Laurent Moret-Bailly
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 1d awarded Enlightened 1d 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