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Nov 14, 2011 at 21:58 comment added Will Sawin This is so since $\mathbb R$ is a vector space over $\mathbb K$. Choosing a basis, a polynomial in $J$ can be decomposed into the sum of polynomials in $I$.
Nov 14, 2011 at 19:15 comment added Emil Jeřábek The set of all polynomials of a fixed degree that vanish on your set of points is defined by a system of linear equations (whose variables are coefficients of the polynomial) with coefficients in K, one for each point. For any linear system over K, a basis of its solution space over K is also a basis of its solution space over any extension field.
Nov 14, 2011 at 18:31 history asked pirignao CC BY-SA 3.0