Noetherianess of a locally noetherian affine scheme without axiom of choice I use the definition of a noetherian ring given by Qiaochu in this:
A commutative ring is noetherian if, for any nonempty collection of ideals $\mathcal{I}$, there is some $I \in \mathcal{I}$ which is not properly contained in any $J \in \mathcal{I}$. 

Can we prove the following theorem without using Axiom of Choice?

Theorem
Let $A$ be a commutative ring with unity.
Let $f_1,\cdots, f_n$ be elements of $A$ such that $A = (f_1,\cdots, f_n)$.
Suppose that each $A_{f_i}$ is noetherian.
Then $A$ is noetherian.

As for why I think this question is interesting, please see(particularly Pete Clark's answer):
Why worry about the axiom of choice?

Remark
Under the condition of the theorem, we can prove without the axiom of choice
that $A$ satisfies the ascending chain condition(see for example Hartshorne).
However, we need the axiom of dependent choice to prove that $A$ is noetherian in our sense using this fact.
Notation
Let $A$ be a commutative ring with unity.
Let $I$ be an ideal of $A$.
Let $f$ be an element of $A$.
We denote by $I_{(f)}$ the inverse image of $IA_f$ by the canonical homomorphism $A \rightarrow A_f$.
Namely $I_{(f)} = \{x \in A | \ $ There exists an integer $n \gt 0$ such that $f^n x \in I  \}$.

Let $\mathcal{I}$ be a nonempty collection of ideals of $A$.
We denote by $\mathcal{I}_{(f)}$ the set $\{I_{(f)}\ | \ I \in \mathcal{I} \}$.

Lemma 1
Let $A$ be a commutative ring with unity.
Let $f$ be an element of $A$.
Let $\mathcal{I}$ be a nonempty collection of ideals of $A$.
Let $J$ be an element of $\mathcal{I}$.
Suppose $JA_f$ is maximal in the set $\{IA_f \ |\ I \in \mathcal{I} \}$.
Then $J_{(f)}$ is maximal in $\mathcal{I}_{(f)}$.

Proof:
Let $I$ be an element of $\mathcal{I}$ such that $J_{(f)} \subset I_{(f)}$.
Then $J_{(f)}A_f \subset I_{(f)}A_f$.
Hence $JA_f \subset IA_f$.
By the assumption, $JA_f = IA_f$.
Hence $J_{(f)} = I_{(f)}$.
QED

Lemma 2
Let $A$ be a commutative ring with unity.
Let $f_1,\cdots, f_n$ be elements of $A$ such that $A = (f_1,\cdots, f_n)$.
Let $I, J$ be ideals of $A$ such that $I \subset J$.
Suppose $I_{(f_i)} = J_{(f_i)}$ for all $i$.
Then $I = J$.

Proof:
Let $x \in J$.
It suffices to prove that $x \in I$.
Since $J \subset J_{(f_i)}$ for all $i$, $J \subset I_{(f_i)}$ for all $i$, 
Hence there exists an integer $m \gt 0$ such that $f_i^m x \in I$ for all $i$.
By this question, we can prove without the axiom of choice that $A = (f_1^m,\cdots, f_n^m)$.
Hence there exists $g_1,\cdots, g_n \in A$ such that $1 = g_1f_1^m + \cdots + g_n f_n^m$.
Then $x = g_1f_1^m x + \cdots + g_n f_n^m x \in I$ as desired.
QED

Proof of the theorem
We define inductively sets of nonempty ideals $\mathcal{I}_0, \mathcal{I}_1, \cdots, \mathcal{I}_n$
and ideals $J_1, \cdots, J_n$ as follows.
We define $\mathcal{I}_0 = \mathcal{I}$.
Let $k$ be an integer such that $1 \le k \le n$.
Suppose $\mathcal{I}_{k-1}$ is defined.
Let $J_k$ be an element of $\mathcal{I}_{k-1}$ such that 
${J_k}_{(f_k)}$ is a maximal element of $\mathcal{I_{k-1}}_{(f_k)}$ which exists by Lemma 1.
We define $\mathcal{I}_k = \{I \in \mathcal{I}_{k-1} \ |\ I_{(f_k)} = {J_k}_{(f_k)}\}$.
We claim that $J_n$ is a maximal element of $\mathcal{I}$.
Let $I \in \mathcal{I}$ be such that $J_n \subset I$.

We first show by induction that $I \in \mathcal{I}_k$ for $k = 0, 1, \cdots, n$.
Since $\mathcal{I}_0 = \mathcal{I}$, $I \in \mathcal{I}_0$.
Let $k$ be an integer such that $1 \le k \le n$.
Suppose $I \in \mathcal{I}_{k-1}$.
Since $J_n \in \mathcal{I}_k$, ${J_n}_{(f_k)} = {J_k}_{(f_k)}$.
Since ${J_n}_{(f_k)} \subset I_{(f_k)}$, ${J_k}_{(f_k)} \subset I_{(f_k)}$.
Since $I \in \mathcal{I}_{k-1}$, ${J_k}_{(f_k)} = I_{(f_k)}$.
Hence $I \in \mathcal{I}_k$ as desired.

