I am struggling with understanding this proof. I am confused as to why they took $a\_{n+1}$ and how they got $a\_{n+1}= b\_1a\_1+...+ b\_na\_n$, $a\_1 = ... = a\_0 = 0,$ and $a\_{n+1} = 1$ gives $1 = 0$ in $\mathbb{R}$. Please help with explaining the steps of this proof in more detail. I am also not sure how to tie in that the ideals are infinitely generated and a proper susbset of each other.

Let $S = \mathbb{R}[a\_0,a\_1,a\_2...]$ be a polynomial ring over R in inﬁnitely many variables. Prove $S$ is non-Noetherian.

**Proof:** We want to show that S in non-Noetherian.Take the inﬁnite chain of ideals: $<a\_0>\subset<a\_0, a\_1>\subset...\subset<a\_0,...,a\_n>\subset ...$. Suppose that for any $n, a\_{n+1}$ were an element of $<a\_1,...,a\_n>$. In other words, there exist polynomials $b\_1,...,b\_n$ such that $a\_{n+1} = b\_1a\_1+...+ b\_na\_n$. Setting $a\_1 = ... = a\_0 = 0,$ but $a\_{n+1} = 1$ gives $1 = 0$ in $\mathbb{R}$, a contradiction. So, S is non-noetherian.