maximal ideals of $k[x_1,x_2,...]$ - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T23:48:50Z http://mathoverflow.net/feeds/question/41262 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/41262/maximal-ideals-of-kx-1-x-2 maximal ideals of $k[x_1,x_2,...]$ Martin Brandenburg 2010-10-06T12:22:03Z 2010-10-06T14:31:08Z <p>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 <em>infinite</em> set. Kernels of evaluation homomorphisms yield an injective map</p> <p>$\overline{k}^I / Aut(\overline{k}/k) \to \text{Spm } k[\{x_i\}_{i \in I}]$.</p> <p>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?</p> http://mathoverflow.net/questions/41262/maximal-ideals-of-kx-1-x-2/41275#41275 Answer by Tony Scholl for maximal ideals of $k[x_1,x_2,...]$ Tony Scholl 2010-10-06T14:16:33Z 2010-10-06T14:16:33Z <p>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.</p>