If $f:X\to S$ is a universal homeomorphism, is $f':X\times_S X\to X$ always a nil-immersion? This seems to be easy, yet possibly I miss something. Should I give references to this fact in a paper?
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asked Jul 16, 2010 at 8:46
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$\begingroup$ what is a universal homomorphism? $\endgroup$– Martin BrandenburgCommented Jul 16, 2010 at 8:50
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1$\begingroup$ Do you mean "homeomorphism"? $\endgroup$– S. Carnahan ♦Commented Jul 16, 2010 at 11:49
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$\begingroup$ Sorry! Certainly, I wanted to say "homeomorphism". $\endgroup$– Mikhail BondarkoCommented Jul 16, 2010 at 12:27
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3$\begingroup$ Mikhail, your terminology seems non-standard. I agree it's obvious that the map becomes an isomorphism on underlying reduced schemes, but it need not be an immersion and hence should not be called a nil-immersion (which I take to mean "closed immersion defined by quasi-coherent ideal sheaf whose local sections are nilpotent"). One already sees this for a non-trivial finite purely inseparable extension of fields (or any other finite flat radiciel covering of degree > 1). $\endgroup$– BCnrdCommented Jul 16, 2010 at 12:54
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In general, the answer is no. As Brian Conrad points out in the comments above, purely inseparable field extensions do not have this property. However (and maybe this is what you were getting at?), it is true that if $X\to S$ is a universal homeomorphism, then the diagonal $X\to X\times_SX$ is a nilimmersion.
