# Why are these curves not defined over a smaller field

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.

-
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 Knaf 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

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...)

-
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.

-