Timeline for 2 questions on Nagata's counterexample; $k[f_1,...,f_r]=k[g_1,...,g_s]$ vs. $k(f_1,...,f_r)=k(g_1,...,g_s)$

6 events

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Apr 13, 2017 at 12:19	history	edited	CommunityBot		replaced http://math.stackexchange.com/ with https://math.stackexchange.com/
Dec 23, 2012 at 21:20	comment	added	InvisiblePanda		@Ralph: What I forgot to say and just remembered: feel free to post this as an answer, since I'm fine now with the other question about a "more direct way", and I'll accept it!
Dec 15, 2012 at 10:22	history	edited	InvisiblePanda	CC BY-SA 3.0	added 172 characters in body
Dec 15, 2012 at 10:09	comment	added	InvisiblePanda		@Ralph: Thanks, I was already becoming confused, so at least there's nothing wrong in the stuff I did to show these equalities.
Dec 15, 2012 at 10:02	comment	added	Ralph		Q2: Showing equalities is quite similar for $[]$ and $()$. For, if $F$ is a $k$-algebra and $S,T$ are subsets of $F$, then $k[S]$ is the least $k$-subalgebra of $F$ that contains $S$ (i.e. $k[S]$ is the intersection of all $k$-subalgebras of $F$ containing $S$). Hence $S \subseteq k[T]$ implies $k[S] \subseteq k[T]$. Similarly, if $F$ is a field, then $k(S)$ is the least subextension of $F\|k$ that contains $S$. Hence, $S \subseteq k(T)$, also implies $k(S) \subseteq k(T)$.
Dec 15, 2012 at 9:06	history	asked	InvisiblePanda	CC BY-SA 3.0