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My question is the following: I encountered the statement

The residue field of an Artin local $\mathbb{C}$-algebra is $\mathbb{C}$

I understand that if the algebra $A$ is finitely generated over $\mathbb{C}$ this is true: $\mathbb{C} \subset A/\mathfrak{m}_A$ would be a field extension where the latter is finitely generated over $\mathbb{C}$ and thus it is a finite algebraic extension by the Zariski Lemma. However without the assumption for $A$ to be finitely generated it seems to me that the statement is not true. $$\mathbb{C} \subset \mathbb{C}(x) = A$$ It seems to me to be a counterexample.

Thank you for your help!

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Yes, $\mathbb C(x)$ is Artinian, local and the residue is not isomorphic to $\mathbb C$. And your comments on finite generation are on-mark.

At nLab I find this:

A local Artin algebra with residue field $\mathbb{K}$ is a finitely generated $\mathbb{K}$-algebra $A$.

So it seems the term "local Artin algebra" also assumes that $A$ is finitely generated over $\mathbb K$, and your example does not include that.

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    $\begingroup$ Yes, the subtle difference is 'Artin algebra' vs. 'artinian algebra'. It is very easy to stumble over this, though. $\endgroup$ – MooS Aug 25 '17 at 13:07
  • $\begingroup$ @MooS I was quite unaware of it myself, until I had a hunch this might be the case. $\endgroup$ – rschwieb Aug 25 '17 at 13:08
  • $\begingroup$ I didn't know it, thank you! $\endgroup$ – Federico Aug 25 '17 at 13:26

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