I know how to prove that A noetherian implies B noetherian using Hilbert's basis theorem. However, I wasn't able to produce an answer to the converse which probably is false.
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Let $k$ be a field, and let $R=k[x,y]$.
Let $M$ be the set of monomials $m \in R\;$such that $\deg(m,x) < \deg(m,y)$.
Define subrings $A,B$ of $R$ by $$A = k[M]$$ $$B = A[x]$$ Then $B\;$is finitely generated over $A$.
Also, $B=R$, so $B\;$is noetherian.
But $A\;$is not noetherian, since the ideal of $A\;$generated by $M$ is not finitely generated.
quasi
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Thank you. I was thinking of something along the lines $A=K[x,xy,xy^2,...]$ and $B=A[y]$. Would it be correct? – B. Rivas Aug 15 '17 at 14:23
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1Yes, that would work. It's similar, in spirit, to the example I gave. – quasi Aug 15 '17 at 14:27
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Let $A$ be any non-Noetherian ring, let $\mathfrak{m}$ be a maximal ideal of $A$, and set $B := A/\mathfrak{m}$. Then $B$ is a finitely generated $A$-algebra and it is Noetherian since it is a field.
Minseon Shin
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