$\operatorname{Aut_{I}(R)} = \operatorname{End_{I}(R) \cap Aut(R)}$ is a group or not?

Question

I found an answer for part $(c)$ of this question:

Let $R$ be a ring and $I \subset R$ a two-sided ideal, with quotient homomorphism $\pi : R \rightarrow R/I.$ Let $\operatorname{End_{I}(R)}$ be the set of $\varphi \in \operatorname{End(R)}$ such that $\varphi(I) \subset I,$ and let $\operatorname{Aut_{I}(R)} = \operatorname{End_{I}(R) \cap Aut(R)}.$

$(a)$ Given $\varphi \in \operatorname{End_{I}(R)},$ show that there exists $\bar{\varphi} \in \operatorname{End(R/I)}$ such that $\bar{\varphi} \pi = \pi \varphi.$

$(b)$ Given $\varphi, \psi \in \operatorname{End_{I}(R)},$ show that $\overline{\varphi \psi} = \bar{\varphi} \bar{\psi}.$

$(c)$ Given $\varphi \in \operatorname{Aut_{I}(R)},$ show that $\varphi(I) = I$ and $\bar{\varphi} \in \operatorname{Aut(R/I)}.$\ Hint: Use part $(b).$

But as can be seen from this link why $(\varphi \circ \varphi^{-1})(I) = I$ implies that $\varphi^{-1}(I) \subset I$? it seems like the answer is only correct iff $\operatorname{Aut_{I}(R)} = \operatorname{End_{I}(R) \cap Aut(R)}$ is a group and I was told that in general it is not a group and it is a group only if $R$ is a Noetherian ring, is this claim correct? can anyone give me an example showing that it is not a group in general?

Answer 1

What we need is an example of a non-Noetherian ring with a ring automorphism that "shifts" an infinite ascending chain of ideals, i.e., maps each ideal to the next in the sequence. Let's take a polynomial ring in infinitely many variables (which is a prototypical example of a non-Noetherian ring) and see if we can find such an ideal and an automorphism. To make "shifting" easier, we'll index the variables by integers.

Let $k$ be a field, and let $R = k[\dots, x_{-2}, x_{-1}, x_0, x_1, x_2, \dots]$. For each $n \in \mathbb{Z}$, let $I_n = (x_n, x_{n-1}, x_{n-2}, \dots)$. We have an infinite chain of ideals $$\dots \subset I_{-2} \subset I_{-1} \subset I_0 \subset I_1 \subset I_2 \subset \dots$$ where all the inclusions are proper. Can you think of a ring automorphism of $R$ that maps $I_0$ to $I_0$ but whose inverse doesn't have this property?

(Example in spoilers below.)

It's enough to find an automorphism $\varphi \colon R \to R$ such that $\varphi(I_1) = I_0$, since then $\varphi^{-1}(I_0) = I_1$. Define $\varphi(x_n) = x_{n-1}$ for all $n \in \mathbb{Z}$.