What can be said about the structure of maximal ideals of $R=k[\{x_i\}_{i \in I}]$, or geometric properties of $\text{Spm } k[\{x_i\}_{i \in I}]$? Here $k$ is an arbitrary field and $I$ is an infinite set. Kernels of evaluation homomorphisms yield an injective map

$\overline{k}^I / Aut(\overline{k}/k) \to \text{Spm } k[\{x_i\}_{i \in I}]$.

The image consists of those maximal ideals whose residue field is algebraic over $k$. If $I$ is finite, every residue field is algebraic (Noether Normalization). However, if $I$ is infinite and $|I| \geq |k|$, for example $k(t)$ is a residue field which is not algebraic. What happens if $|k| > |I|$? Is there a description in the general case?

  • $\begingroup$ Is this the one where $\operatorname{Spec}(R)\overset{homeo}{\cong} \omega + 1$? $\endgroup$ – Harry Gindi Oct 6 '10 at 12:56
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    $\begingroup$ @Harry: This would mean that $R$ has many many idempotents. $\endgroup$ – Martin Brandenburg Oct 6 '10 at 13:49
  • $\begingroup$ Yeah, I didn't really look at the problem for anything more than a second. Just notation-wise, it looks like a problem I did where you adjoin a whole bunch of idempotents. $\endgroup$ – Harry Gindi Oct 6 '10 at 16:14

If $|k| > |I|$ then the usual cheap proof of Nullstellensatz still works: let $K$ be a residue field. Then $\dim_k K \le \dim_kR = |I|$, but if $t\in K$ is transcendental over $k$, the elements $1/(t-a)$ for $a\in k$ are $k$-linearly independent. So $K/k$ is algebraic.

  • $\begingroup$ Nice! I accept your answer, since I think it's unlikely that there is a description in the general case. However, special cases such as $\text{Spm} \mathbb{Q}[x_1,x_2,...]$ would be interesting. $\endgroup$ – Martin Brandenburg Oct 6 '10 at 14:46
  • $\begingroup$ @Martin: thanks! Your example for $|I|\ge|k|$, strikingly illustrates another reason to be very careful working with non-noetherian schemes. $\endgroup$ – Tony Scholl Oct 6 '10 at 18:33

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