The theorem and proof are from these notes : https://ocw.mit.edu/courses/18-782-introduction-to-arithmetic-geometry-fall-2013/a60e5c7aad91910265ee1fc686308413_MIT18_782F13_lec26.pdf
Theorem 26.7.Let $C/k$ be a curve of genus one.Then $j(C)\in k$.
Proof. Let us pick $O\in C(L)$, where $L$ is some finite Galois extension $L/k$, and let $E/L$ be the elliptic curve $(C, O)$. Then $E$ is isomorphic to the base extension of $C$ to $L$, so let $\varphi: C\longrightarrow E$ be the isomorphism (which is defined over $L$). For any $\sigma \in$ Gal$(L/k)$ there is an isomorphism $\varphi^{\sigma}:C^{\sigma}\longrightarrow E^{\sigma}$. But $C$ is defined over $k$, so $C^{\sigma} = C$, and therefore $E^{\sigma}\cong E$ over $L$,so $j(E^{\sigma}) = j(E)$. But then $j(E)^{\sigma} = j(E^{\sigma}) = j(E)$ for all $\sigma \in $Gal$(L/k)$, so $j(E) \in k$.
I do not understand why if $C$ is defined over $k$,then $C^{\sigma}=C$ and how the argument follows from there.Can someone explain this?
In general, for a sub-variety $V\subset\mathbb P^n$ defined by the ideal $I(V)\subset \overline k[X_0,\dots,X_n]$, the twist $V^\sigma$ for $\sigma\in\mathrm{Gal}(\overline k/k)$ is defined by the ideal $I(V^\sigma):=\{f^\sigma:f\in I(V)\}$, where for $f=\sum a_{i_0\cdots i_n}X_0^{i_0}\cdots X_n^{i_n}$, $f^\sigma:=\sum \sigma(a_{i_0\cdots i_n})X_0^{i_0}\cdots X_n^{i_n}$.
In particular, if $V$ is defined over $k$, then $I(V)$ has generators $(f_1,\dots,f_k)$ with $f_i\in k[X_0,\dots,X_n]$, so $f_i^\sigma=f_i$ and thus $V^\sigma=V$.
The crux of the argument is then that for an elliptic curve $E$ defined over $L$, it has some defining equation $y^2=x^3+ax+b$, and $j(E)$ is defined as some rational function in $a$ and $b$. Now $E^\sigma$ has defining equation $y^2=x^3+\sigma(a)x+\sigma(b)$, so $j(E^\sigma)=\sigma(j(E))$. We know that $j(E^\sigma)=j(E)$ since $j$ is only depends on the isomorphism class of $E$, so $j(E)\in L^{\mathrm{Gal}(L/k)}=k$.