Let \[\begin{align*} R=\mathbb Z[x]/(x^2+x+1). \end{align*}\]
Answer the following questions with suitable justification.
Is \(R\) a Noetherian ring?
Is \(R\) an Artinian ring?
Prove that \(R\) is an integrally closed domain.
Solution
1. The ring\[ R=\mathbb Z[x]/(x^2+x+1) \]is a Noetherian integral domain.
The ring \(\mathbb Z\) is Noetherian, so by the Hilbert basis theorem \(\mathbb Z[x]\) is Noetherian. Every quotient of a Noetherian ring is Noetherian, hence \(R\) is Noetherian.
Also \(x^2+x+1\) is irreducible over \(\mathbb Q\), since its discriminant is \(-3\), not a square in \(\mathbb Q\). Being primitive, it is irreducible in \(\mathbb Z[x]\) by Gauss’s lemma. Since \(\mathbb Z[x]\) is a UFD, irreducible elements are prime, so \((x^2+x+1)\) is prime and the quotient \(R\) is a domain.
2. The ring \(R\) is not Artinian.
The image of \(2\) in \(R\) is not a unit: indeed, \[ R/(2)\cong \mathbb F_2[x]/(x^2+x+1) \] is a nonzero ring. Consider the descending chain of ideals \[ (2)\supseteq(2^2)\supseteq(2^3)\supseteq\cdots. \] Each inclusion is strict. If \((2^n)=(2^{n+1})\), then \[ 2^n=2^{n+1}r \] for some \(r\in R\). Since \(R\) is a domain by 1, cancellation gives \(1=2r\), contradicting that \(2\) is not a unit. Thus the descending chain condition fails, so \(R\) is not Artinian.
3. Let\[ \omega=\frac{-1+\sqrt{-3}}2. \]Then\[ R\cong\mathbb Z[\omega], \qquad \operatorname{Frac}(R)=\mathbb Q(\sqrt{-3}). \]
The element \(\omega\) satisfies \[ \omega^2+\omega+1=0. \] Since this polynomial is irreducible over \(\mathbb Q\), evaluation at \(\omega\) induces an injective homomorphism \[ \mathbb Z[x]/(x^2+x+1)\longrightarrow\mathbb C \] with image \(\mathbb Z[\omega]\). Its fraction field is \[ \mathbb Q(\omega)=\mathbb Q(\sqrt{-3}). \]
4. Every element of \(\mathbb Q(\sqrt{-3})\) that is integral over \(\mathbb Z\) lies in \(\mathbb Z[\omega]\).
Let \(\alpha\in\mathbb Q(\sqrt{-3})\) be integral over \(\mathbb Z\). Its conjugate \(\bar\alpha\) is also integral, so its trace and norm \[ T=\alpha+\bar\alpha, \qquad N=\alpha\bar\alpha \] are rational algebraic integers and therefore integers.
Write \[ \alpha=\frac{T+q\sqrt{-3}}2 \] for some \(q\in\mathbb Q\). Since \[ T^2-4N=(\alpha-\bar\alpha)^2=-3q^2, \] we have \(3q^2\in\mathbb Z\). Write \(q=a/b\) in lowest terms. Then \(b^2\mid3a^2\); coprimality gives \(b^2\mid3\), hence \(b=1\). Thus \(q\in\mathbb Z\).
Moreover, \[ 4N=T^2+3q^2. \] Modulo \(4\), this forces \(T\) and \(q\) to have the same parity: if exactly one were odd, the right side would be congruent to \(1\) or \(3\) modulo \(4\). Hence \[ \alpha =\frac{T+q\sqrt{-3}}2 =\frac{T+q}{2}+q\omega\in\mathbb Z[\omega]. \]
5. The domain \(R\) is integrally closed.
Let \(\alpha\in\operatorname{Frac}(R)\) be integral over \(R\). By 3, \[ \operatorname{Frac}(R)=\mathbb Q(\sqrt{-3}). \] The ring \(R=\mathbb Z[\omega]\) is integral over \(\mathbb Z\) because \(\omega\) satisfies the monic polynomial \(x^2+x+1\). Since \(\alpha\) is integral over \(R\), transitivity of integrality implies that \(\alpha\) is integral over \(\mathbb Z\). By 4, \(\alpha\in\mathbb Z[\omega]=R\). Therefore \(R\) is integrally closed.