Whenever $R_{1}\hookrightarrow R_{2}$ is an injection of rings, \begin{equation*} \operatorname{Hom}(R,R_{1})\hookrightarrow\operatorname{Hom}(R,R_{2}) \end{equation*} is an injection for any ring $R$. Thus by dualizing \begin{equation*} \operatorname{Hom}(\operatorname{Spec}(R_{1}),\operatorname{Spec}(R))\hookrightarrow\operatorname{Hom}(\operatorname{Spec}(R_{2}),\operatorname{Spec}(R)) \end{equation*} is an injection as well, and if $X$ is any scheme, by the usual gluing argument so is \begin{equation*} \operatorname{Hom}(\operatorname{Spec}(R_{1}),X)\hookrightarrow\operatorname{Hom}(\operatorname{Spec}(R_{2}),X)\text{.} \end{equation*} It follows from this that the moduli functor \begin{eqnarray*} \mathbf{F}_{1}: & \mathcal{Schemes} & \longrightarrow\mathcal{Sets}\\ & S & \longmapsto\text{$S$-isomorphism classes of genus-$1$ curves over $S$} \end{eqnarray*} is not representable: If it were, $\mathbf{F}_{1}(K)\longrightarrow\mathbf{F}_{1}(L)$ would be an injection for any field extension $K\hookrightarrow L$. But if $K$ is a field that is not closed under square roots, i.e. $\exists d\in K:z^{2}-d\in K[z]\text{ is irreducible}$, then for any $a,b\in K$ the elliptic curves $y^{2}=x^{3}+ax+b$ and $dy^{2}=x^{3}+ax+b$ are not isomorphic over $\operatorname{Spec}(K)$ but over $\operatorname{Spec}(L)$, where $L:=K[z]/(z^{2}-d)$ or any field extension thereof. Hence $\mathbb{F}_{1}(K)\rightarrow\mathbb{F}_{1}(L)$ is not injective.
QUESTION: How do I generalize this argument in order to show that for any $g\in\mathbb{N}$ the functor \begin{eqnarray*} \mathbf{F}_{g}: & \mathcal{Schemes} & \longrightarrow\mathcal{Sets}\\ & S & \longmapsto\text{$S$-isomorphism classes of genus-$g$ curves over $S$} \end{eqnarray*}
Is there a genus $g$ analogon to "quadratic twist"?