Let \(I\) and \(J\) be ideals in the polynomial ring \(R = k[x_1, \dots, x_n]\) where \(k\) is a field.
- Show that \(\sqrt{\sqrt{I}} = \sqrt{I}\).
- Show that \(\sqrt{I \cap J} = \sqrt{IJ} = \sqrt{I} \cap \sqrt{J}\).
- Is \(x^2 - y^2 \in \sqrt{\langle x^2 + x, x^2 - y \rangle}\) in \(k[x, y]\)?
- Is \(x^2 + y^2 \in \sqrt{\langle x + y, x^2 - y \rangle}\) in \(k[x, y]\)?
Solution
1. One has \(\sqrt{\sqrt I}=\sqrt I\).
The inclusion \(\sqrt I\subseteq\sqrt{\sqrt I}\) is immediate from \(I\subseteq\sqrt I\). Conversely, if \(f\in\sqrt{\sqrt I}\), then \(f^m\in\sqrt I\) for some \(m\ge1\). Hence \((f^m)^r=f^{mr}\in I\) for some \(r\ge1\), so \(f\in\sqrt I\).
2. One has\[ \sqrt{I\cap J}=\sqrt{IJ}=\sqrt I\cap\sqrt J. \]
Since \(IJ\subseteq I\cap J\), we have \[ \sqrt{IJ}\subseteq\sqrt{I\cap J}. \] If \(f\in\sqrt{I\cap J}\), then \(f^m\in I\cap J\) for some \(m\), hence \(f\in\sqrt I\cap\sqrt J\). Finally, if \(f\in\sqrt I\cap\sqrt J\), choose \(a,b\ge1\) with \(f^a\in I\) and \(f^b\in J\). Then \[ f^{a+b}=f^af^b\in IJ, \] so \(f\in\sqrt{IJ}\). The three containments give equality.
3. For\[ I_1=\langle x^2+x,\ x^2-y\rangle, \]one has \(x^2-y^2\in I_1\), hence certainly \(x^2-y^2\in\sqrt{I_1}\).
In \(R/I_1\) we have \[ y=x^2=-x. \] Therefore \[ y^2=x^2, \] so the class of \(x^2-y^2\) in \(R/I_1\) is zero. Equivalently, \(x^2-y^2\in I_1\).
4. For\[ I_2=\langle x+y,\ x^2-y\rangle, \]the answer depends on \(\operatorname{char}k\):\[ x^2+y^2\in\sqrt{I_2} \quad\Longleftrightarrow\quad \operatorname{char}k=2. \]
Modulo \(I_2\) we have \(y=-x\) and \(x^2=-x\), hence \[ x^2+y^2=2x^2=-2x. \] If \(\operatorname{char}k=2\), this class is zero, so \(x^2+y^2\in I_2\).
Assume now \(\operatorname{char}k\ne2\). The quotient is \[ R/I_2\cong k[x]/(x^2+x)=k[x]/(x(x+1)), \] under \(y\mapsto -x\). Since \(x\) and \(x+1\) are coprime, \[ (x(x+1))=(x)\cap(x+1) \] is radical. Thus \(I_2\) is radical. The class of \(x^2+y^2\) is \(-2x\), which is nonzero in this quotient (for example its image modulo \((x+1)\) is \(2\ne0\)). Therefore \[ x^2+y^2\notin I_2=\sqrt{I_2}. \]