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I want to know how to show that the ring $\mathbb{C}[x,e^x]$ is Noetherian (I know the answer is yes, it is Noetherian, but I'm unable to prove it!).

My initial thoughts were to construct some isomorphism $\mathbb{C}[x,y]\rightarrow \mathbb{C}[x,e^x]$ such as $f(x,y)\mapsto f(x,e^x)$, and use the fact that $\mathbb{C}[x,y]$ is Noetherian by the Hilbert Bases Theorem. Such a mapping is a well-defined surjective ring homomorphism, but I'm struggling to prove injectivity in this case: if $f$ is in the kernel then $f(x,e^x)$ is identically zero, and hypothetically we could substitute in $y=e^x$ to give us that $f(x,y)$ identically zero for $y>0$ - the problem being that this is only true for $y>0$!

Please could someone help me figure this out? (Or let me know if I'm barking up at the wrong tree entirely!) Thanks :-)

Martin
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2 Answers2

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Hint: Quotients of noetherian rings are noetherian.

Jonas Linssen
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  • I cannot believe I missed that! So I can just apply the first isomorphism theorem to the map I gave above so that $\mathbb{C}[x,e^x]$ is isomorphic to $\mathbb{C}[x,y]$ quotiented by the kernel and therefore Noetherian? That is so simple, thank you! – Martin May 07 '20 at 08:05
  • Exactly :) You are welcome! – Jonas Linssen May 07 '20 at 08:06
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To prove your ring is Noetherian, PrudiiArca provided an answer.

If you want to know how to prove that the kernel of your map is trivial: this is in fact equivalent to proving that $e^X$ is transcendental over $\mathbf{C}[X]$. See this or this answer.

rae306
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