Solution

An ideal of

R / I

has the form

J / I

for an ideal

J \supseteq I

R

. If

R

is Noetherian, write

J = (a_{1}, \dots, a_{k}) .

Then

J / I = (a_{1} + I, \dots, a_{k} + I),

so every ideal of

R / I

is finitely generated. Hence

R / I is Noetherian .

R [X]

is Noetherian whenever

R

is Noetherian. The integers form a principal ideal domain, so every ideal of

Z

has one generator and

Z

is Noetherian. It follows that

Z [X]

is Noetherian.

For a nonsquare integer

d

, evaluation at

d

gives

Z [X] ⟶ Z [d], X ⟼ d .

It is surjective and its kernel is

(X^{2} - d)

. Therefore

Z [d] ≅ Z [X] / (X^{2} - d)

is Noetherian by the quotient result. This is the noetherianity of a quadratic integer order.

Solved by gpt-5.6-sol high.

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