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In this comment Noam Elkies remarked that it is possible to

construct an algebraic integer $\alpha$ in an appropriate polyquadratic extension such that $\alpha$ is not a unit and every factor of the norm of $\alpha$ must be split. (This is a natural generalization of the elementary proof that there are infinitely many primes congruent to $1 \mod 4$ , or to $1 \mod 3$ .)

Question: Could somebody elaborate the proposed construction of this $\alpha$ in more details and clarify in which sense it is generalizing the quoted "elementary" proof on infinitude of primes congruent to $1 \mod 4$ or to $1 \mod 3$ á la Elkies? I'm just wondering which techniques are going to be crucially involved there...

Also, I'm not completely sure to exactly which "elementary proof" Elkies refered there (...indeed there are several proofs using only elementary tools) on existence of infinitely many primes with prescribed congruence condition, but I guess it should be some kind of variant of Euclid's classical proof for existence of infinitely many primes; compare with proofs discussed here.
But still not sure in which sense the construction of algebraic number with described properties generalize it.

Ideas: Elkies suggested to consider polyquadratic extensions, so by def those with $[L:\Bbb Q]=2^d$ and such that $L$ is obtained by sequentially adjoining square roots, ie $L=\Bbb Q(\sqrt{a_1}, ..., \sqrt{a_{d-1}})$ with $ a_k \in \Bbb Q(\sqrt{a_1}, ..., \sqrt{a_{k-1}})$, which is of course Galois.

But although the above $L$ is what usually called "polyquadric" extension I strongly conjecture - as the proof should base of this prime counting result - that presumably Elkies intended to consider a bit more "elementary" $L'$, namely of shape $L'=\Bbb Q(\sqrt{a_1}, ..., \sqrt{a_{d-1}})$ with $a_i \in \Bbb Q$ (!) instead with $a_i$ appearing recursively as $a_k \in \Bbb Q(\sqrt{a_1}, ..., \sqrt{a_{k-1}})$.

The reason why I think so is because the explicit formulas for norms of elements of extensions of type $L$ are less controlable then those of type $L'$ with $a_i \in \Bbb Q$.

Edit/Update/ Refinement: I think that arguments bellow by Lepidopterist and Aphelli essentially answer the question , but there is still a little nitpick I not got resolved: The suggested approaches in comments suggest that the extension $L$ consisting such algebraic number $\alpha$ with desired properties can already be choosen to be quadratic extension, but Elkies mentioned the word "polyquadratic" suggesting that maybe one quadratic extension alone not suffice.

So the only unsolved - admitedly pedantic -rest of the question is if there is a serious reason why Elkies insisted that such algebraic $\alpha$ should live in "polyquadratic" extension and not only just "quadratic" extension (as $\Bbb Q(i)$ or $\Bbb(\sqrt{3})$). Is there a reason for making this requirement or is it just a red herring?

user267839
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    In $\mathbb{Z}[i]$ a prime splits iff it is $1$ modulo $4$. If we have a set of such primes $A={p_1,\dots,p_k}$, then $N(2p_1p_2\cdots p_k + i)=(2p_1p_2\cdots p_k)^2 + 1$. Only primes congruent to $1$ mod $4$ can divide this sum, but not a prime in $A$; therefore there are infinitely many such primes. You can do something similar in the integer ring of $\mathbb Q(\sqrt{-3})$ to show there are infinitely many primes congruent to $1$ modulo $3$. – Lepidopterist Mar 31 '25 at 20:46
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    In his comment, Elkies just needs infinitely many primes congruent to $1$ modulo some integer $N$. It is a standard exercise that any prime divisor of a value of the $N$-th cyclotomic polynomial either divides $N$ or is congruent to $1$ mod $N$. – Aphelli Mar 31 '25 at 22:28
  • @Aphelli: just to avoid confusions, in this exercise you are refering to, by "values of $N$-th cyclotomic polynomials" you mean only integer values, ie $\Phi_N(a), a \in \Bbb Z$, right? – user267839 Mar 31 '25 at 23:51
  • @user267839: yes, sorry for being unclear, this is what I meant. – Aphelli Apr 01 '25 at 00:12
  • @Aphelli: ah ok thanks for clarification, this appears plausible/doable ...for instance the statement is clear for $\Phi_N(1)$; for other integers will try next days, should be not so hard. But had before bad feeling on evaluating in fractions :) – user267839 Apr 01 '25 at 00:20
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    Brief update (for future readers): The discussion about classical result which Aphelli quotes about integer values of $N$-th cyclotomic poly can be found eg here. There is only adressed the case $N$ prime, but the arguments can be easily adapted to arbitrary $N$. – user267839 Apr 01 '25 at 21:59

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