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).
 A: 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 $k$.
