Let $F$ be a field of characteristic $0$. Show that $F(x^2) \cap F(x^2-x)=F$.
Could you give me some hints how I could do that??
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1This is a near duplicate of this question from 2½ years ago. Replacing $x$ with $-x$ interchanges the two questions. – Jyrki Lahtonen Jan 06 '15 at 15:50
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Hint: If $z$ is in the intersection, there exists $R,S\in F(u)$ such that $z=R(x^2)=S(x^2-x)$. Show that $\displaystyle S(x^2-x)=S((x-\frac{1}{2})^2-\frac{1}{4})=S((x+\frac{1}{2})^2-\frac{1}{4})$, and put $T(x)=S((x-\frac{1}{2})^2-\frac{1}{4})$. What can you say of $T(x+1)$ ?
Kelenner
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Nice. You do leave somewhat open the reason, why we need the assumption about characteristic. That might be somewhat uninteresting if it weren't for the fact that the claim is false without it :-) – Jyrki Lahtonen Jan 06 '15 at 18:58