Post Undeleted by David Speyer
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You

I will try to convince you that no Euclidean proof can possibly show that there are looking for the notion of divisionsinfinitely many primes which are $2 \mod 5$. After giving some definitions, I will explain what I will actually show:

Let $G$ be a finite group. Let $g$ be an element of $G$, with order $n$. We define an equivalence relation on $G$ by $g \sim h$ if there exists the (cyclic) subgroups generated by $f \in G$ g$and an integer$k$with$\mathrm{GCD}(k,n)=1$such that by$f g^k f^{-1} = h$are conjugate. (Exercise: this is an equivalence relation.) The equivalence classes for this relation are called divisions. Note Observe that this is a coarser map of groups always respects this equivalence relationthan conjugacy classes. My answer A more precise (but not yet precise) claim is:For $S \subseteq (\mathbb{Z}/N)^*$, we can prove, by "Euclidean" means, Consider any Euclidean proof that there are infinitely many primes$p \in S$if and only if some subset$S$contains a divison of $\mathbb{Z}/N^*$. Then the abelian group set$(\mathbb{Z}/N)^*$S$ contains a complete division.

Of course, since

Since $(\mathbb{Z}/N)^*$\mathbb{Z}/N^*$ is an abeliangroup, the conjugation conjugacy part of the definition of division is irrelevantabelianness doesn't come directly into this theorem, but it will arise in the proof. So, what is a Euclidean proof? Well, at some step in the proof, I include it because must build a prime$p$and say "$p$has the property$P$, and therefore$p$is in$S$." In every Euclidean proof I plan have seen,$P$is one of two things: (1) There is a number$M$such that$M$is not in some subgroup$G$of $(\mathbb{Z}/N)^*$. So$M$has a prime divisor which is not in$G$, and we choose$p$to tell be that prime factor. So an allowable step in a Euclidean-proof is showing that there are infinitely many primes not in some subgroup$G$of $(\mathbb{Z}/N)^*$ (2) Some polynomial$f(X)$factors in a particular way modulo$p$, and therefore$f$lies in some class modulo$N$. (Generally, the observation is that$f$has a root modulo$p$, but I'll allow more general storythings like $f$has a quadratic factor" as well. For a polynomial$(\mathbb{Z}/13)^*$, the divisions are f$ of degree $$\{1 \} \cup \{ 2,6,7,11 \} \cup \{ 3, 9 \} \cup \{ 4, 10 \} n, and a partition \lambda of n, let D(f, \cup lambda) be the set of primes such that the irreducible factors of f have degrees (\lambda_1 \{ 5,8 lambda_2, \} ldots, \cup lambda_r). And let D(f, \{ 12 lambda, N) be the image of D(f, \}.$$

I lambda)$modulo$N$. So an allowable step in an Euclidean-proof will present a Euclidean proof be showing that there are infinitely many primes in each divisionsome union of$D(f, \lambda, N)$'s. So, what I will then show actually be proving is Any subgroup$G$of $(\mathbb{Z}/N)^*$, and any$D(f, n, \lambda)$in $(\mathbb{Z}/N)^*$, is a union of divisions. So, suppose that certain kinds$S$is a set which does not contain any division, and let$T$be disjoint from$S$, but contain an element representing every division class in$S$. If we have a Euclidean proof, it will construct some prime$p$. We may know that$p$is not in various subgroups of arguments cannot $(\mathbb{Z}/N)^*$, or that$p$is in some union of$D(f, \lambda, N)$'s. But that information can't distinguish two primes whether$p$is in the same division$S$or in$T$, so our proof can't show that$p$is in$S$. OK, my last boxed claim is a precise statement. I want Let's prove it. It is easy to make the point see that a subgroup of $(\mathbb{Z}/N)^*$ is a union of divisions (since we are in an abelian group, while presenting this proof for the conjugacy part of the definition is irrelevant). The complement of a general subgroup is likewise such a union. The interesting thing is$(S, D(f, \lambda, N)$involves . Let$K$be a Galois theoryfield where$f$and$x^N-1$both split. Let$G$be the corresponding Galois group, presenting so$G$comes equipped with a map to $(\mathbb{Z}/N)^* \cong \mathrm{Gal}(\mathbb{Q}(\zeta_N)/\mathbb{Q})$. Now, as you probably know, for$p$a particular prime unramified in$S$and K$, the factorization of $N$ does f$modulo$p$is determined by the Frobenius conjugacy class of$p$in$G$. What you may or may not . Thereforeknow is that it is actually determined by the division class of$p$! (For example, I will run$x^5-1$has the same factorization modulo primes which are$2 \mod 5$and primes which are$3 \mod 5$.) To see thisargument both as I would explain it , just look at the recipe for reading the factorization off from the Frobenius class. It may or may not help to prove the following lemma: If$g \sim h$, and$X$is a professional mathematician set with a$G$action, then there is an order preserving bijection between the$g$and as the$h$orbits in$X$. So, the possible Frobenius classes in$G$of primes in$D(f, \lambda)$are unions of divisions (I would present it am implicitly using the Cebotarov density theorem here). But then$D(f, \lambda, N)$is just the projection to a PROMYS student who wanted $(\mathbb{Z}/N)^*$ of the possible Frobenius classes in$G$of primes in$D(f, \lambda)$. Since maps of groups take divisions to prove divisions, this shows that$D(f, \lambda, N)$is a union of divisions. Frobenius's density Theorem states that there were are infinitely many primes congruent to with Frobenius in every division. (And, more precisely, that their Dirichlet density is the size of the division divided by the order of$2$G$.) It is significantly easier than Cebatarov's, using only the material from a first course in algebraic number theory and a first course in analytic number theory.

Cebatarov's density theorem is the "union" of Frobenius's and Dirichlet's theorems. What I am suggesting is that Euclidean methods, at best, can only get at their intersection.

I am not sure whether or not I think Euclidean proofs can get that far. If you think they can, give me a Euclidean proof that there are infinitely many primes which are in $4$ modulo \{ 3, 5 \}$ mod$7$. For the professional: Let I can show infinitely many in $K = \{ 3,5,6 \mathbb{Q}(\zeta_7)$, the }$, and infinitely many in $7$th cyclotomic field. So \{ 2,3,4,5 \}$, but I can't get the Galois group of$K/\mathbb{Q}$is$intersection.

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