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How can be proved that $$X=\{( t,\sin t):t\in\mathbb{R}\}\subset \mathbb{R^2}$$ is not an algebraic set? (there isn't a subset $T$ of $\mathbb{R}[x,y]$ such that X is the zero locus of $T$)

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up vote 3 down vote accepted

Assume there is such $T$. Then for each $p \in T$, the polynomial $p(x,0)\in\mathbb R[x]$ has the infinitely many roots $k\pi$, hence must be the zero polynomial. Therefore $p(x,y)=yq(x,y)$ for some polynomial $q$. But then all points $(t,0)$ are in the zero locus of $T$.

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The line $y= 0$ intersects the set in infinitely many points, which is not possible for algebraic sets by Bezout theorem.

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