MHB Noetherian Rings - R Y Sharp - Chapter 8 - exercise 8.5

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The discussion focuses on proving that the subring \(\mathbb{Z}[\sqrt{-5}]\) is Noetherian. It references Hilbert's basis theorem, which states that since \(\mathbb{Z}\) is Noetherian, the polynomial ring \(\mathbb{Z}[x]\) is also Noetherian. A canonical isomorphism is established between \(\mathbb{Z}[x]/(x^2 + 5)\) and \(\mathbb{Z}[\sqrt{-5}]\), allowing the conclusion that \(\mathbb{Z}[\sqrt{-5}]\) inherits Noetherian properties from the quotient. Thus, the exercise is successfully addressed by demonstrating the Noetherian nature of the subring. The solution strategy effectively utilizes established theorems in commutative algebra.
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I am reading R. Y. Sharp: Steps in Commutative Algebra, Chapter 5 - Commutative Noetherian Rings

Exercise 8.5 on page 147 reads as follows:

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8,5 Exercise.

Show that the subring \mathbb{Z} [ \sqrt{-5} ] of the field \mathbb{C} is Noetherian.

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Can someone please help me get started on this problem and give me a solution strategy.

Peter

[Note: This has also been posted on MHF]
 
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From Hilbert's basis theorem, we know that since $\Bbb Z$ is Noetherian, $\Bbb Z[x]$ is also Noetherian. There is a canonical isomorphism $\Bbb Z[x]/(x^2+5) \cong \Bbb Z[\sqrt{-5}]$ given by $x \mapsto \sqrt{-5}$. As Noetherianty is invariant under taking quotients, $\Bbb Z[x]/(x^2 + 5)$ is Noetherian, hence $\Bbb Z[\sqrt{-5}]$ is also Noetherian $\blacksquare$
 
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