I am trying to solve the following problem as part of a question from the Fall 2021 UCLA algebra qualifying exam: is the fiber product $\mathbb{Q} \times_{\mathbb{Q}[x]/(x^2-x)} \mathbb{Q}[x]$ in the category of commutative $\mathbb{Q}$ algebras Noetherian, where $\mathbb{Q} \rightarrow \mathbb{Q}[x]/(x^2-x)$ is the inclusion and $\mathbb{Q}[x] \rightarrow \mathbb{Q}[x]/(x^2-x)$ is the projection?
This problem leads me to the following question: is it known when fiber products of rings are Noetherian? More specifically, is it known when fiber products of Noetherian algebras over a field are Noetherian? Specifically, I'm looking for an answer to this question which can hint towards a solution to the above problem.
The fiber product is $\{(a, b) : b - a \in (x^2 - x)\}$. First note that $b$ completely determines $a$ since $\mathbb{Q} \rightarrow \mathbb{Q}[x] / (x^2-x)$ is injective. Write $b$ as $p (x^2-x) = q$, with $\deg(q) = 1$. Then, we must have $q = a$. Furthermore, note that $q$ cannot have degree 1.
Therefore, we are looking at $\{b \in \mathbb{Q}[x] : b \bmod x^2-x \text{ has degree 0}\}$. Warning: a subring of a Noetherian ring might not be Noetherian, see here.