Every field has an algebraic closure

Yeah. That’s right. Every fieldYesterday we showed how we could add a root of a polynomial to a field and that would make it bigger. So we can just keep adding roots over and over again until we get them all, right? Well, it’s not that simple. See, it’s conceivable that when I add a new root, now I have a lot of new polynomials whose roots I need to add. And if I add all of them, I could get even more polynomials. Will it ever end?

Luckily, we have a handy tool. The careful reader will have noticed that “Every field has an algebraic closure” kind of looks like “Every vector space has a basis.” The choice of titles was not a coincidence. The proofs both rely on Zorn’s lemma. Once again, this is here for the sake of completeness. If you’re comfortable taking this on faith, I won’t hold it against you. Without further ado…

Proof:

Take all the fields that contain a given field k, and order them with E\ge F if E is a field extension of F. Some things might not be comparable. That’s fine. Then neither will be bigger than the other (this is called a partially ordered set, or poset for short). Last time we showed we can always add on a root to a polynomial (or at least waved our hands in that direction). So if we can produce a maximal field, that means that it must already have all the roots. This is exactly what an algebraically closed field is.

Zorn’s lemma gives us a maximal element under certain conditions, so we’ll just have to satisfy those conditions. The conditions are that “every chain has an upper bound.” What’s a chain? It’s what we call a poset that is totally ordered (I voted for toset, but no one else seemed to like that). If you draw a picture, you’ll see why. Everything lies in one line.

So suppose we have a bunch of fields in a chain. That means that if I pick any two fields, one must contain the other. Let’s call them F_\alpha, for \alpha in some indexing set \mathcal A. It’s important to note that if \mathcal A is finite, this not hard. One of the F_\alphas is a maximum. But if it’s infinite, or really infinite, things aren’t so clear. In any event, like many Zorn’s lemma arguments, unions come to save the day.

\displaystyle\bigcup_{\alpha\in\mathcal A}F_\alpha is an upper bound for every F_\alpha.

Well, duh. The big union contains everything. That was the point. We do however need to check that it’s a field. (Left as an exercise to the reader.) The trick is to take notice that we’re only allowed finite sums/products/etc., and we already know how to do those, since they always come from one of the F_\alphas.

Okay, so any chain has an upper bound. We invoke Zorn’s lemma and we have a maximal element. Like I said, maximal means that we can’t extend it any more (so it must have the root of every polynomial).

\square

One can also show that algebraic closures are unique, though I don’t care to do so. We can actually talk about the algebraic closure of a field k, instead of just talking about an algebraic closure. We’ll write \overline k for the algebraic closure of k. For example, \mathbb C=\overline{\mathbb R}

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One Response to Every field has an algebraic closure

  1. Pingback: Global choice and algebraic closures « Andy Soffer

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