Consider the ring $R=\mathbb{Z}+x \cdot \mathbb{Q}[x]$. I need to check whether it is noetherian or not.
A well known $\mathbb{Q}+x\cdot \mathbb{R}[x]$ is not noetherian, so I think my ring $R$ is also. Please verify my attempt below.
Attempt
Consider the ideal $I := x \cdot \mathbb{Q}[x]$. Suppose it is finitely generated, e.g. $I=\langle f_1, f_2, ..., f_n \rangle$.
Now let's look at how 1-st degree terms can be obtained when we multiply two polynomials.
$(a_0+a_1x+...+a_mx^m)(b_0+b_1x+...+b_kx^k)=a_0b_0+(a_1b_0+a_0b_1)x+...$
It's clear that first degree term can be generated only when the free term of one polynomial multiplies by first degree term of another.
I want to show that with $f_1, ..., f_n$ we can't generate $\mathbb{Q} \cdot x \subset I$ and hence we can't generate $I$.
Consider all first degree terms in $f_1, ... ,f_n$. Their coefficients are some rationals, so let's represent them as $p_i/q_i$ for $f_i$
It's worth to notice that in our ideal $I$ all free terms of polynomials are zero.
When we multiply $g \in R$ by $f \in I$ we obtain first degree term only if $g$ has non-zero free term, e.g. $g$ must be outside of $I$.
Thus, the coefficients of first degree terms generated by all $f_i$ are of type $$z_1 \frac{p_1}{q_1} + z_2 \frac{p_2}{q_2} + ... + z_n \frac{p_n}{q_n} \equiv \frac{z_1 p_1 q_2 q_3 ... q_n + z_2 p_2 q_1 q_3...q_n+... + z_n p_n q_1...q_{n-1}}{q_1 q_2 ... q_n}$$ where $z_i \in \mathbb{Z}$.
As you can see, we "iterate" only through coefficients with denominator $q_1...q_n$. And obviously there are elements in $\mathbb{Q}$ which can not be represented as fractions with that denominator.
Is my solution correct? Do you see where we can optimize it?