Let $S$ be the ring of Cauchy sequences of $\mathbb{Q}$, i.e. $S=\{(a_n)\in\mathbb{Q}^{\mathbb{N}}|(a_n)\, \text{is a Cauchy rational sequence in the ordinary distance} \}$, $S$ is a subring of $\mathbb{Q}^{\mathbb{N}}$. Denote $R$ to the ring $\mathbb{Q}^{\mathbb{N}}$.
Sorry. {Then $S$ is a local ring with the maximal ideal $\mathfrak{m}$ constisting of the sequences converging to zero. $S$ is a reduced ring, so the krull dimension of $S$ $>0$.} Sadly,it is wrong..:(
It is not Noetherian, since the ideal $\bigoplus \mathbb{Q}$ is not finitely generated.
……
Anyway, this ring is reduced, right? And $(1/n)_n\neq(1/n^2)_nx$ for any $x\in S$, it is not absolutely flat, so dim>0.
The obvious maximal ideals are : the maximal ideal $\mathfrak{m}_0$ consisting of the sequences converging to zero, ideal $\mathfrak{m}_i$ consisting of the elements whose $i$-th component is zero. That is to say $S/\mathfrak{m}_0\cong\mathbb{R}$ and $S/\mathfrak{m}_i\cong\mathbb{Q}$.
I am trying to find a non-maximal prime ideal explicitly. What is the krull dimension of $S$ ? More properties of $S$ ?
Is there maximal ideal $\mathfrak{p}$ such that $S/\mathfrak{p}\cong\mathbb{Q}[\sqrt{2}]$ or is isomorphic to any other intermediate field?
More observation, if $\mathfrak{m}$ is a maximal ideal of $R$ such that $R/\mathfrak{m}$ is an algebraic number field, then $\mathfrak{m}\cap S$ is a maximal ideal. However, not every maximal ideal of $S$ comes from $\operatorname{Spec}R$, e.g. $\mathfrak{m}_0$ is not in the image of $\operatorname{Spec}R$.
There are many many things here I am not clear, denote $\phi$ to the map from $\operatorname{Spec}R$ to $\operatorname{Spec}S$, is image $\phi$ contained in $\operatorname{MaxSpec}R$? What do the points in
$\operatorname{Spec}R$ look like? What is the set of the residue field at points in $\operatorname{Spec}R$ ? etc..
OK, I failed too, the maximal ideals of $R$ come from two way, one are ideals $\mathfrak{m}_i=\{(a_j)\in\prod \mathbb{Q}|a_i=0\}$, the other maximal ideal contains the ideal $\oplus\mathbb{Q}$. Let $\mathfrak{m}$ be a maximal ideal containing $\oplus\mathbb{Q}$, then the cardinality of $R/\mathfrak{m}$ will be $c$ the cardinality of continuum. That is to say the residue field of $R$ is either $\mathbb{Q}$ or a non-algebraic field extension of $\mathbb{Q}$.
Proposition Let $\mathfrak{m}$ be a maximal ideal of $S$ containing $\oplus\mathbb{Q}$, then the cardinality of $S/\mathfrak{m}$ is $c$.
Proof. Define map $f$ from the set $\{0,1\}^{\mathbb{N}}$ to $S/\mathfrak{m}$ by sending $(a_i)$ to $(\sum_{k=1}^ia_k/3^k)+\mathfrak{m}$. Then $f$ is injective. We are done.
So there is not prime ideal $\mathfrak{p}$ of $S$ such that $S/\mathfrak{p}\cong \mathbb{Q}[\sqrt{2}]$.
Right. As this post points out that it is easy to see that $\mathfrak{m}_0$ is the unique maximal ideal containing $\oplus\mathbb{Q}$, and the set of $\operatorname{MaxSpec}S$ consists of $\mathfrak{m}_0$ and $\mathfrak{m}_i$s.
Thus the proposition is trivial, but the method of the proof can be available for $R$.
Just for fun.
Thanks.
Best Answer
The map $\phi:$ Spec $R \to$ Spec $S$ is injective and dominant, but not surjective. For more details, see this answer on MO.