---- > [!theorem] Theorem. ([[strong Nullstellensatz]]) > Let $k \subset \Omega$ [[field|fields]] with $\Omega$ [[algebraically closed]], and $\mathfrak{a}$ be an [[ideal]] of the [[polynomial 4|polynomial ring]] $k[T_{1},\dots,T_{n}]$. [[algebraic set|Then]] $I\big( V(\mathfrak{a}) \big)=\sqrt{ \mathfrak{a} },$ the [[radical of an ideal|radical]] of $\mathfrak{a}$. > (One inclusion is elementary. The other follows from the [[weak Nullstellensatz]].) ^theorem > [!proof]- Proof. ([[strong Nullstellensatz]]) > Let $f \in \sqrt{ \mathfrak{a} }$, say, $f^{\ell} \in \mathfrak{a}$. So $f^{\ell}(x)=0$ for all $x \in V(\mathfrak{a})$, hence $f(x)=0$ since $\Omega$ is an [[integral domain]]. That is, $f \in I(V(\mathfrak{a}))$. > > Conversely, take $f \in I(V(\mathfrak{a})) \subset k[T_{1},\dots,T_{n}]$. We wish to show some power of $f$ lies in $\mathfrak{a}$. Successive wants: > - Put $A:=k[T_{1},\dots,T_{n}] / \mathfrak{a}$, and let $\overline{f}=f+\mathfrak{a}$ be the image of $f$ in $A$. Done if we can show $\overline{f}$ [[nilpotent element of a ring|nilpotent]]. > - Equivalently, if we can [[localization|show]] $A_{\overline{f}}=\{ \overline{f}^{m}:m \geq 0 \} ^{-1} A=(0)$. > - Recall that we can realize localization as a quotient. $A_{\overline{f}}$ is the [[quotient ring|quotient]] $\frac{k[T_{1},\dots,T_{n}, T_{n+1}]}{\underbrace{ \mathfrak{a}^{e}+ (T_{n+1}f-1) }_{ := \mathfrak{b} }}$.[^1] So our task is then to show $k[T_{1},\dots,T_{n+1}]=\mathfrak{b}$, i.e., $1 \in \mathfrak{b}$. > - By the [[weak Nullstellensatz]], it is equivalent to show that $V(\mathfrak{b})=\emptyset$ > > > Okay, so let $\boldsymbol x=(x_{1},\dots,x_{n+1}) \in V(\mathfrak{b})$. Then $\boldsymbol x_{0}:=(x_{1},\dots,x_{n}) \in V(\mathfrak{a}^{e}) \subset V(\mathfrak{a})$, and so $f(\boldsymbol x_{0})=0$ since $f \in I(V(\mathfrak{a}))$. So $\underbrace{ (T_{n+1}f-1 }_{ \in \mathfrak{b} })(\boldsymbol x)=-1 \neq 0$, a contradiction. > > ---- #### [^1]: To see this calculus: initially have $A_{\overline{f}} \cong \frac{A[T]}{\langle \overline{f}T - 1 \rangle}$, cf. discussion in [[localization|localization as a quotient]]. Substituting the definition of $A$, this is $A_{\overline{f}}\cong \frac{\frac{k[T_{1},\dots,T_{n}]}{\mathfrak{a}}[T]}{\langle \overline{f} T - 1 \rangle } .$ Recall the general rule for [[quotient ring|ring quotients]] that there is an [[isomorphism]] $\begin{align} \frac{R}{I}[T] &\xrightarrow{\cong} \frac{R[T]}{I^{e}} \\ (f + I)[T] & \mapsto f[T] + I^{e}. \end{align}$ Here, the notation $I^{e}$ (also sometimes written $IR[T]$) refers to the [[extension of an ideal|extension]] $I^{e}$ of the [[ideal]] $I$ under the map $R \to R[T]$. In our case, that gives $A_{\overline{f}} \cong \frac{\frac{k[T_{1},\dots, T_{n}][T]}{\mathfrak{a}^{e}}}{\langle \overline{f}T - 1 \rangle}.$ Now absorb the extra [[indeterminate]] $T$ into $k[T_{1},\dots,T_{n}]$ as $T_{n+1}$: $A_{\overline{f}} \cong \frac{\frac{k[T_{1},\dots, T_{n}, T_{n+1}]}{\mathfrak{a}^{e}}}{\langle \overline{f}T_{n+1} - 1 \rangle}.$ Under the quotient map $k[T_{1},\dots,T_{n},T_{n+1}] \to \frac{k[T_{1},\dots,T_{n},T_{n+1}]}{\mathfrak{a}^{e}}$, $[\overline{f}T_{n+1}-\overline{1}]=[(f+\mathfrak{a})T_{n+1}-(1 + \mathfrak{a})]=[(fT_{n+1}-1)+\mathfrak{a}]=(fT_{n+1}-1)+\mathfrak{a}+\mathfrak{a}^{e}=(fT_{n+1}-1)+\mathfrak{a}^{e}$ lifts to $fT_{n+1}-1$. ----- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```