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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.

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    Are you sure your equation for $X$ is correct? If so, why have two constants when one would do? – 2017-01-22
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    Also, the RHS should likely be cubic in $x$... – 2017-01-22
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    Yes you are right. I made a stupid typo. I edited it now. Thanks – 2017-01-22

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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.

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    I will try it. I got the formula for $2P$. Thank you very much. – 2017-01-22
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    The formula I am using is $x'=-2x+\left(\frac{12x²-a}{8y}\right)^2$ and $y'=-y+\frac{12x²-a(x-x')}{2y}$. For $2P=(x',y')$ However, after expanding I get $y'=0=-y+\frac{36x^3-3ax}{y}- \text{the polynomial of degree} \ 6$, which is somehow not monic but has the coefficient $1728$ for $x^6$. Do you know where it went wrong? – 2017-01-22
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    I just found out that the coefficient $1728$ is not a problem, since we can just factor out the $1728$ and the coefficient are still in $\mathbb{Q}$. The only thing that bothers me is the remaining part $\frac{36x^3-3ax-y^2}{y}$ – 2017-01-22
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    I made it to solve it. By using the equation for the elliptic curve, one gets a common denominator and thus the wished polynomial. Thanks for your help :) – 2017-01-22
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    @Deavor: good work, I'm glad you figured it out! I looked back at my computation of the polynomial of degree 6, and it was for an elliptic curve in Weierstrass form, hence the discrepancy. As you noted, of course, this isn't a problem. :) – 2017-01-22