Next we prove that ${J_n}_{(f_k)} = I_{(f_k)}$ for all $k$.
Since $J_n \in \mathcal{I}_k$, ${J_n}_{(f_k)} = {J_k}_{(f_k)}$.
On the other hand, since $I \in \mathcal{I}_k$, ${J_k}_{(f_k)} = I_{(f_k)}$.
Hence ${J_n}_{(f_k)} = I_{(f_k)}$ as desired.

By Lemma 2, $J_n = I$.
Hence $J_n$ is a maximal element of $\mathcal{I}$.
QED
Notation
Let $A$ be a commutative ring with unity.
Let $I$ be an ideal of $A$.
Let $f$ be an element of $A$.
We denote by $I_{(f)}$ the inverse image of $IA_f$ by the canonical homomorphism $A \rightarrow A_f$.
Namely $I_{(f)} = \{x \in A | \ $ There exists an integer $n \gt 0$ such that $f^n x \in I  \}$.

Let $\mathcal{I}$ be a nonempty collection of ideals of $A$.
We denote by $\mathcal{I}_{(f)}$ the set $\{I_{(f)}\ | \ I \in \mathcal{I} \}$.

Lemma 1
Let $A$ be a commutative ring with unity.
Let $f$ be an element of $A$.
Let $\mathcal{I}$ be a nonempty collection of ideals of $A$.
Let $J$ be an element of $\mathcal{I}$.
Suppose $JA_f$ is maximal in the set $\{IA_f \ |\ I \in \mathcal{I} \}$.
Then $J_{(f)}$ is maximal in $\mathcal{I}_{(f)}$.

Proof:
Let $I$ be an element of $\mathcal{I}$ such that $J_{(f)} \subset I_{(f)}$.
Then $J_{(f)}A_f \subset I_{(f)}A_f$.
Hence $JA_f \subset IA_f$.
By the assumption, $JA_f = IA_f$.
Hence $J_{(f)} = I_{(f)}$.
QED

Lemma 2
Let $A$ be a commutative ring with unity.
Let $f_1,\cdots, f_n$ be elements of $A$ such that $A = (f_1,\cdots, f_n)$.
Let $I, J$ be ideals of $A$ such that $I \subset J$.
Suppose $I_{(f_i)} = J_{(f_i)}$ for all $i$.
Then $I = J$.

Proof:
Let $x \in J$.
It suffices to prove that $x \in I$.
Since $J \subset J_{(f_i)}$ for all $i$, $J \subset I_{(f_i)}$ for all $i$, 
Hence there exists an integer $m \gt 0$ such that $f_i^m x \in I$ for all $i$.
By this question, we can prove without the axiom of choice that $A = (f_1^m,\cdots, f_n^m)$.
Hence there exists $g_1,\cdots, g_n \in A$ such that $1 = g_1f_1^m + \cdots + g_n f_n^m$.
Then $x = g_1f_1^m x + \cdots + g_n f_n^m x \in I$ as desired.
QED

Proof of the theorem
We define inductively sets of nonempty ideals $\mathcal{I}_0, \mathcal{I}_1, \cdots, \mathcal{I}_n$
and ideals $J_1, \cdots, J_n$ as follows.
We define $\mathcal{I}_0 = \mathcal{I}$.
Let $k$ be an integer such that $1 \le k \le n$.
Suppose $\mathcal{I}_{k-1}$ is defined.
Let $J_k$ be an element of $\mathcal{I}_{k-1}$ such that 
${J_k}_{(f_k)}$ is a maximal element of $\mathcal{I_{k-1}}_{(f_k)}$ which exists by Lemma 1.
We define $\mathcal{I}_k = \{I \in \mathcal{I}_{k-1} \ |\ I_{(f_k)} = {J_k}_{(f_k)}\}$.
We claim that $J_n$ is a maximal element of $\mathcal{I}$.
Let $I \in \mathcal{I}$ be such that $J_n \subset I$.

We first show by induction that $I \in \mathcal{I}_k$ for $k = 0, 1, \cdots, n$.
Since $\mathcal{I}_0 = \mathcal{I}$, $I \in \mathcal{I}_0$.
Let $k$ be an integer such that $1 \le k \le n$.
Suppose $I \in \mathcal{I}_{k-1}$.
Since $J_n \in \mathcal{I}_k$, ${J_n}_{(f_k)} = {J_k}_{(f_k)}$.
Since ${J_n}_{(f_k)} \subset I_{(f_k)}$, ${J_k}_{(f_k)} \subset I_{(f_k)}$.
Since $I \in \mathcal{I}_{k-1}$, ${J_k}_{(f_k)} = I_{(f_k)}$.
Hence $I \in \mathcal{I}_k$ as desired.

Next we prove that ${J_n}_{(f_k)} = I_{(f_k)}$ for all $k$.
Since $J_n \in \mathcal{I}_k$, ${J_n}_{(f_k)} = {J_k}_{(f_k)}$.
On the other hand, since $I \in \mathcal{I}_k$, ${J_k}_{(f_k)} = I_{(f_k)}$.
Hence ${J_n}_{(f_k)} = I_{(f_k)}$ as desired.

By Lemma 2, $J_n = I$.
Hence $J_n$ is a maximal element of $\mathcal{I}$.
QED