Proposition

If $A[x]$ is Noetherian, is $A$ necessarily Noetherian?

Solution

Yes.

Let $\phi: A[x] \rightarrow A$ be a ring homomorphism such that $a \mapsto a$ for all $a \in A$ and $x \mapsto 1$. This uniquely determines the homomorphism because every element in $A[x]$ is a polynomial in $x$ with coefficients in $A$. Moreover, $\phi$ is surjective because $\forall a \in A$, the constant polynomial $a$ is in $A[x]$.

By Proposition 7.1 [Atiyah], $A$ is Noetherian.