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visits member for 4 years, 2 months
seen May 29 at 15:58

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?
Jun
4
comment Understanding the conductor ideal of a ring.
(PS: Happy Birthday!)
Jun
4
answered Understanding the conductor ideal of a ring.
Nov
2
asked Casual book on abstract algebra
Sep
9
comment Is a finite type affine scheme spec of a finitely-generated ring?
Thank you, that is most helpful.