[Math] Maximal ideals in a polynomial ring over the real numbers.

ac.commutative-algebraag.algebraic-geometrypolynomials

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 $x_i-a_i$.

Best Answer

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 $$(x_1^2+a_1, x_2-a_2, \ldots, x_n-a_n), \quad a_i \in \mathbf{R}, \quad a_1 >0.$$

This follows from the generalized weak Nullstellensatz, 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 $2$.

For a reference, see Arrondo's notes A geometric introduction to commutative algebra, in particular Example 0.6 and Corollary 5.14.