Let \(k\) be an algebraically closed field and \(R=k[x_1,\ldots,x_n]\). Define the “radical” of an ideal \(I\trianglelefteq R\) and prove it is an ideal. Prove that the ideal of all polynomials vanishing on the zero set of \(I\) is \(\sqrt{I}\).
Solution
The radical is \[ \sqrt I=\{f\in R:f^m\in I\text{ for some }m\ge1\}. \]
It is an ideal. If \(f^m\in I\) and \(g^n\in I\), then every term in the binomial expansion of \((f+g)^{m+n}\) contains either \(f^m\) or \(g^n\), so \[ (f+g)^{m+n}\in I. \] Also, for \(r\in R\), \[ (rf)^m=r^mf^m\in I. \] Thus \(f+g,rf\in\sqrt I\).
Let \[ Z(I)=\{a\in k^n:h(a)=0\text{ for every }h\in I\}, \] and let \(I(Z(I))\) be the ideal of polynomials vanishing on \(Z(I)\).
First, \(\sqrt I\subseteq I(Z(I))\): if \(f^m\in I\), then for every \(a\in Z(I)\), \[ f(a)^m=0, \] so \(f(a)=0\).
For the reverse inclusion, let \(f\in I(Z(I))\). In \[ k[x_1,\ldots,x_n,y] \] consider \[ J=I+(1-yf). \] There is no common zero of \(J\): a common zero would have its first \(n\) coordinates in \(Z(I)\), hence \(f=0\), while simultaneously \(1-yf=0\), impossible. By the weak Nullstellensatz, \(J=(1)\). Hence \[ 1=\sum_i a_i h_i+b(1-yf),\qquad h_i\in I. \] Substitute \(y=f^{-1}\) in the localization \(R_f\). The last term vanishes, so \(1\in IR_f\). Clearing a power of \(f\) gives \[ f^N\in I \] for some \(N\), hence \(f\in\sqrt I\).
Therefore \[ \boxed{I(Z(I))=\sqrt I}. \]