Let $P=(x,y)$ be a point on the elliptic curve $X=\{(x,y)\in\mathbb{C}^2 \ | \ y^2=4x^3-ax-b \ (a,b \in \mathbb{Q})\}$ with order $4$. Show that $x$ is algebraic over $\mathbb{Q}$ with degree at most $6$.
I do not really know how to connect the order of a point with being algebraic. Only things I noticed were, that $2P=(x',0)$, since $2P$ has order $2$. Furthermore, $3P=-P$. However, I do not know if this is helpful for the question. Any help is appreciated. Thanks.
What you've done so far is quite helpful. There is an explicit formula for the coordinates of $2P$ in terms of the coordinates of $P = (x, y)$; in particular, the $y$-coordinate of $2P$ is a rational function whose numerator is a polynomial in $x$ of degree $6$ with rational coefficients. Hence, if the second coordinate of $2P$ is zero, then the aforementioned numerator is zero, which exactly means that $x$ is algebraic over $\mathbb{Q}$ with degree at most $6$. As a reference, Silverman and Tate's beautiful book ''Rational Points on Elliptic Curves'' describes how one might carry out this computation on page 31.