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.

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
