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Let $R\subset A\subset R[X]$ be commutative rings and suppose $A$ is Noetherian. Is $R$ Noetherian?

I guess the answer is yes. Can we say from this relation that $A[X]=R[X]$? If yes, then by Hilbert Basis Theorem, $A[X]$ is Noetherian, hence $R[X]$ is Noetherian $\Longrightarrow R$ is Noetherian.

user26857
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Rohan
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1 Answers1

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I'll add, because you seem skeptical why you can be sure $A[X]=R[X]$, that $R\subset A$ implies $R[X]\subset A[X]$, while $A\subset R[X]$ implies $A[X]\subset R[X][X]=R[X]$. So they are in fact equal

TY Mathers
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    Be careful with that. The hefty point is that your $A[X]$ (i.e. the smallest subring of $R[X]$ which contains $A$ and $X$) is not necessarily isomorphic to the ring of polynomials with coefficients in $A$. It's noetherian because, being finitely generated as a $A$-algebra, it's a quotient of a polynomial ring, hence noetherian. –  Feb 24 '16 at 17:11
  • @G.Sassatelli could you explain the bit about the non isomorphism a little more? An example would be great. – coldnumber Feb 24 '16 at 17:22
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    @coldnumber I was thinking about something like $\Bbb R\subseteq \Bbb R[X^2]\subseteq \Bbb R[X]$. The ring of polynomials over $\Bbb R[X^2]$ is $\Bbb R[X^2, Y]$, while $\Bbb R[X]$ is an entire extension of $\Bbb R[X^2]$, not a purely trascendental one. –  Feb 24 '16 at 17:42