Can $A[x]$ be isomorphic to $A$ if $A$ is noetherian?

Question

Let $A$ be a nonzero noetherian commutative ring with one, and let $x$ be an indeterminate.

Can the rings $A[x]$ and $A$ be isomorphic?

Of course such a ring would have infinite Krull dimension, but noetherian rings of infinite Krull dimension are well known to exist: see this thread.

If $A[x]$ and $A$ are isomorphic then from $A[x] / (x) \cong A$, we get $A[x] / (x) \cong A[x]$ and this is impossible. — Mohammad Bagheri, Sep 28 '19 at 12:57
@MohammadBagheri - Why is $A[x]/(x)\cong A[x]$ impossible? Do you agree that $B[x_0,x_1,x_2,\dots]\cong B[x_1,x_2,\dots]$ (for any $B$)? In other words you must use the assumption that $A$ is noetherian. — Pierre-Yves Gaillard, Sep 28 '19 at 13:12
In fact, this question and the linked one follow immediately from the property of surjective endomorphisms of noetherian rings to be injective. — user26857, Sep 28 '19 at 17:30
@user26857 - I completely agree. Thanks for your intervention! — Pierre-Yves Gaillard, Sep 28 '19 at 17:34

Minseon Shin · Accepted Answer · 2019-09-28T16:11:39.270

Suppose there is a ring isomorphism $f : A \to A[x]$. Let $g : A[x] \to A$ be the $A$-algebra map sending $x \mapsto 0$; then the composition $\varphi = gf$ is a surjective ring automorphism of $A$ with nonzero kernel. Set $K_{n} := \ker \varphi^{n}$. Then $K_{1} \subseteq K_{2} \subseteq K_{3} \subseteq \dotsb$ is an ascending chain of ideals of $A$. It remains to show that $K_{n} \ne K_{n+1}$. This follows from induction on $n$, using that $K_{n} = \varphi^{-1}(K_{n-1})$ and $K_{n+1} = \varphi^{-1}(K_{n})$.

Can $A[x]$ be isomorphic to $A$ if $A$ is noetherian?

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