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awarded  Enlightened
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awarded  Nice Answer
Apr
27
answered Automorphisms of rings fixing all prime ideals
Apr
27
comment Automorphisms of rings fixing all prime ideals
@abx: the condition in the question is not satisfied at $\mathfrak{q}=0$.
Apr
17
answered Checking smoothness of the components of a highly symmetric scheme via quotient?
Mar
7
answered locus of integral fibres open
Mar
5
comment Exterior power of a torsion-free sheaf on a DVR
Here is an example where $\Lambda^nF$ is not $R$-flat: Assume, say, $\dim X=2$ and let $x$ be a closed point (in the closed fiber). Let $I\subset\mathscr{O}_X$ be the ideal sheaf of $x$, which is torsion-free of rank $1$ on $X$. Observe that $\Lambda^2I$ is nonzero and supported at $x$. Now take $F=I\oplus\mathscr{O}_X$: then $\Lambda^2F$ contains $\Lambda^2I$ as a direct summand, hence it has $R$-torsion.
Feb
25
comment Artin approximation vs implicit function theorem in the class of analytic functions
@tst: What exactly is your definition of "branching"?
Feb
23
comment Morphisms with connected fibres and rational functions
Think of the case where $Y$ is a point.
Feb
22
awarded  Nice Question
Feb
20
awarded  Nice Answer
Feb
11
comment Extension of a valuation on a function field
@giladude: Sorry, my example was wrong. Take $L=K\left(\sqrt{x^2-1}\right)$ and $a=x-\sqrt{x^2-1}$. If $b:=x+\sqrt{x^2-1}$ is the conjugate of $a$, then $ab=1$, hence $w(a)+w(b)=0$. But we cannot have $w(a)=w(b)=0$ since $a+b=2x$. So either $w(a)>0$ or $w(b)>0$. In fact you can check that there are two extensions $w^\pm$, with $w^\pm(a)=\pm1$.
Feb
10
comment Extension of a valuation on a function field
@giladude: Please read all my comment!
Feb
10
comment Extension of a valuation on a function field
$K(x^{1/n})$ won't give you a counterexample because there is only one extension of $v$. I suggest you try $L=K\left(\sqrt{x(x-1)}\right)$ and $a=x-\sqrt{x(x-1)}$.
Jan
10
awarded  Good Answer
Dec
21
comment Is $k(\!(x,y)\!)$ a topological field?
Oh, right. Simple, now you say it ;-). So this answers the second question. Thanks!
Dec
21
comment Does the cohomology comparison part of GAGA hold over the reals?
There are lots of nontrivial $X$'s such that $X(\mathbb{R})$ is empty.
Dec
21
comment Is $k(\!(x,y)\!)$ a topological field?
@YCor: for the topology just defined, $(1/f)k[[x,y]]$ is closed, not open.
Dec
21
asked Is $k(\!(x,y)\!)$ a topological field?
Dec
12
awarded  Popular Question