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\(\textbf{Problem 6. }\)Consider the ring extension \(\mathbb{Z} \subseteq R\), where \(R := \mathbb{Z}[\sqrt[3]{2}]\) denotes the smallest subring of \(\mathbb{R}\) containing \(\mathbb{Z}\) and the real cube root of \(2\).
\(\textbf{Solution 1. }\) We want to find the kernel of the surjective ring homomorphism \(\phi : \mathbb{Z}[x] \to R\). Since \(\ker(\phi)\) is the set of all polynomials \(f(x)\) in \(\mathbb{Z}[x]\) such that \(f(\sqrt[3]{2}) = 0\), we have \[ \ker(\phi) := \{f(x) \in \mathbb{Z}[x] \mid f(\sqrt[3]{2}) = 0\}. \] We know that \(x^3 - 2\) is the minimal polynomial of \(\sqrt[3]{2}\) over \(\mathbb{Z}\). Since \(x^3 - 2\) is irreducible in \(\mathbb{Z}[x]\) by using the Eisenstein's criterion with \(p = 2\), we have \[ \ker(\phi) = \langle x^3 - 2\rangle. \] \(\textbf{Proof. }\) For one direction, we want to show that \(\langle x^3 - 2\rangle \subset \ker(\phi)\). For any \(f(x)\in \langle x^3 - 2\rangle\), we have \(f(x) = (x^3 - 2)g(x)\) for some \(g(x)\in \mathbb{Z}[x]\). Then we have \(f(\sqrt[3]{2}) = ((\sqrt[3]{2})^3 - 2)g(\sqrt[3]{2}) = 0\). Hence, we have \(f(x)\in \ker(\phi)\). For the other direction, we want to show that \(\ker(\phi) \subset \langle x^3 - 2\rangle\). Since \(x^3 - 2\) is irreducible in \(\mathbb{Z}[x]\), and we can know that there is no polynomial with degree less than \(3\) in \(\ker(\phi)\). For the other direction, suppose that \(g(x)\in \mathbb{Z}[x]\) such that \(g(\sqrt[3]{2}) = 0\). Since \(\mathbb{Z}\) is a Unique Factorization Domain by the Fundamental Theorem of Arithmetic, we can know that \(\mathbb{Z}[x]\) is a Unique Factorization Domain. Hence, \(g(x) = p_1(x)^{e_1}\cdots p_k(x)^{e_k}\) for some irreducible polynomials \(p_1(x), \ldots, p_k(x)\) in \(\mathbb{Z}[x]\). We know that \(g(\sqrt[3]{2}) = 0\) if and only if \(p_1(\sqrt[3]{2}) = 0\) or \(\ldots\) or \(p_k(\sqrt[3]{2}) = 0\) since \(\mathbb{Z}[x]\) is an integral domain. Without loss of generality, we can assume that \(p_i(\sqrt[3]{2}) = 0\). Since \(p_i(x)\in\mathbb{Z}[x]\) is irreducible, and \(\sqrt[3]{2}\) is a root of \(p_i(x)\), \(p_i(x)\) is the minimal polynomial of \(\sqrt[3]{2}\) over \(\mathbb{Q}\). We know that the minimal polynomial of \(\sqrt[3]{2}\) over \(\mathbb{Q}\) is \(x^3 - 2\), which is unique. Thus, we have \(p_i(x) = x^3 - 2\). Hence, we have \(x^3 - 2\mid g(x)\) and \(g(x)\in \langle x^3 - 2\rangle\). Therefore, we have \(\ker(\phi) \subset \langle x^3 - 2\rangle\). Hence, we have \(\ker(\phi) = \langle x^3 - 2\rangle\). \(\blacksquare\)
\(\textbf{Solution 2. }\) We want to find primes \(p, q\in \mathbb{Z}\) such that \(pR\) is a prime ideal and \(qR\) is not a prime ideal. Since \(\mathbb{Z}[\sqrt[3]{2}] = \{a + b\sqrt[3]{2} + c(\sqrt[3]{2})^2 \mid a, b, c\in \mathbb{Z}\}\), we have \(\sqrt[3]{2}, \sqrt[3]{2}^2\in R\). Since \(\sqrt[3]{2}\cdot\sqrt[3]{2}^2 = 2\), and \(2\not\mid \sqrt[3]{2}\) and \(2\not\mid \sqrt[3]{2}^2\), we can know that \(2\) is not prime in \(R\). Since \(pR\) is a prime ideal if and only if \(p\) is a prime in \(R\), we can know that \(2R\) is not a prime ideal in \(R\). Now, we want to show that \(7\) is still prime in \(R\). Since \(R = \mathbb{Z}[\sqrt[3]{2}]\), and \(R/\langle 7\rangle = \mathbb{Z}/7\mathbb{Z}[\sqrt[3]{2}]\). Now, we want to show that \(\mathbb{Z}/7\mathbb{Z}[\sqrt[3]{2}]\) is a field. Since \(\mathbb{Z}/7\mathbb{Z}\) is a field for \(7\) is a prime. Since there is no \(\alpha\in \mathbb{Z}/7\mathbb{Z}\) such that \(\alpha^3 - 2 = 0\). We can know that \(x^3 - 2\) is the minimal polynomial of \(\sqrt[3]{2}\) over \(\mathbb{Z}/7\mathbb{Z}\). Since \(x^3 - 2\) is irreducible in \(\mathbb{Z}/7\mathbb{Z}[x]\), we can know that \(\mathbb{Z}/7\mathbb{Z}[\sqrt[3]{2}]\) is a field. Hence, we can know that \(\langle 7\rangle\) is a maximal ideal, which implies it is a prime ideal. Hence, \(7\) is prime in \(R\).