Grillet's “Abstract Algebra”, p. 148, ex. 3: Noetherian subring of a ring

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.

• 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. – Eric Wofsey Apr 3 at 3:13
• 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". – user26857 Apr 3 at 8:52
• @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. – Jxt921 Apr 3 at 13:45
• @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 $1 \in R$. – Jxt921 Apr 3 at 13:48

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