Greg Muller
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 Aug 26 awarded Yearling Dec 23 awarded Popular Question Dec 2 awarded Enlightened Dec 2 awarded Nice Answer Aug 11 awarded Yearling Jul 2 awarded Curious Aug 21 awarded Great Answer Aug 11 awarded Yearling May 18 awarded Caucus Jan 31 awarded Taxonomist Dec 6 comment Is this a property of an integral domain that is not a field? $a$ was specified not to be a unit. Dec 3 comment Factor $1+x+x^2+x^3+…+x^{14}$ $-1$ is not a root of your equation. Plugging it into the equation gives $1$, not $0$ as desired. Nov 29 comment What are normal schemes intuitively? Careful, Serre's criterion only works for locally Noetherian schemes. It is far from true in general. Nov 29 answered What are normal schemes intuitively? Nov 28 comment Prime spectrum of a ring, understanding geometry Second, the algebra now has prime ideals which are neither maximal nor zero. It is still true that the maximal ideals of $\mathbb{R}[X,Y]/Y^2-X^3+X+1$ are the same as points in the variety, and the zero ideal is a generic point which is everywhere. What do the other prime ideals correspond to? $\mathbb{Z}/2$ acts on the variety of complex solutions. The remaining prime ideals in the real algebra correspond to faithful $\mathbb{Z}/2$-orbits in the variety of complex solutions. See `Geometry of Schemes' by Eisenbud and Harris, Section II.2 for an explanation of why. Nov 28 comment Prime spectrum of a ring, understanding geometry Several new things happen over $\mathbb{R}$. First, the variety of solutions is now a subset of $\mathbb{R}^2$, and so we can draw it completely... or ask WolframAlpha to draw it: wolframalpha.com/input/?i=plot+Y%5E2-X%5E3%2BX%2B1%3D0 Nov 27 answered Prime spectrum of a ring, understanding geometry Aug 11 awarded Yearling Jul 30 accepted Are bounded open regions in $\mathbb{R}^n$ determined by their boundary? Jul 30 asked Are bounded open regions in $\mathbb{R}^n$ determined by their boundary?