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Oct 6, 2019 at 1:28 comment added David E Speyer Thanks! I don't know if it helps anyone else, but I find it helpful to write this more symmetrically as $k[x_1, x_2, y, z]/(x_2-x_1 y,\ x_1 - x_2 z)$. The point is that $x_1$ divides $x_2$ and $x_2$ divides $x_1$, but there is no unit $u$ with $x_1 = u x_2$.
Oct 6, 2019 at 1:17 vote accept David E Speyer
Oct 5, 2019 at 20:44 history edited Jeremy Rickard CC BY-SA 4.0
I believe myself now!
Oct 5, 2019 at 9:46 history answered Jeremy Rickard CC BY-SA 4.0