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I am studying Chapter 8 of Hungerford for my qualifying exam.

In Chapter 8.1., he gives an exercise concerning the ring that is right Noetherian (resp. Artinian) but not left Noetherian (resp. Artinian).

The ring of all $2\times 2$ matrices $\begin{pmatrix}a&b\\0&c\end{pmatrix}$ such that $a$ is an integer and $b,c$ are rational is right Noetherian but not left Noetherian.

Basically, this question has been answered several times: for example this post and this one.

Showing that the above ring, say $R$, is not left Noetherian is quite straightforward; we simply construct an infinitely ascending chain of left ideals in $R$.

Almost every sources (including this site) argue that $R$ is right Noetherian by showing that a right ideal $I$ is finitely generated. In particular, they separate cases where (i) $I$ contains an element of the form $\begin{pmatrix}a&b\\0&c\end{pmatrix}$ with $a\neq 0$, and (ii) $I$ consists of elements of the form $\begin{pmatrix}0&b\\0&c\end{pmatrix}$. But I don’t see why the following (simpler) argument can be applied:

A ring $R$ is a right Noetherian if and only if for some ideal $I$ of $R$, both $I$ and $R/I$ are Noetherian. Define a map $\phi:R\rightarrow \mathbb{Z}\times\mathbb{Q}$ by $\begin{pmatrix}a&b\\ 0&c\end{pmatrix}\mapsto (a,c)$. Then $\phi$ is a surjective ring homomorphism with kernel $I=\left\{\begin{pmatrix}0&b\\0&0\end{pmatrix}:b\in\mathbb{Q}\right\}$. $R/I\cong\mathbb{Z}\times\mathbb{Q}$ is Noetherian since both $\mathbb{Z}$ and $\mathbb{Q}$ are. The only submodule contained in $I$ is $0$ and $I$ itself, since if $J\subset I$ is a nontrivial submodule of $I$, then for some nonzero $\begin{pmatrix}0&b\\0&0\end{pmatrix}\in J$ we have $\begin{pmatrix}0&1\\0&0\end{pmatrix}=\begin{pmatrix}0&b\\0&0\end{pmatrix}\begin{pmatrix}0&0\\0&1/b\end{pmatrix}\in J$. It follows that both $R/I$ and $I$ are Noetherian.

Any help is appreciated. Thanks!

Jkim
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1 Answers1

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I did precisely that in one of the solutions you linked, although I chose different submodules to prove it with, ones which link with the classification of right ideals of the ring.

The thing that deserves more explanation in what you said is "$\mathbb Z\times \mathbb Q$ is Noetherian." Remember, you have to show that it's Noetherian as a right $R$ module. It's tempting to just say "the two things are Noetherian rings so the product is a Noetherian ring" and that is true, but a third party could legitimately question why this Noetherian ring should be a Noetherian right module over $R$.

rschwieb
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