 Solution

For an algebraically closed field $k$, the Weak Hilbert Nullstellensatz says that every maximal ideal of $k[x_1,\dots ,x_n]$ isfor a unique $a\in k^n$. Equivalently, every proper ideal has a common zero. The Strong Hilbert Nullstellensatz says that for every ideal $I$,

To deduce the strong form, let $f$ vanish on $V(I)$ and introduce a variable $y$. The equations in $I$ together with $1-yf$ have no common zero: a common zero would satisfy both $f=0$ and $yf=1$. The weak theorem therefore givesSubstitute $y=f^{-1}$ in the localization $k[x_1,\dots ,x_n,f^{-1}]$. The last term vanishes, and clearing a power of $f$ yields $f^N\in I$. Thus $f\in \sqrt{I}$. The reverse inclusion is immediate, completing the Rabinowitsch trick proof.