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Consider the extension Q(a,b) of the field of rationals, where a,b are algebraically independent transcendentals. To Q(a,b) adjoin the roots of the polynomials x^5+a^5=1 and y^5+b^5=1. The resulting field Q(a,b)[x,y] is a radical extension of Q(a,b).

Is it true that the only solutions to the equation X^5+Y^5=1 in the field Q(a,b)[x,y] are {0,1},{a,x}, {b,y}, {1/a,-x/a) and (1/b, -y/b)?

Comment: See FC's answer to my previous question.

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Can I recommend that you rewrite your title so that it asks a question? –  Theo Johnson-Freyd Nov 8 '09 at 1:07
Why don't you mimic what FC told you about the 1-dimensional case and see how far it gets and then tell us? –  Kevin Buzzard Nov 8 '09 at 7:44
to FC: i marked your answer as correct. do you think mimicking your answer (to the first question) might work to answer the second question? Let me know what you think...you can email me at bmk@math.cornell.edu, and I will tell you why I am interested in these questions and perhaps this might lead to an interesting work. –  Bakh Nov 9 '09 at 17:11

4 Answers 4

The complete set of solutions consists of $$(1,0), (0,1),$$ $$(a,x), (x,a), (1/a,-x/a), (-x/a,1/a), (1/x,-a/x), (-a/x,1/x),$$ $$(b,y), (y,b), (1/b,-y/b), (-y/b,1/b), (1/y,-b/y), (-b/y,1/y).$$

Let $C$ be the affine curve $X^5+Y^5=1$ over $\mathbf{Q}$. Because your field $K:=\mathbf{Q}(a,b)[x,y]$ is the function field of the $\mathbf{Q}$-variety $C \times C$, the set $C(K)$ is in bijection with the set of rational maps $C \times C \to C$. So the only fact needed beyond what was in FC's answer to your earlier question is the geometric fact that every non-constant rational map $C \times C \to C$ is a composition consisting of the first or second projection $C \times C \to C$ followed by a birational automorphism of $C$. More generally,

If $X,Y,Z$ are curves over a field $k$, and the genus of $Z$ is at least $2$, then every rational map $X \times Y \to Z$ factors through the first or second projection.

Proof: Extend the rational map to a rational map $X \times Y \to Z \times Y$ by using the projection $X \times Y \to Y$ as the second coordinate map. Restrict this to the fibers above the generic point of $Y$ to get a rational map $X_L \to Z_L$, where $L$ is the function field of $Y$. If this is constant, i.e., factors through the structure map to $\operatorname{Spec L}$, then the original map factors through the projection to $Y$. On the other hand, the genus hypothesis implies that, up to powers of Frobenius in characteristic $p$, there are only finitely many non-constant rational maps from $X$ to $Z$ over any field, and hence finitely many of bounded degree in any characteristic, so there are no algebraic families of such maps, so the rational map $X_L \to Z_L$ must be the base extension of a rational map $X \to Z$, which means that the original rational map factors through the projection to $X$. $\square$

@Bakh: Both FC and I used the general observation that if $X$ and $Y$ are curves over a field $k$, and $K$ is the function field of $Y$, then $X(K)$ is in bijection with the set of rational maps from $Y$ to $X$. If you have not studied enough algebraic geometry yet to understand this, the following example may be helpful:

Consider the case where $X$ is the plane curve $f(x,y)=0$ and $Y$ is the affine line $\mathbf{A}^1$, so $K=k(t)$. To give a point in $X(K)$ is to give rational functions $r(t)$ and $s(t)$ such that $f(r(t),s(t))=0$ identically. If we substitute a particular element of $k$ for $t$, then $(r(t),s(t))$ specializes to a point on $X(k)$, and hence we get a map $\mathbf{A}^1(k) \to X(k)$ except that we must avoid the finitely many poles of $r(t)$ and $s(t)$. The same holds for points over any field extension of $k$, and in fact we get a rational map $\mathbf{A}^1 \to X$.

More scheme-theoretically, one can say that an element of $X(K)$ is a $k$-morphism from the generic point of $Y$ to $X$, and such a morphism spreads out to a morphism from some Zariski dense open subscheme of $Y$ to $X$, i.e., to a rational map from $Y$ to $X$.

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Given that Bakh asked this question last Nov 7, and last logged in Nov 15, there's a chance he/she will never see this answer Bjorn. If one of your motivations for doing the exercise I suggested to Bakh was to make their life easier, then you might want to consider emailing them alerting them to the fact you've done it! –  Kevin Buzzard Mar 25 '10 at 8:48
OK, thank you, Kevin; I'll do that. –  Bjorn Poonen Mar 25 '10 at 9:25

You will have more points than just the ones listed. To start, notice that $x$ and $y$ are both invertible. This is because $1 - a^5$ is invertible, and this equals $x^5$. This lets you write $1/x$ as $x^4/(1 - a^5)$. For this reason, you'll at least have extra solutions like $$(1/x, -a/x).$$

As to whether this amended list is complete, I do not know. I would believe that it is, but cannot say at the moment.

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No. If you adjoin all the roots of $x^5=1-a^5$ to $\mathbb{Q}(a, b)$, then you obtain a field that contains a primitive 5th root of unity. This will lead to more solutions of $X^5+Y^5=1$ then you have listed here.

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I think the intention was to adjoin exactly one root of each of the two polynomials. –  S. Carnahan Nov 8 '09 at 8:16

Here are my comments to some of the suggestions and answers:

(1) I dont quite know some of the facts (theorems, formulas, definitions) mentioned in FC's answer in order to mimic his answer in this case. Those facts might assume certain things not satisfied by the fields I am looking at.

(2) i am adjoining exactly one root of each of the polynomials.

(3) My guess is one might use the fact that a and b are algebraically independent (over Q) and that x and y are algebraic over a and b, respectively. Hence, there might be a reasoning that states that any polynomial equation p=0 that uses both something from {x,a} and something from {y,b} should be a trivial one. Therefore, the case might be reduced to the case of the first question (answered by FC). Cant make this work due to the lack of knowledge in the area at this stage.

Cheers, -Bakh.

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