TITLE: Existence of prime ideals and Axiom of Choice.
QUESTION [11 upvotes]: One of the must obvious equivalences of Axiom of Choice is the converse of Krull Theorem.
Bernhard Banaschewski in the Article titled by A New Proof that “Krull implies Zorn” showed a very simple proof of the following statement.
Theorem: the existence of maximal ideals in a ring with unity is equivalent to Axiom of choice.
This means that every attempt to prove the existence of maximal ideals is related to apply the  Axiom of Choice.
Another important theorem in commutative algebra is  Cohen's theorem, which tells us that if $R$ is a commutative ring with unity and  $I$ is an ideal of $R$ disjoint from a multiplicative closed subset $S\subset R$, then there exists a prime ideal $P$ so that $I \subset P$ and $P\cap S=\varnothing$. 
Cohen's theorem implies that In a commutative ring with unity there exists a prime ideal.
Notice that this prime ideal need not be a maximal ideal but we need to apply Zorn's Lemma to show the existence of it. Now Here are my Questions:

Is it true that  For Showing the existence of prime ideal in a commutative ring with unity we need the Axiom of choice or we can show the existence of it without applying this Axiom?
If the Answer of above Question is negative, what kind of Axiom weaker than Axiom of choice is needed to show the existence of prime ideals in a commutative ring?
What kind of relation is between the Axiom of countable choice and The existence of prime ideals in a commutative ring with unity?

REPLY [2 votes]: In light of Laurent Moret-Bailly's comment under Martin's answer, let us supplement that answer by showing how existence of prime ideals in finitely generated rings may be proven in ZF without a choice principle (not even the ultrafilter principle). In fact we will show that countable rings $R$ have prime ideals, using the weak König lemma. 
(Recall the weak König lemma: if $T$ is an infinite subtree of the infinite binary tree $2^\ast$ [the nodes of $2^\ast$ are finite words in letters $0, 1$; the parent of any nonempty word is obtained by omitting its last letter], then $T$ has an infinite branch. This is easily proven in ZF; see e.g. here.)  
Fix an enumeration $a_0, a_1, a_2, \ldots$ of elements of $R$. Build a tree $T$ whose nodes $w$ are labeled by ideals $P_w$ of $R$, by induction. $T_k$ will denote the set of nodes of length $k$, so $T_0$ consists of just the empty word $e$, which we label with the zero ideal. Assuming $T_{k-1}$ and the ideals $P_w$ for words $w \in T_{k-1}$ are given, define $T_k$ and the ideals there according to two cases; for this it is convenient to write $k = 2l + m$ where $m \in \{0, 1\}$ and $l$ is used to code a pair of natural numbers $(i, j)$ by the usual trick, viz. $l = \binom{i+j+1}{2} + j$. 
Case $m = 0$: for each $w \in T_{k-1}$ such that $a_i a_j$ belongs to $P_w$, put $w0, w1$ in $T_k$, and define $P_{w0} := P_w + \langle a_i \rangle$, and $P_{w1} := P_w + \langle a_j \rangle$. For any other $w \in T_{k-1}$, put $w0$ in $T_k$, and define $P_{w0} = P_w$. 
Case $m = 1$: for each $w \in T_{k-1}$ such that $1 \notin P_w$, put $w0$ in $T_k$, and define $P_{w0} := P_w$. If $1 \in P_w$, then put neither $w0$ nor $w1$ in $T_k$. 
$T$ is infinite: it is easy to prove by induction that for each $k$ there is $w \in T_k$ such that $1 \notin P_w$. If $b$ is an infinite branch through $T$, then put $P = \bigcup_{w \in b} P_w$. Then $P$ is a proper ideal by the recipe of case $m=1$, and is prime by the recipe of case $m=0$.