Although godelian has already answered, let me give a more direct answer (with more proofs instead of references; perhaps they coincide). First, notice that the existence of prime ideals in $R$ disjoint from multiplicative subsets $S$ and containing a given ideal $I$ is actually (quantified over $R$) equivalent to the existence of prime ideals in $R$ (quantified over $R$, of course $\neq 0$). The non-trivial direction just uses $S^{-1} (R/I)$. So we actually have only one statement, the existence of prime ideals.

I claim that the Compactness Theorem (for propositional logic) implies the existence of prime ideals: For each $a \in R$ let $p_a$ be a new variable. Consider the theory whose axioms are $p_0$, $\neg p_1$, $p_a \wedge p_b \longrightarrow p_{a+b}$, $p_a \longrightarrow p_{ab}$, $p_{ab} \longrightarrow p_{a} \vee p_{b}$ for all $a,b \in A$. A model of this theory is precisely a prime ideal of $R$. Since finitely generated rings are noetherian (Hilbert) and noetherian rings have maximal and therefore prime ideals, the theory is finitely consistent. Hence, it is consistent.

On the other hand, the Compactnass Theorem is weaker than the Axiom Choice.

PS: Now I've realized that the whole question is a dublicate of this one. See especially the answer by Chris Phan.

Although godelian has already answered, let me give a more direct answer (with more proofs instead of references; perhaps they coincide). First, notice that the existence of prime ideals in $R$ disjoint from multiplicative subsets $S$ and containing a given ideal $I$ is actually (quantified over $R$) equivalent to the existence of prime ideals in $R$ (quantified over $R$, of course $\neq 0$). The non-trivial direction just uses $S^{-1} (R/I)$. So we actually have only one statement, the existence of prime ideals.

I claim that the Compactness Theorem (for propositional logic) implies the existence of prime ideals: For each $a \in R$ let $p_a$ be a new variable. Consider the theory whose axioms are $p_0$, $\neg p_1$, $p_a \wedge p_b \longrightarrow p_{a+b}$, $p_a \longrightarrow p_{ab}$, $p_{ab} \longrightarrow p_{a} \vee p_{b}$ for all $a,b \in A$. A model of this theory is precisely a prime ideal of $R$. Since finitely generated rings are noetherian (Hilbert) and noetherian rings have maximal and therefore prime ideals, the theory is finitely consistent. Hence, it is consistent.

On the other hand, the Compactness Theorem is weaker than the Axiom Choice.

PS: Now I've realized that the whole question is a duplicate of this one. See especially the answer by Chris Phan.

Although godelian has already answered, let me give a more direct answer (with more proofs instead of references; perhaps they coincide). First, notice that the existence of prime ideals in $R$ disjoint from multiplicative subsets $S$ and containing a given ideal $I$ is actually (quantified over $R$) equivalent to the existence of prime ideals in $R$ (quantified over $R$, of course $\neq 0$). The non-trivial direction just uses $S^{-1} (R/I)$. So we actually have only one statement, the existence of prime ideals.

I claim that the Compactness Theorem (for propositional logic) implies the existence of prime ideals: For each $a \in R$ let $p_a$ be a new variable. Consider the theory whose axioms are $p_0$, $\neg p_1$, $p_a \wedge p_b \longrightarrow p_{a+b}$, $p_a \longrightarrow p_{ab}$, $p_{ab} \longrightarrow p_{a} \vee p_{b}$ for all $a,b \in A$. A model of this theory is precisely a prime ideal of $R$. Since finitely generated rings are noetherian (Hilbert) and noetherian rings have maximal and therefore prime ideals, the theory is finitely consistent. Hence, it is consistent.

On the other hand, the Compactnass Theorem is weaker than the Axiom Choice.

PS: Now I've realized that the whole question is a dublicate of this one. See especially the answer by Chris Phan.

Although godelian has already answered, let me give a more direct answer (with more proofs instead of references; perhaps they coincide). First, notice that the existence of prime ideals in $R$ disjoint from multiplicative subsets $S$ and containing a given ideal $I$ is actually (quantified over $R$) equivalent to the existence of prime ideals in $R$ (quantified over $R$, of course $\neq 0$). The non-trivial direction just uses $S^{-1} (R/I)$. So we actually have only one statement, the existence of prime ideals.

I claim that the Compactness Theorem (for propositional logic) implies the existence of prime ideals: For each $a \in R$ let $p_a$ be a new variable. Consider the theory whose axioms are $p_0$, $\neg p_1$, $p_a \wedge p_b \longrightarrow p_{a+b}$, $p_a \longrightarrow p_{ab}$, $p_{ab} \longrightarrow p_{a} \vee p_{b}$ for all $a,b \in A$. A model of this theory is precisely a prime ideal of $R$. Since finitely generated rings are noetherian (Hilbert) and noetherian rings have maximal and therefore prime ideals, the theory is finitely consistent. Hence, it is consistent.

On the other hand, the Compactnass Theorem is weaker than the Axiom Choice.

PS: Now I've realized that the whole question is a dublicate of this one. See especially the answer by Chris Phan.