[proofplan] For Part (1), the forward direction ($1 \in \mathfrak{a} \implies V(\mathfrak{a}) = \varnothing$) is immediate. The reverse uses that a proper ideal is contained in a maximal ideal $\mathfrak{m}$; by Zariski's lemma, $k[T_1,\dots,T_n]/\mathfrak{m}$ is a finite extension of $k$ that embeds into the algebraically closed field $\Omega$, yielding a point of $V(\mathfrak{a})$. For Part (2), the inclusion $\sqrt{\mathfrak{a}} \subset I(V(\mathfrak{a}))$ follows from the fact that $\Omega$ is an integral domain. The reverse inclusion uses the Rabinowitsch trick: given $f \in I(V(\mathfrak{a}))$, introduce a new variable $T_{n+1}$ and the ideal $\mathfrak{b} = \mathfrak{a}^e + (T_{n+1}f - 1)$ in $k[T_1,\dots,T_{n+1}]$, show $V(\mathfrak{b}) = \varnothing$ in $\Omega^{n+1}$, apply the weak Nullstellensatz to conclude $1 \in \mathfrak{b}$, and deduce that $f$ is nilpotent modulo $\mathfrak{a}$. [/proofplan]

[step:Prove the easy direction of the weak Nullstellensatz: $1 \in \mathfrak{a}$ implies $V(\mathfrak{a}) = \varnothing$] If $1 \in \mathfrak{a}$, then $\mathfrak{a} = k[T_1,\dots,T_n]$, so every polynomial belongs to $\mathfrak{a}$. For any $\underline{x} = (x_1,\dots,x_n) \in \Omega^n$, the constant polynomial $1 \in \mathfrak{a}$ satisfies $1(\underline{x}) = 1 \neq 0$, so $\underline{x} \notin V(\mathfrak{a})$. Therefore $V(\mathfrak{a}) = \varnothing$. [/step]

[step:Prove the reverse direction of the weak Nullstellensatz: $1 \notin \mathfrak{a}$ implies $V(\mathfrak{a}) \neq \varnothing$ using Zariski's lemma]Suppose $1 \notin \mathfrak{a}$, so $\mathfrak{a}$ is a proper ideal. By Zorn's lemma, $\mathfrak{a}$ is contained in some maximal ideal $\mathfrak{m}$ of $k[T_1,\dots,T_n]$. The quotient $K := k[T_1,\dots,T_n]/\mathfrak{m}$ is a field (since $\mathfrak{m}$ is maximal) that is finitely generated as a $k$-algebra (generated by the images $\bar{T}_1,\dots,\bar{T}_n$). By [Zariski's Lemma](/theorems/???), a field that is finitely generated as an algebra over a field $k$ is a finite (algebraic) extension of $k$. Therefore $[K : k] < \infty$. Since $\Omega$ is algebraically closed and contains $k$, and $K/k$ is a finite extension, there exists a $k$-algebra embedding $\sigma \colon K \hookrightarrow \Omega$. (The minimal polynomial of any generator of $K$ over $k$ splits in $\Omega$, so we can build the embedding inductively on a tower of simple extensions.) Define $\varphi \colon k[T_1,\dots,T_n] \to \Omega$ by $\varphi = \sigma \circ \pi$, where $\pi \colon k[T_1,\dots,T_n] \twoheadrightarrow K$ is the quotient map. Then $\varphi$ is a $k$-algebra homomorphism with $\ker \varphi = \ker \pi = \mathfrak{m}$. Set $\underline{x} = (\varphi(T_1),\dots,\varphi(T_n)) \in \Omega^n$. For any $f \in \mathfrak{a} \subset \mathfrak{m} = \ker \varphi$: \begin{align*} f(\underline{x}) = f(\varphi(T_1),\dots,\varphi(T_n)) = \varphi(f) = 0. \end{align*} Therefore $\underline{x} \in V(\mathfrak{a})$, so $V(\mathfrak{a}) \neq \varnothing$.[/step]

