# Prime ideals and localizations of the ring $\mathbb Z[\{\sqrt p: p \text{ prime}\}]$

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.

• 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.
– YCor
Nov 5, 2020 at 21:50
• @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! Nov 5, 2020 at 22:32
• 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.
– YCor
Nov 5, 2020 at 23:14

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 $$R=\mathbb{Z}$$). 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 $$Q\cap\mathbb{Z}=Q\cap R$$ 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 $$\epsilon$$.