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  1. Is $\mathbb{Z}[\sqrt{17}]$ a Noetherian domain?
  2. Is $\mathbb{C}[x^2, x^3]$ a Dedekind domain?

Can't seem to make any headway with these. Any help appreciated

2 Answers 2

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  1. $\mathbb{Z}[\sqrt{17}]$ is isomorphic to the quotient of $\mathbb{Z}[x]$ modulo the ideal $(x^2-17)$. Is $\mathbb{Z}[x]$ Noetherian? Is a quotient of a Noetherian ring a Noetherian ring?

  2. Dedekind domains are integrally closed; can you find a monic polynomial with coefficients in $\mathbb{C}[x^2,x^3]$ that has solutions in $\mathbb{C}(x^2,x^3)$ but not in $\mathbb{C}[x^2,x^3]$?

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    @Dylan: Yes... which would show that it is Noetherian as a $\mathbb{Z}$-module; and then argue that this implies that it is Noetherian as a ring as well (which is true, but you do have to "jump between structures").2012-02-23
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  1. $\mathbb {Z}[\sqrt{17}]$ is a Noetherian ring, as $\mathbb Z$ is clearly Noetherian, so by Hilbert's basis theorem $\mathbb Z[x]$ is, and since the homomorphic image of a Noetherian ring is Noetherian we have that $\mathbb{Z}[x]/(x^2-17)\cong \mathbb Z[\sqrt{17}]$ is Noetherian. Additionally, $(x^2-17)$ is a prime ideal (why?), so $\mathbb {Z}[\sqrt{17}]$ is a domain.

  2. $\mathbb C[x^2,x^3]$ is not integrally closed, so not a Dedekind domain. I'll let you find an example of an element integral over it but not in it.

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    another way to see that $\mathbb{Z}[\sqrt{17}]$ is a domain is to note that it is a subring of $\mathbb{R}$ which is a field, and hence a domain. A subring of an integral domain is an integral domain.2012-02-23