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Let $R$ be a Noetherian subring of a commutative ring $S$. Suppose that $S = (R\cup\{b_1,...,b_m\})$ for some $b_1,...,b_m \in S_n$. Then $S$ is Noetherian.

I'm not sure how to approach this exercise. One idea was to take an ideal $J$ of $S$ and to look at the ideal $I = \{r \in R \mid s_1r + s_2b_1 + \cdots + s_{m+1}b_m$ for some $s_1,s_2,...,s_{m+1} \in S \}$ of $R$. More generally, it seems we need to represent an ideal of $J$ of $S$ as an ideal $I$ of $R$ plus some additional data.

Of course, $S = (1)$, hence $(1) = (R\cup\{b_1,...,b_m\})$, hence there are $r_1,...,r_n \in R$ and $s_1,...,s_{n+m}$ so that $1 = s_1r_1 + \cdots + s_nr_n + s_{n+1}b_1 + \cdots + s_{n+m}b_m$, but I'm not sure how we can use this.

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    $\begingroup$ What does $S = (R\cup\{b_1,...,b_m\})$ mean? You seem to be interpreting the parentheses as meaning "ideal generated by", but that doesn't make sense because $1\in R$ and so trivially $R$ already generates all of $S$ as an ideal. $\endgroup$ – Eric Wofsey Apr 3 at 3:13
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    $\begingroup$ It's a personal notation for $S=R[b_1,\dots,b_m]$. Grillet says: "S is generated by R and finitely many elements of S". $\endgroup$ – user26857 Apr 3 at 8:52
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    $\begingroup$ @user26857 Hmm, it settles it then: I misunderstood the question. But I don't think Grillet introduced any other notion of being generated with respect to a ring other than an ideal generated by a subset. But maybe I missed it. $\endgroup$ – Jxt921 Apr 3 at 13:45
  • $\begingroup$ @EricWofsey You are right, I interpreted "a ring generated by" as "a ring generated by as its own ideal". It was pretty silly of me to know that $(1) = S$ but not to notice that $(R) = S$ always as $1 \in R$. $\endgroup$ – Jxt921 Apr 3 at 13:48
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Hint: try finding a surjective ring homomorphism $R[x_1,\dots,x_m]\to S$, and then use the fact that $R[x_1,\dots,x_m]$ is Noetherian by the Hilbert basis theorem

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