Concrete instances of evaluation rings of ranking 2.

Allow $A$ be an evaluation ring of rank 2. After that $A$ offers an instance of a commutative ring such that $\mathrm{Spec}(A)$ is a noetherian topological room, yet $A$ is non-noetherian. (Without a doubt, or else $A$ would certainly be a distinct evaluation ring.)

Exists a concrete instance of such a ring $A$?

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2019-05-04 16:28:41
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Answers: 2

Take the ring of official power collection (over $\mathbb{C}$, claim) with backers in $\mathbb{Z}^2$ under lex order.

Modify : As Robin Chapman states, one have to take care concerning what this suggests. The specific building and construction for any kind of entirely gotten abelian team is defined in the Wikipedia article.

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2019-05-08 04:19:16
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Qiaochu's solution is audio in concept, yet in technique one requires to be extra mindful with the definition of the ring. The quotient area $K$ of $A$ contains official Laurent collection of the kind

$$f=\sum_{r=-r_0}^\infty x^r\sum_{s=-s_0(r)}^\infty a_{r,s}y^s.$$

Here $r_0$ is an integer and also for each and every integer $r$, $s_0(r)$ is an integer (relying on $r$ ). So these are the power collection where the powers of $x$ are bounded listed below and also for each and every integer $r$ the coefficient of $x^r y^s$ is absolutely no for all $s$ listed below a bound relying on $r$. This complicated-looking problem makes certain that the item of 2 components of $K$ is additionally a component of $K$ (note that can not increase 2 basic Laurent collection ).

After that $A$ will certainly contain all such collection with the added problems that $r_0=0$ and also $s_0(0)=0$. The evaluation of a component $f$ is the least $(r,s)$ under lexicographic getting with $a_{r,s}\ne0$. Below the getting is $(r,s)<(r',s')$ if $r < r'$ or $r=r'$ and also $s < s'$.

An even more high-brow analysis of the problem for memebership of $K$ is that the assistance of $f$, the setof $(r,s)$ for which $a_{r,s}\ne0$, need to be well-ordered, that is each part of the assistance has a the very least component. (With repsct to this lexicographic getting certainly. )

By taking into consideration a variation of this building and construction in $n$ variables one can construct clearly a ring with an evaluation of ranking $n$.

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2019-05-07 22:29:11
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