$R\subset A\subset R[X]$, $A$ is Noetherian. Is $R$ Noetherian?

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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.

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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