Past exam of the mathematics course of the University of Cambridge / 2015 / ii / Paper 2 / 21F (Algebraic Geometry)

The radical of an ideal $I$ in a commutative ring $A$ is $\sqrt{I}=\{a\in A:a^m\in I\text{ for some }m\ge 1\}$. It is an ideal: if $a^m,b^n\in I$, every term of $(a+b{)}^{m+n}$ has either at least $m$ factors $a$ or at least $n$ factors $b$, so the sum belongs to $I$. Also $(ca{)}^m\in I$ for every $c\in A$, and additive inverses preserve the defining condition.

If $f\in \sqrt{I}$, a power $f^m$ vanishes at every point of $Z(I)$, so $f$ itself vanishes there. This gives $\sqrt{I}\subseteq I(Z(I))$.

For the reverse inclusion, take $f$ vanishing on $Z(I)$ and use the Rabinowitsch trick. In $k[x_1,\dots ,x_n,t]$ form $J=(I,1-tf)$. A common zero would have $x\in Z(I)$, whence $f(x)=0$ and $1-tf(x)=1$, a contradiction. The assumed Weak Hilbert Nullstellensatz gives $J=(1)$. Thus there is a finite expressionIf $f=0$ the conclusion is immediate. Otherwise substitute $t=f^{-1}$ in the localized polynomial ring. The last term vanishes, and clearing a sufficiently high power of $f$ gives $f^N\in I$. Hence $f\in \sqrt{I}$, proving the Strong Hilbert Nullstellensatz:

If a power of $a$ lies in $I\cap J$, then $a\in \sqrt{I}\cap \sqrt{J}$. Conversely, if $a^m\in I$ and $a^n\in J$, the power $a^{\max (m,n)}$ lies in both ideals. Therefore $\sqrt{I\cap J}=\sqrt{I}\cap \sqrt{J}$ for any commutative ring.

The equality is false. In $A=k[x,y]$, take $I=(y)$ and $J=(y-x^2)$. Both are prime ideals, since their quotient rings are isomorphic to $k[x]$, so both are radical. ButThe element $x$ belongs to the latter ideal and not the former. This also shows why taking the radical of an ideal can enlarge a sum of already radical ideals.