[guided]The idea is to find a point in $\Omega^n$ at which all polynomials in $\mathfrak{a}$ vanish. Since $\mathfrak{a}$ is proper, it is contained in a maximal ideal $\mathfrak{m}$, and the quotient $K = k[T_1,\dots,T_n]/\mathfrak{m}$ is a field. The images $\bar{T}_1,\dots,\bar{T}_n$ of the variables generate $K$ as a $k$-algebra, so $K$ is a finitely generated $k$-algebra that is also a field. This is exactly the setup of [Zariski's Lemma](/theorems/???): a field finitely generated as an algebra over a field $k$ must be a finite (algebraic) extension of $k$. The lemma gives $[K : k] < \infty$. Why does finiteness help? Because $\Omega$ is algebraically closed: any algebraic element over $k$ has a minimal polynomial that splits in $\Omega$. A finite extension $K/k$ can be built as a tower of simple algebraic extensions $k = K_0 \subset K_1 \subset \cdots \subset K_d = K$, and at each stage the minimal polynomial of the adjoined element splits in $\Omega$, so we can define a $k$-algebra embedding $K_i \hookrightarrow \Omega$ extending the previous one. This yields $\sigma \colon K \hookrightarrow \Omega$. Composing $\sigma$ with the quotient map $\pi \colon k[T_1,\dots,T_n] \twoheadrightarrow K$ gives $\varphi = \sigma \circ \pi \colon k[T_1,\dots,T_n] \to \Omega$ with $\ker \varphi = \mathfrak{m} \supset \mathfrak{a}$. The point $\underline{x} = (\varphi(T_1),\dots,\varphi(T_n))$ satisfies $f(\underline{x}) = \varphi(f) = 0$ for all $f \in \mathfrak{a}$ (since $\varphi$ is a $k$-algebra homomorphism, evaluation at $\underline{x}$ is the same as applying $\varphi$). Therefore $\underline{x} \in V(\mathfrak{a})$.[/guided]

[step:Prove the easy inclusion of the strong Nullstellensatz: $\sqrt{\mathfrak{a}} \subset I(V(\mathfrak{a}))$] Let $f \in \sqrt{\mathfrak{a}}$, so $f^\ell \in \mathfrak{a}$ for some $\ell \geq 1$. For any $\underline{x} \in V(\mathfrak{a})$, we have $f^\ell(\underline{x}) = 0$, i.e., $f(\underline{x})^\ell = 0$ in $\Omega$. Since $\Omega$ is an integral domain (it is a field), $f(\underline{x})^\ell = 0$ implies $f(\underline{x}) = 0$. Therefore $f \in I(V(\mathfrak{a}))$. [/step]

[step:Prove the hard inclusion $I(V(\mathfrak{a})) \subset \sqrt{\mathfrak{a}}$ via the Rabinowitsch trick]Let $f \in I(V(\mathfrak{a}))$, so $f(\underline{x}) = 0$ for all $\underline{x} \in V(\mathfrak{a})$. We must show $f \in \sqrt{\mathfrak{a}}$, i.e., $f^\ell \in \mathfrak{a}$ for some $\ell \geq 1$. Introduce a new variable $T_{n+1}$ and define the ideal \begin{align*} \mathfrak{b} = \mathfrak{a}^e + (T_{n+1} f - 1) \trianglelefteq k[T_1,\dots,T_{n+1}], \end{align*} where $\mathfrak{a}^e$ denotes the extension of $\mathfrak{a}$ to $k[T_1,\dots,T_{n+1}]$ (the ideal generated by $\mathfrak{a}$ in the larger polynomial ring). [claim:$V(\mathfrak{b}) = \varnothing$ in $\Omega^{n+1}$] Suppose for contradiction that $\underline{y} = (y_1,\dots,y_n,y_{n+1}) \in V(\mathfrak{b})$. Since $\mathfrak{a}^e \subset \mathfrak{b}$, every polynomial in $\mathfrak{a}$ (viewed in $k[T_1,\dots,T_{n+1}]$) vanishes at $\underline{y}$. In particular, every $g \in \mathfrak{a}$ satisfies $g(y_1,\dots,y_n) = 0$ (since $g$ does not involve $T_{n+1}$), so $(y_1,\dots,y_n) \in V(\mathfrak{a})$. By hypothesis, $f \in I(V(\mathfrak{a}))$, so $f(y_1,\dots,y_n) = 0$. But $T_{n+1}f - 1 \in \mathfrak{b}$, so evaluating at $\underline{y}$: \begin{align*} y_{n+1} \cdot f(y_1,\dots,y_n) - 1 = y_{n+1} \cdot 0 - 1 = -1 \neq 0. \end{align*} This contradicts $\underline{y} \in V(\mathfrak{b})$. Therefore $V(\mathfrak{b}) = \varnothing$. [/claim] [proof] The argument above establishes $V(\mathfrak{b}) = \varnothing$. [/proof] By the weak Nullstellensatz (Part 1) applied to $\mathfrak{b}$ in $k[T_1,\dots,T_{n+1}]$, the condition $V(\mathfrak{b}) = \varnothing$ implies $1 \in \mathfrak{b}$. Therefore there exist polynomials $g_1,\dots,g_r \in \mathfrak{a}$ and $h_1,\dots,h_r, q \in k[T_1,\dots,T_{n+1}]$ such that \begin{align*} 1 = \sum_{i=1}^{r} h_i \cdot g_i + q \cdot (T_{n+1} f - 1) \quad \text{in } k[T_1,\dots,T_{n+1}]. \end{align*} Substitute $T_{n+1} = \tfrac{1}{f}$ in the fraction field $k(T_1,\dots,T_n)$ (where $f \neq 0$ as an element of $k[T_1,\dots,T_n]$ unless $f = 0$, in which case $f \in \sqrt{\mathfrak{a}}$ is immediate). The term $q(T_1,\dots,T_n,\tfrac{1}{f})(T_{n+1}f - 1)$ evaluated at $T_{n+1} = \tfrac{1}{f}$ vanishes, giving \begin{align*} 1 = \sum_{i=1}^{r} h_i\!\left(T_1,\dots,T_n, \frac{1}{f}\right) g_i(T_1,\dots,T_n) \quad \text{in } k(T_1,\dots,T_n). \end{align*} Each $h_i(T_1,\dots,T_n,\tfrac{1}{f})$ is a polynomial in $T_1,\dots,T_n$ and $\tfrac{1}{f}$, hence a rational function of the form $\tfrac{p_i(T_1,\dots,T_n)}{f^{d_i}}$ for some polynomial $p_i$ and non-negative integer $d_i$. Setting $d = \max(d_1,\dots,d_r)$ and multiplying both sides by $f^d$: \begin{align*} f^d = \sum_{i=1}^{r} p_i'(T_1,\dots,T_n) \cdot g_i(T_1,\dots,T_n) \quad \text{in } k[T_1,\dots,T_n], \end{align*} where $p_i' = f^{d - d_i} p_i \in k[T_1,\dots,T_n]$. Since each $g_i \in \mathfrak{a}$, the right-hand side lies in $\mathfrak{a}$, so $f^d \in \mathfrak{a}$. Taking $\ell = d$, we have $f^\ell \in \mathfrak{a}$, which gives $f \in \sqrt{\mathfrak{a}}$.[/step]

