An interesting mathoverflow question was one due to Philipp Lampe that asked whether a non-surjective polynomial function on an infinite field can miss only finitely many values. In my interpretation of the question, if $k$ is a starting field and $f$ is a polynomial, you could ask what happens if you repeatedly adjoin a root of $f(x)-a$, except for a finite set of values $a \in S \subset k$ for which you hope a root never appears. You have to adjoin a root for all $a \in \tilde{k} \setminus S$, where $\tilde{k}$ is the growing field. Either a root of $f(x)-a$ for some $a \in S$ will eventually appear by accident, or $f$ as a polynomial over the limiting field $\tilde{k}$ is an example.

(Edit: I call this an interpretation rather than a construction, because in generality it is equivalent to Philipp's original question. I also don't mean to claim credit for the idea; it was already under discussion when I posted my answer then. Maybe an answer to the question below was already implied in the previous discussion, but if so, I didn't follow it.)

For some choices of $f$ and a non-value $a$, you can know that you are sunk at the first stage. For instance, suppose that $f(x) = x^n$. When you adjoin a root of $x^n-a$, you also adjoin a root of $x^n-b^na$ for every $b \in k$. You cannot miss $a$ without also missing every $b^na$, which is then infinitely many values when $k$ is infinite.

So let $k$ be an infinite field, and let $f \in k[x]$ be a polynomial. Define an equivalence relation on those elements $a \in k$ such that $f(x)-a$ is irreducible. The relation is that $a \sim b$ if adjoining one root of $f(x)-a$ and $f(x)-b$ yield isomorphic field extensions of $k$. Is any such equivalence class finite? What if $k$ is $\mathbb{Q}$ or a number field?

In my partial answer to the original MO question, I calculated that if $f$ is cubic and the characteristic of $k$ is not 2 or 3, then the equivalence classes are all infinite.

  • $\begingroup$ The cubic case is where I got stuck too. Another related problem is that you can add the root not immediately but after several extensions (i.e., taking b to be in the extension generated by a root of the polynomial with the free term that is in the extension generated by a root of, etc.). It can happen that this iterative procedure will add the roots that cannot be added in a single extension step but I don't know how often that may happen... $\endgroup$ – fedja Dec 27 '09 at 16:34

The answer is that over a number field $k$, an equivalence class can be finite, and in fact it is usually so for $f$ of moderately large degree. Consider $f(x):=x^7+x$ over $\mathbf{Q}$, for example. If the equations $f(x)=1$ and $f(x)=t$ for some other $t \in \mathbf{Q}$ yield the same degree $7$ extension, then in particular the discriminant $D(t)$ of the polynomial $f(x)-t$ in $x$ must equal $D(1)$ times a square. In fact, $D(t) = -823543 t^6 - 46656$, so the necessary condition is $-823543 t^6 - 46656 = -870199 u^2$. This defines a genus $2$ curve, so Faltings' theorem implies that this equation has only finitely many rational solutions.

Moreover, for a typical $f$ of slightly higher degree, it is reasonable to expect that all the equivalence classes are singletons, although proving such a statement would seem to require understanding the rational points on surfaces of general type, which is probably beyond the current state of knowledge.

Reasoning along these lines suggests to me that the iterative procedure you and others have proposed should succeed in constructing a field with a nonsurjective polynomial as in Philipp Lampe's question, even if we can't prove this.

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    $\begingroup$ Excellent! Maybe you should post something to the other question too. $\endgroup$ – Greg Kuperberg Dec 27 '09 at 16:38
  • $\begingroup$ Also: If you can get this far with number fields, can you get further with function fields over a finite field? You could even make use of higher transcendence degree. $\endgroup$ – Greg Kuperberg Dec 27 '09 at 17:36

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