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Apr
25
asked The lattice points in the real cone of some semigroups are just the integer cone of that semigroup.
Apr
13
answered Vakil's Foundations of Algebraic Geometry, Exercise 7.3.F
Apr
8
accepted Being a regular embedding is an open condition for locally Noetherian schemes
Apr
7
awarded  Revival
Apr
7
revised Being a regular embedding is an open condition for locally Noetherian schemes
Added necessary hypothesis
Apr
7
comment Being a regular embedding is an open condition for locally Noetherian schemes
mathoverflow.net/questions/129242/…
Apr
7
comment Being a regular embedding is an open condition for locally Noetherian schemes
@AreaMan, Hoot, Ah yes, I had meant to put a locally Noetherian hypothesis in the question, it was just a transcription error. The next remark in the notes is that this isn't true without the hypothesis. You make a good point Areaman, and in fact you can see in the link below a counter example to the given statement, so not only is the lemma false, but there's no way to prove the theorem that avoids it. Thanks for the correction!
Apr
6
answered Being a regular embedding is an open condition for locally Noetherian schemes
Apr
6
asked Being a regular embedding is an open condition for locally Noetherian schemes
Mar
31
answered The equivalence of two definitions of closed subscheme, Vakil's Ex 8.1.K
Mar
24
answered free module implies surjective map of affine schemes
Mar
23
comment Describe all ring homomorphisms from $\mathbb{R}[T] \rightarrow \mathbb{R}[T]$
It looks like your attempt is very good, and I don't think that there's a swifter way to approach it. You're right in that $\mathbb{R}$ maps to $\mathbb{R}$, I would use the fact that units map to units (which is not so hard to prove) and that the only units of $\mathbb{R}[T]$ are the non-zero constants.
Mar
22
revised To show a morphism of affine k-varieties which is surjective on closed points is surjective
added 429 characters in body
Mar
22
answered To show a morphism of affine k-varieties which is surjective on closed points is surjective
Mar
6
comment Question on morphism locally of finite type
@AlexSaad You're very welcome. In answer to your question, that's exactly what I mean, it does indeed come from the diagram (because the maps of schemes commute) and it's the statement we need to say that $A$ being finitely generated over one is equivalent to being finitely generated over the other. I'm glad it was helpful!
Mar
5
answered Question on morphism locally of finite type
Mar
3
awarded  Nice Answer
Feb
27
reviewed Looks OK conjugates of upper triangular matrices
Feb
27
reviewed Close What is the minimum of this expression
Feb
27
reviewed Close Taylor Series of $f(x) = \sqrt{x}$ about $c = 1$