is just $\mathbf C$. So we can just take the algebraic closure of $\mathbf Q$ in $\mathbf C$, that is, the set $L$ of all $\alpha \in \mathbf C$ which are algebraic over $\mathbf Q$. Since $\mathbf C$ is algebraically closed, it follows that $L$ is an algebraic closure of $\mathbf Q$.

If you asked a friend across the planet to pick an algebraic closure of $\mathbf Q$, there is a good chance that they would pick the same field $L$ without even thinking twice about it. This is surprising, if you ponder on the fact that Artin's construction, on the other hand, requires making an infinite number of choices, and taking the limit of an essentially infinite process!

  • $\begingroup$ In the paper provided, it does not involve degree of polynomial. So I can also remove parts which involve degree of polynomial? $\endgroup$ – Idonknow Oct 27 '13 at 12:34
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    $\begingroup$ Dear @Idonknow Well, two things: first, you cannot add all the roots of a polynomial at once. You can only add one root at a time. Otherwise, why couldn't you just say "Take the field which contains all the roots of all polynomials", and be done with it? Second, that is correct, you don't need the restriction on the degree. The crucial thing is really that for each polynomial, you can only add one root of it at a time. This is why the process needs many steps. Does that make sense? $\endgroup$ – Bruno Joyal Oct 27 '13 at 14:40
  • $\begingroup$ Okay. I agree that we cannot add all roots of a polynomial at once. But here I consider adding all roots of polynomial of degree less than or equal to $n$. Wouldn't it make a difference? $\endgroup$ – Idonknow Oct 29 '13 at 19:34
  • $\begingroup$ @Idonknow I don't think so, unfortunately. We cannot even construct the splitting field of a single polynomial in one step. We really can only add one root at a time, until we have them all. As far as I can see, "adding all of the roots" doesn't even make sense, in the sense that I wouldn't know of any procedure which amounts to this. Where do these roots live before you add them? You can add a root of $f$ by considering the extension $Q[T]/f(T)$, but there is no such direct construction of the splitting field. $\endgroup$ – Bruno Joyal Oct 29 '13 at 19:58
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    $\begingroup$ Dear Bruno - On top of this being a great answer, what a cool avatar. My avatar wants to meet yours. Best regards, $\endgroup$ – user12802 Mar 18 '15 at 16:29

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