[guided]This is the **Rabinowitsch trick** (also called the "trick of Rabinowitsch"): to show $f \in \sqrt{\mathfrak{a}}$, we introduce an auxiliary variable that "forces" $f$ to be invertible, derive a contradiction with the weak Nullstellensatz, and then extract the nilpotency exponent. The ideal $\mathfrak{b} = \mathfrak{a}^e + (T_{n+1}f - 1)$ in $k[T_1,\dots,T_{n+1}]$ encodes the condition "all elements of $\mathfrak{a}$ vanish AND $f$ is nonzero" (since $T_{n+1}f = 1$ forces $f \neq 0$ and provides a formal inverse). Any common zero of $\mathfrak{b}$ in $\Omega^{n+1}$ would give a point $(y_1,\dots,y_n) \in V(\mathfrak{a})$ with $f(y_1,\dots,y_n) \neq 0$. But $f \in I(V(\mathfrak{a}))$ means $f$ vanishes on all of $V(\mathfrak{a})$, so no such point exists: $V(\mathfrak{b}) = \varnothing$. The weak Nullstellensatz (already proved) then gives $1 \in \mathfrak{b}$. Writing $1 = \sum_i h_i g_i + q(T_{n+1}f - 1)$ and substituting $T_{n+1} = 1/f$ kills the last term and expresses $1$ as a combination of the $g_i$ with coefficients involving powers of $1/f$. Clearing the denominators (multiplying by a sufficiently high power of $f$) yields $f^d \in \mathfrak{a}$. Why does the substitution $T_{n+1} = 1/f$ make sense? We work in the fraction field $k(T_1,\dots,T_n)$, where $f$ is a nonzero element (if $f = 0$ then $f \in \sqrt{\mathfrak{a}}$ holds vacuously). The polynomial identity in $k[T_1,\dots,T_{n+1}]$ specialises to a rational function identity in $k(T_1,\dots,T_n)$. After clearing denominators, we return to a polynomial identity in $k[T_1,\dots,T_n]$ that places $f^d$ in $\mathfrak{a}$.[/guided]

[step:Combine both inclusions to conclude $I(V(\mathfrak{a})) = \sqrt{\mathfrak{a}}$] The inclusion $\sqrt{\mathfrak{a}} \subset I(V(\mathfrak{a}))$ was established using the integral domain property of $\Omega$. The inclusion $I(V(\mathfrak{a})) \subset \sqrt{\mathfrak{a}}$ was established via the Rabinowitsch trick and the weak Nullstellensatz. Together they give \begin{align*} I(V(\mathfrak{a})) = \sqrt{\mathfrak{a}}. \end{align*} This completes the proof of the strong Nullstellensatz. [/step]