Fix a number field $K$. For an integer $m$, let $S_1(m,K)$ be the congruence classes $a$ mod $m$ which contain infinitely many primes $p$ such that $p \mid P$ for some prime $P$ of $K$ satisfying $f(P \mid p) = 1$. (That was a mouthful: $p$ is lying below some prime of $K$ with residue field degree 1.) If $K/\Bbb Q$ is Galois, then such $p$ are the primes splitting completely in $K$, up to finitely many exceptions (among the ramified primes). That is, when $K/\Bbb Q$ is Galois, $S_1(m,K)$ is the set of congruence classes mod $m$ containing infinitely many primes which split completely in $K$. (The prime numbers which split completely in a number field are identical to the prime numbers which split completely in its Galois closure over $\Bbb Q$, so attempting to describe such "split sets" by congruence conditions could just as well assume the number field is Galois over $\Bbb Q$. I am working over base field $\Bbb Q$ throughout.)

As Kevin has suggested, it is not obvious at first that these sets $S_1(m,K)$ have much structure, particularly that they contain $1$ mod $m$. By the pigeonhole principle, any $S_1(m,K)$ is certainly a nonempty set, and it is a subset of the unit group $(\Bbb Z/m)^*$ rather than just $\Bbb Z/m$, but this is kind of superficial.
A good reason (the right reason?) that $1$ mod $m$ is in $S_1(m,K)$ is that $S_1(m,K)$ is actually a subgroup of $(\Bbb Z/m)^*$. In fact, under the usual identification of $\mathrm{Gal}(\Bbb Q(\zeta_m)/ \Bbb Q)$ with $(\Bbb Z/m)^*$, $S_1(m,K)$ is the image of the restriction homomorphism $\mathrm{Gal}(K(\zeta_m)/K) \longrightarrow \mathrm{Gal}(\Bbb Q(\zeta_m)/\Bbb Q)$. For a proof, see Theorem 3 at

http://www.math.uconn.edu/~kconrad/blurbs/gradnumthy/dirichleteuclid.pdf

and Theorem 4 there is a generalization where $(\Bbb Z/m)^*$ is replaced with any Galois group of number fields.

Fix a number field $K$. For an integer $m$, let $S_1(m,K)$ be the congruence classes $a$ mod $m$ that contain infinitely many primes $p$ such that $p \mid \mathfrak p$ for some prime $\mathfrak p$ of $K$ satisfying $f(\mathfrak p|p) = 1$. (That was a mouthful: $p$ is lying below some prime of $K$ with residue field degree 1.) If $K/\mathbf Q$ is Galois, then such $p$ are the primes splitting completely in $K$, up to finitely many exceptions (among the ramified primes). That is, when $K/\mathbf Q$ is Galois, $S_1(m,K)$ is the set of congruence classes mod $m$ containing infinitely many primes which split completely in $K$. (The prime numbers that split completely in a number field are identical to the prime numbers that split completely in its Galois closure over $\mathbf Q$, so attempting to describe such "split sets" by congruence conditions could just as well assume the number field is Galois over $\mathbf Q$. I am working over base field $\mathbf Q$ throughout.)

As Kevin has suggested, it is not obvious at first that these sets $S_1(m,K)$ have much structure, particularly that they contain $1$ mod $m$. By the pigeonhole principle, any $S_1(m,K)$ is certainly a nonempty set, and it is a subset of the unit group $(\mathbf Z/m)^\times$ rather than just $\mathbf Z/m$, but this is kind of superficial.

A good reason (the right reason?) that $1$ mod $m$ is in $S_1(m,K)$ is that $S_1(m,K)$ is actually a subgroup of $(\mathbf Z/m)^\times$. In fact, under the usual identification of $\mathrm{Gal}(\mathbf Q(\zeta_m)/ \mathbf Q)$ with $(\mathbf Z/m)^\times$, $S_1(m,K)$ is the image of the restriction homomorphism $\mathrm{Gal}(K(\zeta_m)/K) \longrightarrow \mathrm{Gal}(\mathbf Q(\zeta_m)/\mathbf Q)$. For a proof, see Theorem 3 at

http://www.math.uconn.edu/~kconrad/blurbs/gradnumthy/dirichleteuclid.pdf

and Theorem 4 there is a generalization where $(\mathbf Z/m)^\times$ is replaced with any Galois group of number fields.

