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Aug 16, 2011 at 17:35 vote accept Jesko Hüttenhain
Aug 16, 2011 at 17:35 comment added Jesko Hüttenhain Yes that's what I thought, I hadn't realized the source of my problem at that point. I guess I have to re-think this whole thing, thanks everybody.
Aug 16, 2011 at 16:32 comment added Graham Leuschke The assumption on dimension is automatic from integrality, but not helpful anyway, as David points out.
Aug 16, 2011 at 15:02 comment added David E Speyer You never showed (and it isn't true) that $\mathfrak{a}$ is prime.
Aug 16, 2011 at 14:27 comment added Jesko Hüttenhain Hm. Yea that sounds about right? Where does my argument fail?
Aug 16, 2011 at 13:51 comment added David E Speyer Look at Karl's example. $\mathfrak{q}_1$ is $\langle x^2-y^3 \rangle$, $\mathfrak{q}_2$ is $\langle y \rangle$ so $\mathfrak{a}$ is $\langle x^2-y^3, y \rangle = \langle x^2, y\rangle$, which is not maximal.
Aug 16, 2011 at 13:35 history edited Jesko Hüttenhain CC BY-SA 3.0
added 31 characters in body
Aug 16, 2011 at 13:35 comment added Jesko Hüttenhain $\mathfrak{q}_i$ lies over $(x_i)$, so $\mathfrak{a}:=\sum_i\mathfrak{q}_i$ lies over $\mathfrak{m}$, so by going-up, we know $\dim(\mathfrak{a})=\dim(\mathfrak{m})$. Hence, $\mathfrak{a}$ must be maximal. Oh shoot. Do I need to require that $A$ and $B$ have the same dimension? Maybe I will just add it.
Aug 16, 2011 at 12:54 comment added Graham Leuschke I don't understand the "By going up..." part. Doesn't Karl's example show that the sum of the $\mathfrak{q}$'s need not be maximal?
Aug 16, 2011 at 11:40 history edited Jesko Hüttenhain CC BY-SA 3.0
fixed a typo
Aug 16, 2011 at 11:22 answer added Karl Schwede timeline score: 3
Aug 16, 2011 at 10:07 history asked Jesko Hüttenhain CC BY-SA 3.0