The question whether there are infinitely many primes in some coprime residue class mod m
is stronger than asking whether there are infinitely many primes with given inertia index in the field of mth roots of unity, which is a special case of Chebotarev's density. The possibility of elementary proofs of the latter was discussed, as you know, over
here.
It seems that Dirichlet's technique is perfectly natural for this kind of questions. Before he started using Euler's ideas on zeta functions, he played around with Legendre's approach (Legendre had tried to prove the result in his 2nd edition of his Théorie des nombres since he had used special cases in his "proof" of the quadratic reciprocity law), but without success.
Another attempt at giving an "elementary" proof was made by Italo Zignago:
His proof was, however, incorrect.
BTW, question 1 goes back to the correspondence of Euler and Goldbach; they tried (in vain) to find a sequence of numbers divisible only by primes of the form $4n+3$. Eventually, Euler convinced himself that quadratic forms $x^2 + my^2$ won't do.
I will try to convince you that no Euclidean proof can possibly show that there are infinitely 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. We define an equivalence relation on $G$ by $g \sim h$ if the (cyclic) subgroups generated by $g$ and by $h$ are conjugate. The equivalence classes for this relation are called divisions. Observe that a map of groups always respects this equivalence relation. A more precise (but not yet precise) claim is:
Consider any Euclidean proof that there are infinitely many primes in some subset $S$ of
$\mathbb{Z}/N^*$. Then the set $S$ contains a complete division.
Since $\mathbb{Z}/N^*$ is abelian, the conjugacy part of the definition of abelianness 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 must build a prime $p$ and say "$p$ has the property $P$, and therefore $p$ is in $S$." In every Euclidean proof I 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 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 things like $f$ has a quadratic factor" as well. For a polynomial $f$ of degree $n$, and a partition $\lambda$ of $n$, let $D(f, \lambda)$ be the set of primes such that the irreducible factors of $f$ have degrees $(\lambda_1 \lambda_2, \ldots, \lambda_r)$. And let $D(f, \lambda, N)$ be the image of $D(f, \lambda)$ modulo $N$. So an allowable step in an Euclidean-proof will be showing that there are infinitely many primes in some union of $D(f, \lambda, N)$'s.
So, what I will 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 $S$ 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 $(\mathbb{Z}/N)^*$, or that $p$ is in some union of $D(f, \lambda, N)$'s. But that information can't distinguish whether $p$ is in $S$ or in $T$, so our proof can't show that $p$ is in $S$.
OK, my last boxed claim is a precise statement. Let's prove it.
It is easy to see that a subgroup of $(\mathbb{Z}/N)^*$ is a union of divisions (since we are in an abelian group, the conjugacy part of the definition is irrelevant). The complement of a subgroup is likewise such a union.
The interesting thing is $D(f, \lambda, N)$. Let $K$ be a Galois field where $f$ and $x^N-1$ both split. Let $G$ be the corresponding Galois group, 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 prime unramified in $K$, the factorization of $f$ modulo $p$ is determined by the Frobenius conjugacy class of $p$ in $G$. What you may or may not know is that it is actually determined by the division class of $p$! (For example, $x^5-1$ has the same factorization modulo primes which are $2 \mod 5$ and primes which are $3 \mod 5$.) To see this, 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 set with a $G$ action, then there is an order preserving bijection between the $g$ and the $h$ orbits in $X$.
So, the possible Frobenius classes in $G$ of primes in $D(f, \lambda)$ are unions of divisions (I am implicitly using the Cebotarov density theorem here). But then $D(f, \lambda, N)$ is just the projection to $(\mathbb{Z}/N)^*$ of the possible Frobenius classes in $G$ of primes in $D(f, \lambda)$. Since maps of groups take divisions to divisions, this shows that $D(f, \lambda, N)$ is a union of divisions.
Frobenius's density Theorem states that there are infinitely many primes with Frobenius in every division. (And, more precisely, that their Dirichlet density is the size of the division divided by the order of $G$.) 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 $\{ 3, 5 \}$ mod $7$. I can show infinitely many in $\{ 3,5,6 \}$, and infinitely many in $\{ 2,3,4,5 \}$, but I can't get the intersection.
Best Answer
This post started as some minor observations, but I believe it now contains a full proof that there is no such ring. Throughout, we let $R$ be an example of the kind wanted.
Observation 1: $R$ is indecomposable.