Fix a number field K. For an integer m, let S_1(m,K) be the congruence classes a mod m which contain infinitely many primes p such that p|P for some prime P of K satisfying f(P|p) = 1. (That was a mouthful: p is lying below some prime of K with residue field degree 1.) If K/Q is Galois, then such p are the primes splitting completely in K, up to finitely many exceptions (among the ramified primes). That is, when K/Q is Galois, S_1(m,K) is the set of congruence classes mod m containing infinitely many primes which split completely in K. (The prime numbers which split completely in a number field are identical to the prime numbers which split completely in its Galois closure over Q, so attempting to describe such "split sets" by congruence conditions could just as well assume the number field is Galois over Q. I am working over base field Q throughout.)

As Kevin has suggested, it is not obvious at first that these sets S_1(m,K) have much structure, particularly that they contain 1 mod m. By the pigeonhole principle, any S_1(m,K) is certainly a nonempty set, and it is a subset of the unit group (Z/m)* rather than just Z/m, but this is kind of superficial. A good reason (the right reason?) that 1 mod m is in S_1(m,K) is that S_1(m,K) is actually a subgroup of (Z/m). In fact, under the usual identification of Gal(Q(zeta_m)/Q) with (Z/m) , S_1(m,K) is the image of the restriction homomorphism Gal(K(zeta_m)/K) ---> Gal(Q(zeta_m)/Q). For a proof, see Theorem 3 at

http://www.math.uconn.edu/~kconrad/blurbs/gradnumthy/dirichleteuclid.pdf

and Theorem 4 there is a generalization where (Z/m)* is replaced with any Galois group of number fields.

Fix a number field $K$. For an integer $m$, let $S_1(m,K)$ be the congruence classes $a$ mod $m$ which contain infinitely many primes $p$ such that $p \mid P$ for some prime $P$ of $K$ satisfying $f(P \mid p) = 1$. (That was a mouthful: $p$ is lying below some prime of $K$ with residue field degree 1.) If $K/\Bbb Q$ is Galois, then such $p$ are the primes splitting completely in $K$, up to finitely many exceptions (among the ramified primes). That is, when $K/\Bbb Q$ is Galois, $S_1(m,K)$ is the set of congruence classes mod $m$ containing infinitely many primes which split completely in $K$. (The prime numbers which split completely in a number field are identical to the prime numbers which split completely in its Galois closure over $\Bbb Q$, so attempting to describe such "split sets" by congruence conditions could just as well assume the number field is Galois over $\Bbb Q$. I am working over base field $\Bbb Q$ throughout.)

As Kevin has suggested, it is not obvious at first that these sets $S_1(m,K)$ have much structure, particularly that they contain $1$ mod $m$. By the pigeonhole principle, any $S_1(m,K)$ is certainly a nonempty set, and it is a subset of the unit group $(\Bbb Z/m)^*$ rather than just $\Bbb Z/m$, but this is kind of superficial.
A good reason (the right reason?) that $1$ mod $m$ is in $S_1(m,K)$ is that $S_1(m,K)$ is actually a subgroup of $(\Bbb Z/m)^*$. In fact, under the usual identification of $\mathrm{Gal}(\Bbb Q(\zeta_m)/ \Bbb Q)$ with $(\Bbb Z/m)^*$, $S_1(m,K)$ is the image of the restriction homomorphism $\mathrm{Gal}(K(\zeta_m)/K) \longrightarrow \mathrm{Gal}(\Bbb Q(\zeta_m)/\Bbb Q)$. For a proof, see Theorem 3 at

http://www.math.uconn.edu/~kconrad/blurbs/gradnumthy/dirichleteuclid.pdf

and Theorem 4 there is a generalization where $(\Bbb Z/m)^*$ is replaced with any Galois group of number fields.