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.

  • $\begingroup$ There appear to be several typos in what you've written. Is this what is written in the text? $\endgroup$ – RideTheWavelet May 2 '18 at 13:34
  • $\begingroup$ I don't see how letting $a_{n+1}=1$, and $a_1=\dots =a_n=0$ is valid. Wouldn't this be like saying saying "Assume, $x=y$, now let $x=1$ and $y=0$"? $\endgroup$ – cansomeonehelpmeout May 2 '18 at 13:55
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    $\begingroup$ It is valid by the universal property of the polynomial ring, i.e. you can define a homomorphism out of it by specifying it on the variables. $\endgroup$ – Leon Hendrian May 2 '18 at 13:57
  • $\begingroup$ I'm confused as to how $a_{n-1}=1$ and $a_0 = ... = a_n = 0.$ Can you explain how they were able to make these distinctions? $\endgroup$ – Hannah Diaz May 2 '18 at 14:11
  • $\begingroup$ @LeonHendrian Thank you, you're right. I wasn't thinking about the evaluation homomorphism. $\endgroup$ – cansomeonehelpmeout May 2 '18 at 15:44

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.


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.

  • $\begingroup$ @Hannah Diaz - You're saying that two polynomials are equal. So, they must take the same values for any assignment of values to the variables. In particular, equality must hold if $a_1= \dots a_n=0$ and $a_{n+1}=1.$ $\endgroup$ – Chris Leary May 2 '18 at 14:14
  • $\begingroup$ But how does this suggest it’s not Noetherian? $\endgroup$ – Hannah Diaz May 2 '18 at 15:05
  • $\begingroup$ It proves that $<a_0, \ldots, a_n> \neq <a_0,\ldots,a_{n+1}>$ for all $n$, therefore the chain of ideals is indeed an infinite ascending chain of ideals, which directly contradicts Noetherianness. $\endgroup$ – Leon Hendrian May 2 '18 at 16:32

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