Proof of observation 1: Write $R=S_1\times S_2$, with $S_1,S_2\neq 0$. Suppose, by way of contradiction, that $S_1$ has finite characteristic $m>1$. Let $p$ be any integer prime divisor of $m$. Then $p=(p,p)$ is not irreducible (since $(1,0)(p,p)=0$), contradicting our assumption that the primes of $\mathbb{Z}$ stay irreducible in $R$ (and using your definition of irreducible, which includes not being a zero-divisor).
Thus, each $S_i$ has characteristic $0$. Moreover, the argument above shows that each integer prime cannot be a zero-divisor in any $S_i$.
If an integer prime $p$ is a unit in $S_i$, then $S_i$ contains a copy of $\mathbb{Z}[1/p]$, which is not finitely generated as a $\mathbb{Z}$-module, contradicting the fact that submodules of finitely generated $\mathbb{Z}$-modules are finitely generated.
Also, if $p$ is not irreducible in $S_1$ (or, by symmetry $S_2$), say $p=ab$ where $a,b$ are nonunits, then $(p,p)=(a,p)(b,1)$ is a product into nonunits, contradicting the irreducibility of the primes of $\mathbb{Z}$.
The elements $\{(p,1)\, :\, p>1\text{ prime}\}$ are not units, not zero-divisors, and are easily shown to be irreducible in $R$. They are not associate to each other, nor to the integer primes $(p,p)\in R$. Thus, this would contradict our assumption of having only finitely many new primes (up to associates).
Observation 2: If $r\in R$, then the subring $\mathbb{Z}[r]\subseteq R$ is isomorphic to $\mathbb{Z}[x]/I$ where $I$ is a principal ideal generated by a monic polynomial $f$.
Proof of observation 2: Since $\mathbb{Z}[r]$ is a $\mathbb{Z}$-submodule of $R$, it must be finitely generated. Hence, $r$ satisfies some monic polynomial. Let $f\in \mathbb{Z}[x]$ be the monic polynomial of smallest degree satisfied by $r$.
Now, $\mathbb{Z}[r]\cong \mathbb{Z}[x]/I$ where $I$ is the ideal of polynomials satisfied by $r$. It thus suffices to show that $I=f\mathbb{Z}[x]$. Supposing otherwise, let $g\in I-f\mathbb{Z}[x]$.
Working over $\mathbb{Q}[x]$ for a moment, take $d=\gcd(f,g)$, a monic polynomial. Since $f$ is monic, we know that $d\in \mathbb{Z}[x]$. By the extended Euclidean algorithm, we can write $d$ as a $\mathbb{Q}[x]$-linear combination of $f$ and $g$. Hence $cd\in I$ for some minimal $c\in \mathbb{Z}_{>0}$. We know $c\neq 1$ by the minimality of the degree of $f$ (since $d$ is monic).
Let $p$ be any prime dividing $c$. Then $p\cdot (c/p)d(r)=0$, and hence $p$ is a zero-divisor in $R$, which is a contradiction.
Observation 3: The monic polynomial $f$ is a power of an irreducible polynomial $q\in\mathbb{Z}[x]$.
Proof of observation 3: Assume $f$ is not a power of an irreducible, so $f=gh$ over $\mathbb{Z}[x]$, where $\gcd(g,h)=1$ and $\deg(g),\deg(h)\geq 1$.
By an argument used previously, we can find some constant $c\in \mathbb{Z}_{>0}$ that is a $\mathbb{Z}[x]$-linear combination of $g$ and $h$, say $c=gg'+hh'$.
Let $J=g\mathbb{Z}[x]$ and $K=h\mathbb{Z}[x]$. Let $S_1=\mathbb{Z}[x]/J$ and $S_2=\mathbb{Z}[x]/K$. Consider the map $\varphi\colon \mathbb{Z}[x]/I\to S_1\times S_2$ where $a+I\mapsto (a+J,a+K)$. This is an injective ring homomorphism. Thus, we can view $\mathbb{Z}[r]$ as a subring of this direct product. Moreover the image of $\mathbb{Z}[r]$ contains the elements $(c+J,0+K)$ and $(0+J,c+K)$. (Indeed, $h(r)h'(r)\mapsto (hh'+J,hh'+K) = (c+J,0+K)$.) Since the image of $\mathbb{Z}[r]$ contains $(1+J,1+K)$, we see (by a trivial use of Dirichlet's theorem) that the image of $\mathbb{Z}[r]$ contains elements of the form $(p+J,1+K)$ and $(1+J,p+K)$, with $p$ an integer prime. Thus (suppressing the coset notation) we have $(p,p)=(p,1)(1,p)$ in this image. Back inside $\mathbb{Z}[r]$, write this factorization as $p=ab$. Without loss of generality, $a$ is a unit in $R$, with inverse $a^{-1}$.
