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Let $R$ be a Noetherian integral domain of characteristic $0$, and let $R\langle x_1,\ldots,x_n\rangle$ be the free $R$-ring. Is $R\langle x_1,\ldots,x_n\rangle$ necessarily Noetherian?

1) The proof of Hilbert Basis Theorem fails. Also, even if it works, induction on the number of indeterminate would fail.

2) Another approach that I tried is constructing a short exact sequence, for example $0 \to (x_1x_2-x_2x_1) \to R\langle x_1,x_2\rangle \to R[x_1,x_2] \to 0$. However, I am not sure how to prove $(x_1x_2-x_2x_1)$ is Noetherian as an $R\langle x_1,x_2\rangle$-module (like I cannot construct an explicit ideal that is not finitely generated and I cannot prove it is Noetherian).

Can anyone give me some suggestions? Thank you!

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    $\begingroup$ $$ \large \begin{align} \text{correct notation: } & R\langle x_1,\ldots,x_n\rangle \\ \text{incorrect notation: } & R<x_1,\ldots,x_n> \end{align}$$ (I edited the question accordingly.) $\qquad$ $\endgroup$ – Michael Hardy Jun 22 '17 at 23:37
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It's not Noetherian (if $n \ge 2$). It's not even close to being Noetherian. Consider the left (or right) ideal generated by $$ \{ x_1x_2^kx_1 : k \in \mathbf{N} \}.$$

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