Rigorous proof that certain ideal is not finitely generated

Question

Let $R = F[x_1,x_2,\ldots]$ (polynomials in an infinite number of indeterminates) and let $I = \{f \in R : f(0,0,\ldots) = 0\}$. One can easily see that this is indeed an ideal.

The proof for why this is not finitely generated was presented to me in the following argument. Suppose it is finitely generated. The one can write $I = \langle r_1,\ldots,r_k \rangle$ for some $r_i \in R$. Then pick $\ell \in \mathbb{N}$ such that each $r_i \in F[x_1,\ldots,x_\ell]$. Then $x_{\ell+1} \in I$ so $x_{\ell+1} = \sum f_i r_i$ and hence (here is where this gets messed up) we can write $1 = \sum (f_i/x_{\ell+1}) r_i$ therefore $1 \in I$.

So, why does one have $x_{\ell+1} \mid f_i$ for each $i$?

Answer 1

Note that all of the $r_i$ are in the ideal $I$ and so $r_i(x_1, \ldots, x_l)$ are zero for $(x_1, \ldots , x_l) = (0,\ldots, 0)$. Now in the equality $$x_{l+1} = \sum f_i r_i$$ plug $x_1 = \cdots = x_l = 0$ and $x_{l+1} = 1$ and get $1=0$, not possible.

Answer 2

Each element in $\;R\;$ , and thus also in $\;I\;$, is a polynomial in a finite number of indeterminates (the pol's indeterminates, let's call them).

Thus, we must understand the condition $\;f(0,0,...)=0\;$ as "the polynomial (function, now) vanishes when its indeterminates are substituted, each and all of them, by zero."

Since the polynomial $\;g_{\ell+1}(x_{\ell+1}):=x_{\ell+1}\;$ clearly belongs to $\;I\;$ , it then must be a $\;R-$ linear combination of the $\;r_i'$ s , so

$$x_{\ell+1}=\sum_{i=1}^k f_ir_i\;\;(*)$$

But the polynomials $\;r_i\;$ do not have $\;x_{\ell+1}\;$ as one of their indeterminates! Thus, it must be that $\;x_{\ell+1}\;$ is one of the indeterminates of $\;f_i\;$ -- we assume that in $\;(*)\;$ above only appear relevant $\;f_i$'s, meaning: only those that aren't zero and thus will certainly have $\;x_{\ell+1}\;$ as one of its indeterminates --, and then substituting $\;x_{\ell+1}=0\;$ in $\;(*)\;$ gives us that all the (relevant, of course) $\;f_i$'s are divisible by $\;x_{\ell+1}\;$

Answer 3

The argument you've presented here is wrong; we cannot conclude that $x_{l+1}$ divides each $f_i$. For example, if $r_i=r_j$, we could replace $f_i$ and $f_j$ with $f_i + 1$ and $f_j-1$, and this would ruin such a property.

There are various ways to modify this proof to be correct—by arguing in various ways that we can choose the $f_i$ so that this property holds—but orangeskid has certainly presented the most elegant of these.