most recent 30 from http://mathoverflow.net 2013-05-25T02:14:32Z http://mathoverflow.net/feeds/question/85915 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/85915/maximal-ideals-in-a-polynomial-ring-over-the-real-numbers Miguel 2012-01-17T16:23:30Z 2012-01-17T18:43:09Z

<p>Let $\mathbf{R}$ be the field of real numbers. What are the generators of the maximal ideals of the polynomial ring $\mathbf{R}[x_1, ... , x_n]$? If instead of $\mathbf{R}$ one considers the field $\mathbf{C}$ of complex numbers, then Hilbert's Nullstellensatz implies that each maximal ideal $\mathfrak{m}$ of $\mathbf{C}[x_1, ... , x_n]$ is generated by $n$ generators of the form $x_i-a_i$. </p>

http://mathoverflow.net/questions/85915/maximal-ideals-in-a-polynomial-ring-over-the-real-numbers/85916#85916 Francesco Polizzi 2012-01-17T16:57:31Z 2012-01-17T18:43:09Z

<p>There are two kind of maximal ideals in $\mathbf{R}[x_1, \ldots, x_n]$: the ideals corresponding to real points of $\mathbf{A}^n_{\mathbb{R}}$, i.e. of the form $$(x_1-a_1, \ldots, x_n-a_n), \quad a_i \in \mathbf{R}$$ and the ideals corresponding to pairs of complex-conjugated points, that after a real change of coordinates can be put in the form $$(x_1^2+a_1, x_2-a_2, \ldots, x_n-a_n), \quad a_i \in \mathbf{R}, \quad a_1 >0.$$</p> <p>This follows from the <em>generalized weak Nullstellensatz</em>, saying that if $\mathbf{K}$ is any field and $\mathfrak{m} \subset \mathbf{K}[x_1, \ldots, x_n]$ is a maximal ideal, then the field $\mathbf{K}[x_1, \ldots, x_n]/ \mathfrak{m}$ is a finite extension of $\mathbf{K}$. In particular, when $\mathbf{K}=\mathbf{R}$ this extension must have degree at most $2$.</p> <p>For a reference, see Arrondo's notes <a href="http://www.mat.ucm.es/~arrondo/commalg.pdf" rel="nofollow">A geometric introduction to commutative algebra</a>, in particular Example 0.6 and Corollary 5.14.</p>