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Jan 26, 2014 at 4:27 comment added Jesse Elliott I found this reference, which shows that $A \otimes_k A \longrightarrow A$ is an isomorphism if and only if $k \longrightarrow A$ is an epimorphism of rings. mathoverflow.net/questions/56591/… So the question reduces to that of epimorphisms. Also found this: mathoverflow.net/questions/109/…
Jan 26, 2014 at 1:43 vote accept Jesse Elliott
Jan 26, 2014 at 1:42 comment added Jesse Elliott Ok, thanks. Now I see why flatness might need to be assumed.
Jan 26, 2014 at 1:14 comment added Jason Starr @JesseElliott: No, that does not get rid of those types of counterexamples -- please see my third edit.
Jan 26, 2014 at 0:24 comment added Jesse Elliott I added the assumption that $k \longrightarrow A$ be injective, which gets rid of some easy counterexamples, including the $\mathbb{Z}/2\mathbb{Z}$ example.
Jan 26, 2014 at 0:18 comment added Jason Starr @JesseElliot: "Please add $A\neq 0$ to my assumptions." Why don't you add that to your assumptions, yourself? In fact, I suggest that you just accept one of the two correct answers. If you want to ask another question, you can always do that.
Jan 26, 2014 at 0:13 comment added Jesse Elliott Please add $A \neq 0$ to my assumptions.
Jan 26, 2014 at 0:07 comment added Laurent Moret-Bailly "Which means $A\subset K$": yes assuming e.g. that $k$ is a domain and $A\neq0$. In fact, the zero algebra is a counterexample to many claims. Also, "finitely presented" is irrelevant, I think.
Jan 25, 2014 at 23:23 comment added Jesse Elliott Flatness is sufficient for the given conclusion, but is it necessary? And what is a counterexample when $k \longrightarrow A$ is injective?
Jan 25, 2014 at 20:30 comment added Jason Starr That makes sense to me.
Jan 25, 2014 at 19:54 history answered abx CC BY-SA 3.0