Why are these curves not defined over a smaller field

Question

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.

Answer 1

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...)

Answer 2

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.