Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $K$ be a number field and let $\pi$ be an element in $K$. Assume that $\pi$ is not contained in a subfield of $K$.

Consider the curve $y^2 = x^{2g+1}+\pi$. This defines (after homogenization and normalization) a hyperelliptic curve of genus $g$ over $K$.

Why is this curve not defined over a smaller field $k\subset K$?

That is, why isn't there some transformation of the equation $y^2 = x^{2g+1}+\pi$ such that the coefficients lie in a smaller field?

I might be wrong about this though. That is, maybe these curves ARE defined over a smaller field. I would just like to know how to prove the correct statement rigorously.

share|improve this question
    
Your new hypothesis is still not strong enough to prevent the curve from being defined over a smaller field. For instance, if $\pi$ is of the form $x^{2g+1} \pi'$ with $\pi' \in k$, then a change of variables gives a defining equation with coefficients in $k$. I might recommend the hypothesis: suppose that there is a prime $\mathfrak{p}$ of $k$ such that $\mathfrak{p} Z_K = (\pi)^{[K:k]}$. Here the curve is not obviously defined over $k$... –  Pete L. Clark Jul 30 '12 at 20:08
    
Pete, I don't see this change of variables. Its early in the morning ... maybe this is the reason, –  Hagen Jul 31 '12 at 7:38
    
@Hagen: I don't think I phrased my comment especially well. Let me try again: as $\pi$ varies over $K$, the family of curves $y^2 = x^{2g+1} + \pi$ are all twists of the same hyperelliptic curve $y^2 = x^{2+1} + 1$ over the algebraic closure. But the isomorphism class of the twist depends only on the class of $\pi$ in $K^{\times}/K^{\times 2g+1}$. So if $\pi$ is not in $k$ but there is some perfect $2g+1$ power $a^{2g+1}$ ($x$ was an unfortunate name for this, perhaps!) such that $\pi a^{2g+1} \in k$, then the curve can be defined over $k$. This can certainly happen. –  Pete L. Clark Aug 3 '12 at 20:29
    
@Hagen: By the way, since you didn't use an @ in your comment, I wasn't notified of it. –  Pete L. Clark Aug 3 '12 at 20:30

2 Answers 2

Let $p$ be a prime not dividing the discriminant of $K$. Then it would be acceptable to take $\pi = p$, and of course the curve $y^2 = x^{2g+1} + p$ is defined over $\mathbb{Q}$.

(So perhaps you should give a more precise set of hypotheses. There seems to be an interesting question lurking in here...)

share|improve this answer
    
I gave it some thought. See the edited question for a more precise set of hypotheses. –  Harry Jul 30 '12 at 14:05

Here is an attempt to approach the problem in some generality in the affine case.

Let $V$ be an affine variety over the field $K$, that is $V=\mathrm{Spec}(A)$ for some domain $A$, finitely generated as a $K$-algebra. Thus one has presentations of $A$ of the form

$0\rightarrow Q\rightarrow K[x_1,\ldots ,x_n]\rightarrow A\rightarrow 0$

where $Q$ is a (finitely generated) prime ideal.

We say that $V$ is defined over $k\subseteq K$ if $Q$ possesses a set of generators $f_1,\ldots f_m\in k[x_1,\ldots ,x_n]$ .

Suppose that $k\subseteq K$ is algebraic, then $k[x_1,\ldots ,x_n]\subseteq K[x_1,\ldots ,x_n]$ is an integral ring extension.

(*) Claim: the ideal $P$ of $k[x_1,\ldots ,x_n]$ generated by the $f_k$ is prime, and thus $Q\cap k[x_1,\ldots ,x_n]=P$ and $PK[x_1,\ldots ,x_n]=Q$ hold.

Proof: the extension $k[x_1,\ldots ,x_n]\subseteq K[x_1,\ldots ,x_n]$ is faithfully flat, hence $PK[x_1,\ldots ,x_n]\cap k[x_1,\ldots ,x_n]=P$. $\square$

Tensoring the presentation

$0\rightarrow P\rightarrow k[x_1,\ldots ,x_n]\rightarrow B\rightarrow 0$

with $K$ now yields $A=B\otimes_kK$.

Vice versa: if $A=B\otimes_kK$, then $Q$ posesses a set of generators in $k[x_1,\ldots ,x_n]$.

So we seem to arrive at the following reformulation of ''being defined over a subfield $k\subseteq K$'': the affine $K$-variety $V$ is defined over the subfield $k\subseteq K$, where the extension is assumed to be algebraic, if and only if $V=V_0\times_kK$ for some $k$-variety $V_0$.

This formulation is independent of a particular embedding of $V$ into affine space.

Suppose now that $K$ is algebraic over its prime field -- for example take $K$ to be a number field. Then there are minimal (with respect to inclusion) fields of definition for the variety $V$, but maybe several of them.

Let $C$ be an affine, plane algebraic curve over $K$ given by the polynomial $f\in K[x,y]$. If $C$ is defined over $k\subseteq K$, then $Q=fK[x,y]$ must have a set of generators in $k[x,y]$ and by (*) any set of generators of $P=fK[x,y]\cap k[x,y]$ will do the job. Since $Q$ has height $1$ and $k[x,y]\subseteq K[x,y]$ is integral, $P$ has height $1$ and is thus principal: $P=gk[x,y]$. Consequently $f=cg$ for some $c\in K$.

As a consequence we get: if $f$ is such that one of its coefficients generates $K$ over the prime field, and another coefficient lies in the prime field, then $C$ is not defined over any proper subfield of $K$. This fact can be applied to the affine part of the hyperelliptic curve considered in the original post.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.