Timeline for Restricted extension of prime ideals of the ring of polynomials over $\mathbb{Q}$

7 events

when toggle format	what		by	license	comment
May 31, 2018 at 1:37	comment	added	Sean Sanford		@FilippoAlbertoEdoardo You make a good point. Since the validity of the claim depends on the base ring, and my argument doesn't reference this at all, it cannot possibly be salvaged without explaining why this works for $\mathbb Q$ and not $\mathbb C$.
May 31, 2018 at 1:34	comment	added	Sean Sanford		@AdamPrzeździecki You are right, and I don't see a way of proving it.
May 31, 2018 at 1:28	comment	added	Neil Epstein		@FilippoAlbertoEdoardo Actually when $n=1$ almost every maximal ideal intersects $V$ trivially. But your statement is true when $n\geq 2$.
May 30, 2018 at 20:04	comment	added	Filippo Alberto Edoardo		@Sean Sanford: Observe that the statement is false if $\mathbb{Q}$ is replaced by $\mathbb{C}$: the prime ideal $(0)\subseteq \mathbb{C}[X_1,\dots,X_n]$ has trivial intersection with $V$ but every maximal ideal has non-trivial intersection with $V$ by the Nullstellensatz. On the other hand, your argument "seems" to work in general, and as Adam Przezdziecki points out, I don't see why $\mathfrak{m}$ should be maximal.
May 30, 2018 at 19:52	comment	added	Adam Przeździecki		How do you know you ideal $\frak m$ is maximal in $R$?
May 30, 2018 at 18:54	review	First posts
May 30, 2018 at 19:19
May 30, 2018 at 18:53	history	answered	Sean Sanford	CC BY-SA 4.0