The ring $\mathbb{Z}[r,a^{-1}]$ is a finitely generated $\mathbb{Z}$-module (being a $\mathbb{Z}$-submodule of $R$), and it naturally maps onto $(\mathbb{Z}[x]/J)[1/p]$, which is not a finitely generated $\mathbb{Z}$-module, which is a contradiction.
Observation 4: $\deg(q)=1$.
Proof of observation 4: If $\deg(q)>1$, then there are infinitely many primes $p$ that factor nontrivially in $\mathbb{Z}[x]/(q)$ (by an argument supplied by John Voight) and such a factorization lifts to $\mathbb{Z}[x]/(q^n)\cong \mathbb{Z}[r]$; see this link for the full argument.
Now, if such a prime $p$ is to be irreducible in $R$, one of those factors must become a unit $u$ in $R$. But then $\mathbb{Z}[r,u^{-1}]$ is a commutative ring, mapping to $(\mathbb{Z}[x]/(q))[a^{-1}]$ (where $a$ is that corresponding nontrivial factor, but modulo $q$ rather than $q^n$), which is not a finitely generated $\mathbb{Z}$-module, giving us a contradiction, as before.
Observation 5: If $J$ is the set of nilpotent elements in $R$, then $J$ is a (nilpotent) ideal.
Proof of observation 5: An arbitrary element $r\in R$ satisfies $q^n$, with $q(x)=x-k\in \mathbb{Z}[x]$, and hence $r=k-t$ where $t$ is nilpotent of index $n$. Therefore, $r(k^{n-1}+k^{n-2}t + \cdots + t^{n-1})=k^n$. If $k\neq 0$, then $k^n$ is not a zero-divisor (since all the primes of $\mathbb{Z}$ are not zero divisors), and hence $r$ is not a zero-divisor. Thus, the zero-divisors are exactly the nilpotent elements.
Next, if $t\in R$ is nilpotent, say of index $n\geq 1$, and $r\in R$ is arbitrary then $t^{n-1}(tr)=0$ with $t^{n-1}\neq 0$. Hence $tr$ is a zero-divisor, hence nilpotent. Thus $J$ is closed by right multiplication from $R$; and by left multiplication by a symmetric argument.
In particular $J$ is closed under multiplication. By the paper Rings in which nilpotents form a subring by Janez Ster (arXiv version here), we know that $J$ is also closed under addition, since $R$ satisfies Koethe's conjecture. (Quick argument: The Jacobson radical $J(R)$ is nilpotent, since tensoring up to $\mathbb{Q}$ keeps it in a finite-dimensional $\mathbb{Q}$-algebra.) Thus, $J$ is an ideal.
Observation 6: The type of ring we want cannot exist.
Proof of observation 6: Fix a $\mathbb{Z}$-basis for $J$, say $t_1,\ldots, t_k$. Then a $\mathbb{Z}$-basis for $R$ is $1,t_1,\ldots, t_k$ (since every element of $R$ is an integer shift from a nilpotent, by observation 4).
The units of $R$ are exactly $\pm 1+t$ for some $t\in J$. Thus, the associates of a prime $p\in \mathbb{Z}$ are of the form $\pm p+t'$ where $t'\in J$ is divisible by $p$. Thus $p+t_1$ is not associate to any of the integer primes. A similar argument shows that $p+t_1$ and $p'+t_1$ are not associate, for distinct primes $p,p'$. Further, $p+t_1$ is not a unit, and not a nilpotent (hence not a zero-divisor). It thus suffices to show that $p+t_1$ is irreducible, and then we'll have infinitely many "new" nonassociate irreducibles.
If $p+t_1$ factors as $(a+t)(b+t')$ with $a,b\in \mathbb{Z}$ and $t,t'\in J$, then $ab=p$. Hence, without loss of generality, $a=\pm 1$. Thus, $a+t$ is a unit, so every factorization is trivial.