# Conditions for a curve to be defined over a subfield

I have just finished reading Hartshorne, Chapter 1, Section 6 and have some questions about curves defined over a subfield of an algebraically closed field. For simplicity, let $k$ be a perfect field, $\overline{k}$ a fixed algebraic closure.

1) Let $E$ be a funtion field of transcendence degree 1 over k. Then $\overline{E}=E\cdot\overline{k}$ is a funtion field of transcendence degree 1 over $\overline{k}$, and hence (keeping the notation of Hartshorne) $C_{\overline{E}}$ is isomorphic to a non-singular projective curve, $X$ say. Is $X$ necessarily defined over $k$?

2) if $\varphi:X\rightarrow Y$ is a morhpism of curves and $X$ is defined over $k$, then is $Y$? Same question but with $\varphi$ a birational map.

Any help much appreciated (as always).

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I think it is better to make clear the meaning of being defined over $k$. When you have a curve $X$ over $\bar{k}$, you know it only up to $\bar{k}$-isomorphisms. So the only reasonable definition is that $X$ is isomorphic (over $\bar{k}$) to $Z_{\bar{k}}$ for some curve $Z$ defined over $k$.

(1) Yes. This is because $X=C_{\bar{E}}$ and $(C_E)_{\bar{k}}$ have the same function fields (equal to $\bar{k}E$) and are thus isomorphic over $\bar{k}$.

(2.1) No. Let $Y$ be any curve over $\bar{k}$ not defined over $k$. But it is always defined over a finite (hence separable) extension $L/k$. Enlarging $L$ if necessary, we can suppose $L/k$ is Galois of Galois group $G$. Fix an algebraic closure $\overline{L(Y)}$ of $L(Y)$. Let $F$ be the compositum in $\overline{L(Y)}$ of the conjugates $\sigma(L(Y))$, $\sigma\in G$. Then $F$ is stable by $G$. Let $E=F^G$. This is a function field over $k$ and we have $LE=F$ (Speiser lemma). So $X:=C_F=(C_E)_L$ is defined over $k$. The extension $L(Y)\to F$ induces a morphism $\varphi: X\to Y$ over $L$. But by hypothesis, $Y$ is not defined over $k$.

A concrete example: $k=\mathbb R$; $Y$ is an elliptic curve over $\mathbb C$ defined by an equation $$y^2=x^3+ax+b$$ with $j(Y)\notin\mathbb R$. The (non trivial) conjugate of $Y$ is defined by the equation $$z^2=x^3+\bar{a}x+\bar{b}$$ (here $\bar{a}$ stands for complex conjugation). The compositum is $\mathbb C(x,y,z)$ with the above relations, and $(\mathbb C(x,y,z))^G=\mathbb R(x, y+z, yz, i(y-z))=\mathbb R(x)[y+z]$.

Remark. In fact, we proved that any (smooth projective connected) curve over $\bar{k}$ is dominated by a curve defined over $k$.

(2.2) Yes, if $\varphi$ is birational, then $Y$ is isomorphic to $X$ over $\bar{k}$. Hence by definition (see the preamble), $Y$ is defined over $k$.

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Hi, thanks for your answer. I just have a couple of quick questions (a) To clarify, if $X$ and $Y$ are $\overline{k}$-isomorphic does that mean they are isomorphic as varieties or is it saying that their function fields are isomorphic as $\overline{k}$-algebras? (I'm guessing/hoping the latter as otherwise I don't understand your answer to (2.2)!) (b) With regards to (1) I am not familiar with $(C_E)_{\overline{k}}$. What is this curve please, and why is it defined over $k$? Cheers –  M Davolo Sep 9 '12 at 14:37
@MDavolo (a) Yes, this is true for smooth projective connected curves over any field. This should be in the section in Hartshorne you are reading. (b) $(C_E)_{\bar{k}}$ is the extension of $C_E$ to $\bar{k}$. It is also the projective smooth connected curve over $\bar{k}$ with function field equal to $\bar{k}E$. It is defined over $k$ by definition. –  user18119 Sep 9 '12 at 14:46
Dear QiL, I don't understand what $\overline{k}E$ means, since $\overline{k}$ and $E$ are not contained in a common field. However if $k$ is algebraically closed in $E$ then $\overline{k}\otimes_k E$ is a field (if I'm not mistaken) . Does the notion of "function field" include such an algebraically closed hypothesis? Or do we have to take quotients of the above tensor product? –  Georges Elencwajg Sep 9 '12 at 17:26
Dear @GeorgesElencwajg, I took the notation of OP. If we fix an algebraic closure $\Omega$ of $E$, then $\bar{k}E$ can be defined as the compositum of $E$ and the algebraic closure $\bar{k}$ of $k$ in $\Omega$. Of course this is highly non-canonical, but is harmless here because we only care about isomorphisms classes. As you said, when $k$ is algebraically closed in $E$, then $\bar{k}E$ is isomorphic to $\bar{k}\otimes_k E$ and is a quotient otherwise. In this question, $E$ is regular extension $k$ (so $k$ is algebraically closed in $E$). –  user18119 Sep 9 '12 at 18:40
Hi QiL, apologies but did you misread question (a)? I was basically asking what it means for two curves to be $\overline{k}$-isomorphic. Also, when you say $(C_E)_{\overline{k}}$ is the extension of $C_E$ to $\overline{k}$, again this is something I am unfamiliar with. I am guessing that $C_E$ stands for the discrete valuation rings of $E/k$ and that it can be realized as a curve in $\mathbb{P}^n(k)$ defined by polynomials over $k$ and that $(C_E)_{\overline{k}}$ denotes the same curve but in $\mathbb{P}^n(\overline{k})$. How far off the mark is this?! –  M Davolo Sep 9 '12 at 19:18