Sorry if this is duplicated. I couldn't find an exact answer of my question.
One definition of Noetherian ring is: A ring $R$ is Noetherian if all its ideals are finitely generated. I know there are other definitions. I am just wondering why this definition has to state "all its ideals".
Can there be a ring $R$ that is finitely generated as its own ideal, and has a proper ideal that is not finitely generated?
Are there examples? Thank you!
The question amounts to: ‘Does there exist non-noetherian rings?’
One of the simplest examples is the ring of polynomials in a countable set of indeterminates $\mathbf Z[x_1,\dots, x_n,\dots]$: any finite set of polynomials $f_1,\dots,f_r$ involves only a finite number of indeterminates, say up to $x_N$. Then none of the ideals that involve the subsequent indeterminates $x_{N+1},\dots$ can be expressed with $f_1,\dots,f_r$ for degree reasons.
Indeed, if $(f_1,\dots,f_r)=(x_{N+1})$, there would exist $g_1, \dots, g_r\in \mathbf Z[x_1,\dots, x_n,\dots]$ such that $g_1f_1+ \dots+ g_rf_r=x_{N+1}$. We may as well suppose each $g_i$ can be written as $x_{N+1}h_i$, thus $x_{N+1}= x_{N+1}(h_1g_1f_1+ \dots+ h_rf_r)$ and finally $h_1g_1f_1+ \dots+ h_rf_r=1$. This implies $(f_1,\dots,f_r)=(x_{N+1})=\mathbf Z[x_1,\dots, x_n,\dots]$, and $x_{N+1}$ would be unit in $\mathbf Z[x_1,\dots, x_n,\dots]$.