Challenge Problem Claim:

⟨( \begin{matrix}0 & 1/2 \\ 0 & 0 \end{matrix} )⟩ ⊊ ⟨( \begin{matrix}0 & 1/4 \\ 0 & 0 \end{matrix} )⟩ ⊊ ⟨( \begin{matrix}0 & 1/8 \\ 0 & 0 \end{matrix} )⟩ ⊊ …

is an infinite ascending chain of left ideals.

Certainly,

( \begin{matrix}2 & 0 \\ 0 & 0 \end{matrix} ) ( \begin{matrix}0 & 1/2^{n} \\ 0 & 0 \end{matrix} ) = ( \begin{matrix}0 & 1/2^{n−1} \\ 0 & 0 \end{matrix} ) ,

so

⟨( \begin{matrix}0 & 1/2^{n−1} \\ 0 & 0 \end{matrix} )⟩ ⊆ ⟨( \begin{matrix}0 & 1/2^{n} \\ 0 & 0 \end{matrix} )⟩.

Now suppose that

( \begin{matrix}a & b \\ 0 & c \end{matrix} ) ( \begin{matrix}0 & 1/2^{n−1} \\ 0 & 0 \end{matrix} ) = ( \begin{matrix}0 & 1/2^{n} \\ 0 & 0 \end{matrix} ) .

Then,

( \begin{matrix}0 & a/2^{n−1} \\ 0 & 0 \end{matrix} ) = ( \begin{matrix}0 & 1/2^{n} \\ 0 & 0 \end{matrix} ) ,

so that a/2^{n−1} = 1/2^{n}, for some a ∈ ℤ, which is not true, so

⟨( \begin{matrix}0 & 1/2^{n−1} \\ 0 & 0 \end{matrix} )⟩ ⊊ ⟨( \begin{matrix}0 & 1/2^{n} \\ 0 & 0 \end{matrix} )⟩.

Thus,

( \begin{matrix}ℤ & ℚ \\ 0 & ℚ \end{matrix} )

is not left Noetherian.

To show it is right Noetherian, we determine its right ideals. Let

( \begin{matrix}a & b \\ 0 & c \end{matrix} ) ∈ I,

where I is a right ideal, and suppose that ac ≠ 0.

Then

( \begin{matrix}a & b \\ 0 & c \end{matrix} ) ( \begin{matrix}m & n \\ 0 & p \end{matrix} ) ∈ I ∀m ∈ ℤ, n, p ∈ ℚ.

Thus,

( \begin{matrix}am & an + bp \\ 0 & cp \end{matrix} ) ∈ I ∀m ∈ ℤ, n, p ∈ ℚ.

Thus, I contains all elements of the form

( \begin{matrix}⟨a⟩ & ∗ \\ 0 & ∗ \end{matrix} ) .

Now suppose that a ≠ 0, c = 0. Then,

( \begin{matrix}a & b \\ 0 & 0 \end{matrix} ) ( \begin{matrix}m & n \\ 0 & p \end{matrix} ) ∈ I ∀m ∈ ℤ, n, p ∈ ℚ

implies that

( \begin{matrix}am & an + bp \\ 0 & o \end{matrix} ) ∈ I.

Thus, I contains all elements of the form

\{( \begin{matrix}⟨a⟩ & ∗ \\ 0 & 0 \end{matrix} )\} .

Now suppose that a = 0, c ≠ 0. Then

( \begin{matrix}0 & b \\ 0 & c \end{matrix} ) ( \begin{matrix}m & n \\ 0 & p \end{matrix} ) ∈ I ∀m ∈ ℤ, n, p ∈ ℚ

implies that

( \begin{matrix}0 & bp \\ 0 & cp \end{matrix} ) = p ( \begin{matrix}0 & b \\ 0 & c \end{matrix} ) ∈ I∀p ∈ ℚ.

Note that no ideal generated by an element of the form

( \begin{matrix}0 & b \\ 0 & c \end{matrix} )

can be strictly contained in another ideal of the same form. So the possible chains are

⟨( \begin{matrix}0 & b \\ 0 & c \end{matrix} )⟩ ⊊ ⟨( \begin{matrix}⟨a⟩ & ∗ \\ 0 & ∗ \end{matrix} )⟩ ⊊ ⟨( \begin{matrix}⟨b⟩ & ∗ \\ 0 & ∗ \end{matrix} )⟩ ⊊ …

(This chain stops because a has only finitely many divisors.) OR

⟨( \begin{matrix}0 & b \\ 0 & 0 \end{matrix} )⟩ ⊊ ⟨( \begin{matrix}⟨a⟩ & ∗ \\ 0 & 0 \end{matrix} )⟩ ⊊ ⟨( \begin{matrix}⟨a_{1}⟩ & ∗ \\ 0 & 0 \end{matrix} )⟩ ⊊ … ⟨( \begin{matrix}⟨m⟩ & ∗ \\ 0 & 0 \end{matrix} )⟩ ⊊ ⟨( \begin{matrix}⟨m⟩ & ∗ \\ 0 & ∗ \end{matrix} )⟩ ⊊ ….

(This chain also stops because m has only finitely many divisors.)