91Ó°ÊÓ

A proper ideal is a nonempty subset of a ring that is closed under the operations of addition and multiplication with ring elements, but does not contain all of the elements of the ring. Since \(I=\left(X_{1}-a_{1}, \ldots, X_{n}-a_{n}\right)\) and we can express every element of \(I\) as a linear combination of the given generators, we know that \(I\) is a nonempty subset of \(k[X_1, \ldots, X_n]\) that is closed under the ring operations. However, we need to verify that \(I\) does not contain all of the elements of \(k[X_1, \ldots, X_n]\). Consider the polynomial \(1 \in k[X_1, \ldots, X_n]\). This polynomial cannot be represented as a linear combination of the generators of \(I\). Hence, \(1 \notin I\) and \(I\) is a proper ideal in \(k[X_1, \ldots, X_n]\).

To show that the quotient ring \(k[X_1, \ldots, X_n]/I\) is a field, we must prove that every nonzero element in the quotient ring has a multiplicative inverse. Let \(f+I \in k[X_1, \ldots, X_n]/I\) be a nonzero element, where \(I = (X_1-a_1, \ldots, X_n-a_n)\). Since \(f+I \neq 0+I\), we know that \(f\) is not divisible by any of the generators of \(I\). Now consider the evaluation homomorphism \(\phi: k[X_1, \ldots, X_n] \rightarrow k\) defined by \(\phi(f) = f(a_1, \ldots, a_n)\). Note that \(\phi(I) = 0\), so the kernel of \(\phi\) is a subset of \(I\). Since \(f \notin I\), we have \(\phi(f) \neq 0\). Therefore, \(f(a_1, \ldots, a_n) \in k^{\times}\), which means \(f(a_1, \ldots, a_n)\) has a multiplicative inverse in \(k\). Let \(c\) be the inverse of \(f(a_1,\ldots,a_n)\), i.e. \(c \cdot f(a_1, \ldots, a_n) = 1\). Define the constant polynomial \(g = c \in k[X_1, \ldots, X_n]\). It now follows that \((f+I)(g+I) = fg+I = cf+I = 1+I\) in the quotient ring \(k[X_1, \ldots, X_n]/I\). Thus, every nonzero element in \(k[X_1, \ldots, X_n]/I\) has a multiplicative inverse, and therefore \(k[X_1,\ldots,X_n]/I\) is a field, proving that \(I\) is a maximal ideal.

The natural homomorphism \(\psi: k \rightarrow k[X_1, \ldots, X_n]/I\) is defined by \(\psi(a) = a+I\) for all \(a \in k\). To prove that \(\psi\) is injective, let \(a, b \in k\) such that \(\psi(a) = \psi(b)\). By definition of the homomorphism, \(a+I = b+I\). Thus, \(a - b \in I\). However, since \(I\) is an ideal generated by polynomials of degree 1, only the constant polynomial can be in \(I\), which implies \(a - b = 0\). Therefore, \(a = b\), and the natural homomorphism \(\psi\) is injective.

To prove that the natural homomorphism \(\psi: k \rightarrow k[X_1, \ldots, X_n]/I\) is surjective, we need to show that for every element \(f+I\) in \(k[X_1, \ldots, X_n]/I\), there exists an element \(a \in k\) such that \(\psi(a) = f+I\). Let \(f+I \in k[X_1, \ldots, X_n]/I\). Since \(k[X_1, \ldots, X_n]/I\) is a field, there exists a constant polynomial \(g+I \in k[X_1, \ldots, X_n]/I\) such that \((f+I)(g+I) = 1+I\). Thus, \(fg - 1 \in I\). By definition of \(I\), we can write \(fg - 1\) as a linear combination of the generators: \(fg - 1 = h_1(X_1 - a_1) + \cdots + h_n(X_n - a_n),\) where \(h_i \in k[X_1, \ldots, X_n]\) for \(1 \leq i \leq n\). Considering the degree of the above equation, it's clear that \(f\) and \(g\) must be constant polynomials (in \(k\)). Let \(f = a\) and \(g = b\). If \(\psi(a) = a+I\), then we find that \(fg - 1 = 0+k[X_1,\ldots,X_n]\), implying \(\psi(a) = f+I\). This means the natural homomorphism is surjective. As the homomorphism is both injective and surjective, it is an isomorphism between \(k\) and \(k[X_1, \ldots, X_n]/I\).