Radical in polynomial ring over $\mathbb{Q}$ implies radical in polynomial ring over $\mathbb{C}$? Let $f_1,\ldots,f_s\in\mathbb{Q}[x_1,\ldots,x_n]$ be polynomials such that the ideal $\langle f_1,\ldots,f_s\rangle$ is radical. Viewing each $f_i$ as an element of $\mathbb{C}[x_1,\ldots,x_n]$, is it true that $\langle f_1,\ldots,f_s\rangle$ generated as an ideal in $\mathbb{C}[x_1,\ldots,x_n]$ is also radical?
For example, the ideal generated by $x^2+1$ in $\mathbb{Q}[x]$ is radical (it is prime) and even though the ideal generated by $x^2+1$ in $\mathbb{C}[x]$ is not prime, it is still radical. Indeed, if $f^n\in\langle x^2+1\rangle\subset\mathbb{C}[x]$ then $f^n$ has $x^2+1$ as a factor, so it has $x-i$ and $x+i$ as factors, which is only possible if $f$ had $x-i$ and $x+i$ as factors to begin with. Hence $f$ has $x^2+1$ as a factor so that $f\in\langle x^2+1\rangle\subset\mathbb{C}[x]$.
 A: Yes, this works more generally for any separable field extension.  To be precise, let $A$ be a reduced commutative algebra over a field $K$ and let $L$ be a separable extension of $K$.  Then $L\otimes_K A$ is also reduced.  (To apply this to your question, let $K=\mathbb{Q},L=\mathbb{C},$ and $A=\mathbb{Q}[x_1,\dots,x_n]/(f_1,\dots,f_s)$.  Note that I include the possibility that $L$ is transcendental over $K$, in which case "separable" means there is a transcendence basis $B$ for $L$ over $K$ such that $L$ is separable over $K(B)$.)
To prove this, note that $L$ can be obtained from $K$ by repeatedly adjoining elements that are either transcendental or algebraic and separable.  So, it suffices to show that $L\otimes_K A$ is reduced when $L$ is generated over $K$ by a single element $\alpha$ that is either transcendental or algebraic and separable.  If $\alpha$ is transcendental, then $L\otimes_K A$ is a localization of $K[\alpha]\otimes_K A$ which is just a polynomial ring over $A$ and thus reduced.
Now suppose $\alpha$ is algebraic and separable.  Since $A$ is reduced, it embeds into a product of fields $\prod F_i$ (namely the residue fields of all its prime ideals), and then $L\otimes_K A$ embeds in $\prod L\otimes_K F_i$ (the tensor product distributes over the possibly infinite direct product since $L$ is finite over $K$).  So it suffices to show $L\otimes_K F_i$ is reduced, i.e. we may assume $A$ is a field.  Now $L\otimes_K A\cong A[x]/(g)$, where $g\in K[x]$ is the minimal polynomial of $\alpha$.  Since $g$ is separable by assumption, it is still squarefree over the extension field $A$, and thus $A[x]/(g)$ is reduced, as desired.
