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.
 A: If $S=\mathbb{R}[a_0,a_1,\dots]$ were Noetherian then any chain of ideals $$I_0\subset I_1\subset\dots$$ would stabilize, that is, for some $n$ we would have $$I_n=I_{n+1}=I_{n+2}=\dots$$ To prove that $S$ is not Noetherian, we first assume that it is, and then see if we get a contradiction. 
First we need a chain of ideals, so let's look at the chain: $$(a_0)\subset(a_0,a_1)\subset (a_0,a_1,a_2)\subset\dots$$
If we assume that $S$ is Noetherian, then this chain eventually stabilizes, so we would have, for some $n$, that $$(a_0,a_1,\dots,a_n)=(a_0,a_1,\dots,a_n,a_{n+1})$$
This would mean, in particular, that $a_{n+1}\in (a_0,a_1,\dots,a_n)$ so it should be possible to write $$a_{n+1}=\sum_{i=0}^{n}b_ia_i$$ Now, every polynomial ring has an evaluation homomorphism (this is what first confused me, but was clearified by @Leon Hendrian) that maps each $a_i$ to some element in the ring. (For instance, the evaluation homomorphism for $R[x]$, $R$ is any ring, is $\phi_a:R[x]\rightarrow R$, such that $f(x)\mapsto f(a)$, for some $a\in R$).
Using this homomorphism we can evaluate $a_{n+1}=\sum_{i=0}^nb_ia_i$ at $a_0=a_1=a_2=\dots =a_n=0$ and $a_{n+1}=1$. This gives $1=0$ which is a contradiction. Therefore $(a_0,a_1,\dots,a_n)\neq (a_0,a_1,\dots,a_n,a_{n+1})$ and so $S$ is not Noetherian.
A: 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.
