Let $I\in\mathbb{C}[x_1,\dots,x_n]$ be a an ideal generated by polynomials with rational coefficients. Is $\sqrt{I}$ also generated by polynomials with rational coefficients?
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6$\begingroup$ Yes. Abstractly it comes from the fact that reduction of a scheme is functorial (EGA I,§5.1; so we have a morphism $(X_{\mathbb{C}})_{\operatorname{red}} \to (X_{\operatorname{red}})_{\mathbb{C}}$) and that a reduced scheme over $\mathbb{Q}$ is geometrically reduced (EGA IV,§4.6; so the morphism I just wrote is an isomorphism). Computationally, it comes from the fact that the algorithms used to compute the radical (e.g., Becker & Weispfenning, Gröbner Bases, §8.7) are insensitive to the base field. $\endgroup$– Gro-TsenCommented Dec 17, 2021 at 14:26
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4$\begingroup$ Another way to say this is that if we have a polynomial which lies in the radical, so some power of it lies in the ideal, then all its Galois conjugates do (because the ideal is stable under the Galois group), and taking linear (not necessarily over $\mathbb Q$) combinations of its Galois conjugates one finds rational polynomials, a linear combination of which give the original polynomial (using a basis for the field generated by the coefficients). For transcendentals one can do something similar by specializing. $\endgroup$– Will SawinCommented Dec 17, 2021 at 14:32
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