[proofplan] We establish the bijection by showing that both compositions of the map $\underline{x} \mapsto \mathfrak{m}_{\underline{x}} := (T_1 - x_1, \dots, T_n - x_n)$ and its putative inverse are the identity. Injectivity is immediate from the definition. For surjectivity, we use the fact that singletons $\{\underline{x}\}$ are the minimal nonempty $\Omega$-algebraic sets and that the Nullstellensatz correspondence sends these bijectively to the maximal ideals of $\Omega[T_1, \dots, T_n]$. [/proofplan]

[step:Define the map and verify it is well-defined]For each point $\underline{x} = (x_1, \dots, x_n) \in \Omega^n$, define \begin{align*} \Phi: \Omega^n &\to \operatorname{mspec}(\Omega[T_1, \dots, T_n]) \\ \underline{x} &\mapsto \mathfrak{m}_{\underline{x}} := (T_1 - x_1, \dots, T_n - x_n). \end{align*} We must verify that $\mathfrak{m}_{\underline{x}}$ is indeed a maximal ideal. Consider the evaluation homomorphism \begin{align*} \operatorname{ev}_{\underline{x}}: \Omega[T_1, \dots, T_n] &\to \Omega \\ f &\mapsto f(x_1, \dots, x_n). \end{align*} This is a surjective ring homomorphism (since every constant $c \in \Omega$ is the image of the constant polynomial $c$), and $\ker(\operatorname{ev}_{\underline{x}}) = \mathfrak{m}_{\underline{x}}$: a polynomial $f$ vanishes at $\underline{x}$ if and only if $f \in (T_1 - x_1, \dots, T_n - x_n)$, which follows by writing $f$ as an iterated Taylor expansion in the variables $T_i - x_i$. By the first isomorphism theorem, \begin{align*} \Omega[T_1, \dots, T_n] / \mathfrak{m}_{\underline{x}} \cong \Omega, \end{align*} which is a field. Therefore $\mathfrak{m}_{\underline{x}}$ is a maximal ideal, and $\Phi$ is well-defined.[/step]

[guided]We want to construct a map from points in $\Omega^n$ to maximal ideals of $\Omega[T_1, \dots, T_n]$. The natural candidate sends a point $\underline{x} = (x_1, \dots, x_n)$ to the ideal of all polynomials vanishing at $\underline{x}$. We need to verify this ideal is maximal. For each $\underline{x} \in \Omega^n$, define \begin{align*} \Phi: \Omega^n &\to \operatorname{mspec}(\Omega[T_1, \dots, T_n]) \\ \underline{x} &\mapsto \mathfrak{m}_{\underline{x}} := (T_1 - x_1, \dots, T_n - x_n). \end{align*} To see that $\mathfrak{m}_{\underline{x}}$ is maximal, consider the evaluation homomorphism $\operatorname{ev}_{\underline{x}}: \Omega[T_1, \dots, T_n] \to \Omega$ defined by $f \mapsto f(x_1, \dots, x_n)$. This is a ring homomorphism because evaluation preserves addition and multiplication. It is surjective because any $c \in \Omega$ is the image of the constant polynomial $c$. The kernel of $\operatorname{ev}_{\underline{x}}$ is exactly $\mathfrak{m}_{\underline{x}}$. Indeed, $f \in \ker(\operatorname{ev}_{\underline{x}})$ means $f(x_1, \dots, x_n) = 0$. Writing $f$ as a polynomial in the shifted variables $T_i - x_i$ (which is possible by expanding each $T_i = (T_i - x_i) + x_i$ and collecting terms), every term of $f$ that does not involve any $T_i - x_i$ contributes exactly $f(x_1, \dots, x_n) = 0$, so $f \in (T_1 - x_1, \dots, T_n - x_n)$. The converse inclusion is immediate since each generator $T_i - x_i$ vanishes at $\underline{x}$. By the first isomorphism theorem, $\Omega[T_1, \dots, T_n] / \mathfrak{m}_{\underline{x}} \cong \Omega$. Since $\Omega$ is a field, the quotient is a field, so $\mathfrak{m}_{\underline{x}}$ is a maximal ideal. The map $\Phi$ is therefore well-defined.[/guided]

