Prove that a polynomial ring over R in inﬁnitely many variables is non-Noetherian. 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.

When you write $a_n+1$, I assume that you mean $a_{n+1}$. Actually, this might already solve your confusion. 
The logic of the proof you wrote is as follows: Assume $<a_0, \ldots, a_n> = <a_0, \ldots, a_{n+1}>$. This means that $a_{n+1} \in <a_0, \ldots, a_n>$, which means that $a_{n+1}=\sum_{i=0}^n b_ia_i$. But then, evaluate both sides on $a_0=\ldots=a_n=0, a_{n+1}=1$ to get the contradiction.
