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.

  • 1
    $\begingroup$ 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
    Commented 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
    Commented 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
    Commented Nov 5, 2020 at 23:14

1 Answer 1


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 $\epsilon$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.