[step:Prove injectivity by recovering coordinates from the ideal] Suppose $\Phi(\underline{x}) = \Phi(\underline{y})$, i.e., $(T_1 - x_1, \dots, T_n - x_n) = (T_1 - y_1, \dots, T_n - y_n)$. For each $i \in \{1, \dots, n\}$, the element $T_i - x_i$ belongs to the ideal $(T_1 - y_1, \dots, T_n - y_n)$, and evaluating at $\underline{y}$ gives $(y_i - x_i) = 0$ since every element of $(T_1 - y_1, \dots, T_n - y_n)$ vanishes at $\underline{y}$. Therefore $x_i = y_i$ for all $i$, so $\underline{x} = \underline{y}$, and $\Phi$ is injective. [/step]

[step:Prove surjectivity using the Nullstellensatz correspondence]Let $\mathfrak{m} \in \operatorname{mspec}(\Omega[T_1, \dots, T_n])$ be an arbitrary maximal ideal. We must find $\underline{x} \in \Omega^n$ with $\Phi(\underline{x}) = \mathfrak{m}$. Since $\mathfrak{m}$ is a maximal ideal, $V(\mathfrak{m}) \neq \varnothing$ by the Weak Nullstellensatz (which asserts that every proper ideal of $\Omega[T_1, \dots, T_n]$ has a common zero in $\Omega^n$ when $\Omega$ is algebraically closed). Choose $\underline{x} \in V(\mathfrak{m})$. Since every polynomial in $\mathfrak{m}$ vanishes at $\underline{x}$, we have $\mathfrak{m} \subseteq I(\{\underline{x}\}) = \mathfrak{m}_{\underline{x}}$. Both $\mathfrak{m}$ and $\mathfrak{m}_{\underline{x}}$ are maximal ideals (the former by hypothesis, the latter by the computation in the first step). A maximal ideal is contained in another proper ideal only if they are equal. Therefore $\mathfrak{m} = \mathfrak{m}_{\underline{x}} = \Phi(\underline{x})$. Since $\mathfrak{m}$ was arbitrary, $\Phi$ is surjective.[/step]

[guided]Let $\mathfrak{m} \in \operatorname{mspec}(\Omega[T_1, \dots, T_n])$. We need to show $\mathfrak{m}$ is in the image of $\Phi$, i.e., $\mathfrak{m} = (T_1 - x_1, \dots, T_n - x_n)$ for some $(x_1, \dots, x_n) \in \Omega^n$. The key ingredient is the Weak Nullstellensatz: since $\Omega$ is algebraically closed, every proper ideal of $\Omega[T_1, \dots, T_n]$ has a common zero in $\Omega^n$. In particular, since $\mathfrak{m}$ is a maximal ideal (hence proper), there exists $\underline{x} = (x_1, \dots, x_n) \in \Omega^n$ with $f(\underline{x}) = 0$ for all $f \in \mathfrak{m}$. That is, $\underline{x} \in V(\mathfrak{m})$. Now we use a maximality argument. Every $f \in \mathfrak{m}$ vanishes at $\underline{x}$, so $\mathfrak{m} \subseteq I(\{\underline{x}\})$. We computed in the first step that $I(\{\underline{x}\}) = \ker(\operatorname{ev}_{\underline{x}}) = \mathfrak{m}_{\underline{x}}$. So $\mathfrak{m} \subseteq \mathfrak{m}_{\underline{x}}$. But $\mathfrak{m}$ is maximal, meaning it is not properly contained in any ideal other than $\Omega[T_1, \dots, T_n]$ itself. Since $\mathfrak{m}_{\underline{x}}$ is a proper ideal (it is maximal, hence proper), the inclusion $\mathfrak{m} \subseteq \mathfrak{m}_{\underline{x}}$ forces $\mathfrak{m} = \mathfrak{m}_{\underline{x}} = \Phi(\underline{x})$. This shows every maximal ideal is in the image of $\Phi$, so $\Phi$ is surjective. Since $\Phi$ is both injective and surjective, it is a bijection $\Omega^n \leftrightarrow \operatorname{mspec}(\Omega[T_1, \dots, T_n])$.[/guided]