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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 torsionfree 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 torsionfree 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 $L=K\left(\sqrt{x^21}\right)$ and $a=x\sqrt{x^21}$. If $b:=x+\sqrt{x^21}$ 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 $L=K\left(\sqrt{x(x1)}\right)$ and $a=x\sqrt{x(x1)}$. 
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 