When f(x)-a and f(x)-b yield the same field extension - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T06:08:27Z http://mathoverflow.net/feeds/question/9863 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/9863/when-fx-a-and-fx-b-yield-the-same-field-extension When f(x)-a and f(x)-b yield the same field extension Greg Kuperberg 2009-12-27T06:24:50Z 2009-12-27T16:32:29Z <p>An interesting mathoverflow question was one due to Philipp Lampe that asked whether <a href="http://mathoverflow.net/questions/6820/" rel="nofollow">a non-surjective polynomial function on an infinite field can miss only finitely many values</a>. 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.</p> <p>(<strong>Edit:</strong> 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.)</p> <p>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.</p> <p>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?</p> <p>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.</p> http://mathoverflow.net/questions/9863/when-fx-a-and-fx-b-yield-the-same-field-extension/9889#9889 Answer by Bjorn Poonen for When f(x)-a and f(x)-b yield the same field extension Bjorn Poonen 2009-12-27T16:11:52Z 2009-12-27T16:11:52Z <p>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.</p> <p>Moreover, for a typical $f$ of slightly higher degree, it is reasonable to expect that all the equivalence classes are singletons, although <em>proving</em> 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. </p> <p>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.</p>