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

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
• You are right, that was a mistake. – Mohammad Bagheri Sep 28 '19 at 13:20
• 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

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 $$K_{n+1} = \varphi^{-1}(K_{n})$$.