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I fell a little uncomfortable with this real stuff. The question here is more general, but we can suppose $\mathbb{K}=\mathbb{R}$.

Take a set of (distinct) points in $\mathbb{P}^n$, the complex projective space. Suppose these points have coefficients in a fixed field $\mathbb{K}\subseteq \mathbb{C}$. Now, I can consider three ideals:

  1. the ideal of the points $I$, that is, the set of polynomials in $\mathbb{K}[x_1,x_2,\dotsc,x_k]$ which vanishes on the points, which is finitely generated by the Hilbert Basis Theorem;

  2. the ideal $I\cdot \mathbb{C}[x_1,x_2,\dotsc,x_k]$, which is generated by the same generators of $I$, viewed as polynomials with complex coefficients;

  3. the ideal $J$, which is the ideal of the points in the usual sense in $\mathbb{C}[x_1,x_2,\dotsc,x_k]$.

Question: are the second one and the third one equal?


  1. I guess the answer is yes, but I am not able to prove it.
  2. I know the answer is yes, if I have five points in general position in $\mathbb{P}^3$ and $\mathbb{K}=\mathbb{Q}$.
  3. Another point of view is: the generators of $I$ have no non-real solution in common, right?
  4. The question arised as: if I have a set of real points (or with real coefficients), can I take a minimal free resolution of their ideal in $\mathbb{C}[x_0,x_1,\dotsc,x_k]$ which involves only "real" maps? That is, which involves only matrices with real polynomials?

Thank you in advance!

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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. – Emil Jeřábek Nov 14 '11 at 19:15
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 $I$. – Will Sawin Nov 14 '11 at 21:58

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