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One can define fields using a finite list of axioms that quantify over the field itself. However, the obvious way to define algebraically closed fields involves either an infinite list of axioms, or a more logically complex framework where one can quantify over natural numbers. It would be interesting if one could avoid this, and give a finite list of axioms that quantify only over the field, but my guess is that that is impossible. Are there known theorems in that direction?

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    $\begingroup$ The theory of algebraically closed fields is not finitely axiomatizable: books.google.com/… (page 109 and preceding). $\endgroup$ – Todd Trimble Oct 18 '15 at 14:49
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From Dirk van Dalen's Logic and Structure: the theory of algebraically closed fields is not finitely axiomatizable (see page 109 and preceding).

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Certainly Todd answered the question, but the following "theorem" shows that this sort of problem is not as hard as it originally appears.

Suppose $\Sigma$ is an infinite set of statements such that no finite subset of $\Sigma$ implies all of $\Sigma$. Then the theory $T$ which is the set of consequences of $\Sigma$ is not finitely axiomatizable.

The proof of this is immediate; if there were a finite set of axioms, they would be implied by some finite subset $\Sigma_0$ of $\Sigma$, so by transitivity, $\Sigma_0$ would imply all of $T$, and thus all of $\Sigma$.

Now to think about algebraically closed fields, see that the first set of axioms you come up with satisfies this. We simply say that $F$ is a field, and for every $n\geq 2$, every monic polynomial of degree $n$ has a solution (this is a single sentence for each $n$; you quantify across the coefficients). This works, and no finite subset works (to verify this one does need to show that you can have fields which have no extensions of degree $\leq k$, but has some irreducible polynomial of degree $>k$, which should be elementary).

A lot of other questions along these lines can be answered similarly.

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  • $\begingroup$ Right; your second paragraph is essentially Lemma 4.2.9 in the text I linked to, which was the main observation needed in the proof. $\endgroup$ – Todd Trimble Oct 22 '15 at 2:33
  • $\begingroup$ Ah! To be honest I did not click your link; I just assumed it would be hard, for some reason, since it was a link and not just a paragraph or two. Sorry for the redundancy. $\endgroup$ – Richard Rast Oct 22 '15 at 10:53
  • $\begingroup$ It's no problem; I probably should have written out the underlying argument, which you effectively did, and for which I awarded you an upvote. $\endgroup$ – Todd Trimble Oct 22 '15 at 11:27

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