I am currently studying for my University's preliminary exam in Algebra. I have come across a problem that is practice. I hope I can get some feedback on my proof.
Question:
Consider the ring $\mathbb{Z}$. Let $\mathbb{Z}[x]$ be the polynomial ring in one variable over $\mathbb{Z}$. Define $T \subset \mathbb{Z}[x]$ by $T = \mathbb{Z}[2x, 2x^2 , 2x^3 , \cdots ]$. Determine whether or not $T$ is noetherian.
Proof: Consider $$\pi: \mathbb{Z}[x] \rightarrow \frac{\mathbb{Z}[x]}{2\mathbb{Z}[x]} \cong \mathbb{F}_2[x].$$ Note that $\pi(T) \subset \mathbb{F}_2$. Suppose that, for some $n \geq 1$: $$2x^{n + 1} = t_1 2x + t_2 2x^2 + \cdots + t_n 2x^n \quad \text{ for } t_i \in T.$$ Then in $\mathbb{Z}[x]$ we have that $x^{n + 1} = t_1 x + t_2 x^2 + \cdots + t_n x^n$. By applying $\pi$ we get that $x^{n + 1} = c_1 x + c_2 x^2 + \cdots + c_n x^n$ for $c_i \in \mathbb{F}_2$. This however is a contradiction, since the monomials $x^i$ are linearly independent over the field $\mathbb{F}_2$. Therefore, the ascending chain of ideals in $T$ defined by $$(2x) \subset (2x,2x^2) \subset (2x,2x^2,2x^3) \subset \cdots $$ does not stabilize and thus $T$ is not noetherian.
Am I on the right track? If I am not, can I please have some hint's in the correct direction or let me know where in my proof I have failed.