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I have been trying to study the prime ideals of the ring $R:=\mathbb Z [\{ \sqrt{p_n}\}_{n=1}^\infty]$, where $p_n$ denotes the $n$-th prime. This is how far I got: I could conclude, by means of the first isomorphism theorem and the linear independence of square roots of distinct square-free numbers over $\mathbb Q$, the following isomorphism $$\mathbb Z[\{x_n\}_{n=1}^\infty] / (x_n^2 - p_n : n \in \mathbb N) \cong R.$$ Indeed, we consider the map $\phi: \mathbb Z[\{x_n\}_{n=1}^\infty] \longrightarrow R$ acting as identity on $\mathbb Z$ and sending $x_n$ to $\sqrt{p_n}$ for each natural number $n$. Then $\phi$ is clearly a surjective ring homomorphism and has kernel equal to $I := (x_n^2 - p_n : n \in \mathbb N)$: that this ideal is part of the kernel is easily checked whereas for any $f$ in the kernel, there must exist a natural number $N$ for which $f \in \mathbb Z [x_1, \cdots , x_n]$, whence by inductively performing long division by the polynomials $x_1^2-2, \cdots , x_N^2-p_N$, I could write $f(x_1, \cdots , x_N)$ in the form $g(x_1, \cdots , x_N) + h(x_1, \cdots , x_N)$ where $g(x_1, \cdots , x_N)$ is an element of the ideal $(x_1^2 - p_1, \cdots , x_N^2-p_N) \subset I$ and $h (x_1, \cdots , x_N)$ is a (finite) integer linear combination of $1$ and squarefree monomials in the indeterminates $x_1, \cdots , x_N$. Consequently, we have $h(\sqrt 2, \cdots , \sqrt{p_n}) = 0$, which by the well-known result of $\mathbb Q$-linear independence of the square roots of distinct squarefree positive integers allows us to conclude that each coefficient of $h$ must be zero whereupon $f \in (x_1^2 - p_1, \cdots , x_N^2-p_N) \subset I$ follows, thereby yielding $I = \ker \phi$. The first isomorphism theorem now yields the above isomophism.

As a consequence of this, we see that the prime ideals of $R$ are in one-to-one correspondence with the prime ideals of $\mathbb Z[\{x_n\}_{n=1}^\infty]$ containing $I$. Now I also did observe that by the above argument itself any ideal $\mathfrak p \vartriangleleft \mathbb Z[\{x_n\}_{n=1}^\infty]$ containing $I$ can be generated by $\{x_n^2-p_n : n \in \mathbb N\}$ and squarefree monomials in the variables $\{x_n\}_{n=1}^\infty$. I tried to investigate what other linear combinations of squarefree monomials could a prime ideal $\mathfrak p$ contain, but such an investigation has gotten nowhere as of yet.

More precisely what I want to study are localizations of the ring $R$ at its prime ideals. I know that ideals of the localization at $\mathfrak p$ are in one to one correspondence with those of $A$ saturted by $A \setminus \mathfrak p$. So it again seems to come down to studying (as a first step) the prime ideals of the polynomial ring $\mathbb Z [\{x_n\}_{n=1}^\infty]$ containing $I$. Has this ring $R$ been studied before? I would really appreciate any references, proofs, hints, suggestions or ideas.

Edit: I vaguely recall reading somewhere a long time ago that all localizations of $R$ at prime ideals should be principal, so I would at least like to be able to prove that statement. Unfortunately, I can't recollect the reference, but if I remember correctly, this was stated there as a fact without any justification whatsoever.

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    $\begingroup$ I don't believe that writing this ring as quotient of this huge complicated ring will help a lot except trivial facts (bijection between primes of $R/I$ and primes of $R$ containing $I$ etc.)... you could do the same with any countable commutative ring. $\endgroup$
    – YCor
    Nov 5, 2020 at 21:50
  • $\begingroup$ @YCor Yes, you may be right, I was hoping to get some non-trivial information by doing this (inspired from the much simpler standard problem of describing the prime ideals in the quotient ring $\mathbb Z[x_1, x_2, \cdots ]/(x_1, x_2^2, x_3^3, \cdots )$; of course here the situation is much more complicated in comparison to that but that is why I expected just a little bit of information). Also please see new edit, can you give me any idea about this? Thanks! $\endgroup$
    – asrxiiviii
    Nov 5, 2020 at 22:32
  • $\begingroup$ You can try to check that this is a domain of Krull dimension 1 and that every quotient field is locally finite (= algebraic over a finite field). I even believe (but am not sure) that every quotient field is finite. $\endgroup$
    – YCor
    Nov 5, 2020 at 23:14

1 Answer 1

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For each odd prime $p$, let $F(p)$ be a finite field of order $p^2$. (This is unique up to non-canonical isomorphism, by the standard theory of finite fields.) Now consider a prime $q\neq p$. If $q$ does not have a square root in $\mathbb{Z}/p$, then we can adjoin one to get a field of order $p^2$, which must be isomorphic to $F(p)$. It follows that in all cases $q$ has a square root in $F(p)$. Choose one such root and call it $\alpha_{pq}$. As $p$ is odd we note that the element $-\alpha_{pq}\in F(p)$ is different from $\alpha_{pq}$ and is the unique other square root of $q$ in $F(p)$.

  • The ideal $P(0)$ is prime
  • Now let $P(2)$ be generated by $\sqrt{2}$ together with $\sqrt{p}-1$ for all odd primes $p$. This is the kernel of an evident surjective homomorphism $R\to\mathbb{Z}/2$, so it is maximal (and thus prime).
  • Now let $p$ be an odd prime, and let $\epsilon$ be a system of signs $\epsilon_q$ for all primes $q\neq p$. We then have a homomorphism $\phi\colon R\to F(p)$ such that $\phi(\sqrt{p})=0$ and $\phi(\sqrt{q})=\epsilon_q\alpha_{pq}$ for all $q\neq p$. One can check that this is surjective, and we define $P(p,\epsilon)$ to be the kernel. This is again maximal (and therefore prime). It is not hard to see that $P(p,\epsilon)=P(p',\epsilon')$ iff $p=p'$ and $\epsilon$ and $\epsilon'$ are equivalent under the action of $\text{Aut}(F(p))=C_2$.

I claim that this gives all the prime ideals of $R$. To see this, let $R[k]$ be generated by the square roots of $p_1,\dotsc,p_k$ (so $R[0]=\mathbb{Z}$). Let $Q$ be a nonzero prime. Then $Q$ contains a nonzero element $a\in R[k]$ for some $k\geq 0$. If $k>0$ then we can write $a=b+c\sqrt{p_k}$ for some $a,b\in R[k-1]$, then the element $$ a'=(b+c\sqrt{p_k})(b-c\sqrt{p_k}) = b^2-p_kc^2 $$ is nonzero in $Q\cap R[k-1]$. By iterating this, we see that the group $Q\cap\mathbb{Z}=Q\cap R[0]$ is a nonzero ideal in $\mathbb{Z}$. In fact, this must be a prime ideal, so it must be generated by some prime number $p$. Now choose a prime $q_0\neq p$ that is not a square mod $p$, and let $F'$ be the subring of $R/Q$ generated by $\sqrt{q_0}$. This is a field of order $p^2$ and so is isomorphic to $F(p)$. It follows that for all $q\neq p$ the polynomial $x^2-q$ factors over $F'$, and thus that the image of $\sqrt{q}$ in $R/Q$ is actually in $F'$. It follows that $F'=R/Q$, so $Q$ is the kernel of a map $R\to F'\simeq F(p)$, so $Q=P(p,\epsilon)$ for some $\epsilon$.

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