The form of maximal ideal in the real polynomial ring $\mathbb R[x,y]$ Every maximal ideal of the real polynomial ring $\mathbb R[x,y]$ is of the form $(x-a, y-b)$ for some $a,b \in \mathbb R$. True or false? Any suggestions?
 A: Hint: How about the ideal $(x, y^2+1)$? 
A: In fact the nullstellensatz allows one to deduce the answer also to the question implicit in your title, of classifying all maximal ideals of $\mathbb R[x,y]$, since there is a simple relation between maximal ideals of $\mathbb R[x,y]$ and maximal ideals of $\mathbb C[x,y]$.  According to Th. 1, chap. II.4 of Mumford’s redbook of algebraic geometry, since $\mathbb R[x,y]$ is contained in $\mathbb C[x,y]$, intersecting with $\mathbb R[x,y]$ defines a surjective map from maximal ideals of $\mathbb C[x,y]$ to maximal ideals of $\mathbb R[x,y]$, whose fibers are conjugate pairs of complex ideals.  Thus in addition to the maximal ideals of form (x-a,y-b) where a,b are real, the fixed points of the conjugation action, we also have the intersection of $\mathbb R[x,y]$ with maximal ideals of form (x-a,y-b) where at least one of a,b is non real.  If a’,b’ are the complex conjugates of a,b, and b say is non real, this intersection ideal is generated by the quadratic (y-b)(y-b’) and the (real) linear equation for the line joining the two points (a,b) and (a’,b’).  The example given by Prism is the result of intersecting $\mathbb R[x,y]$ with the complex ideal (x, y-i) or also (x,y+i).  I.e. since x=0 is the line joining (0,i) and (0,-i), we get the generators (x, (y-i)(y+i)) = (x, $y^2+1$).
A: Remember $\rm M\triangleleft R~maximal\iff R/M~field$. Can you obtain $\bf C$ as a quotient?
