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Jun 30, 2010 at 13:10 vote accept TOM
Jun 30, 2010 at 8:00 comment added TOM Thank you !And what about the prime ideal in the case L/k inifinite
Jun 29, 2010 at 20:47 answer added Georges Elencwajg timeline score: 6
Jun 29, 2010 at 18:56 comment added Keenan Kidwell When $L$ is of finite degree, (a) is a straightforward exercise using the primitive element theorem, the CRT, and some basic facts about tensor products. As such, I don't think it's appropriate for the site. For arbitrary algebraic $L$, the same result follows because $L\otimes_kH$ is the union of the reduced rings $F\otimes_kH$ with $F$ a finite subextension of $L/k$ (this follows from flatness and what it means to be algebraic). The statement about prime ideals, at least in the case $L/k$ finite, is a fact about artinian rings.
Jun 29, 2010 at 16:33 history edited Mariano Suárez-Álvarez CC BY-SA 2.5
Texify and fix punctuation.
Jun 29, 2010 at 16:28 answer added Robin Chapman timeline score: 4
Jun 29, 2010 at 16:06 answer added Mariano Suárez-Álvarez timeline score: 6
Jun 29, 2010 at 16:02 history asked TOM CC BY-SA 2.5