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Let $(A,m)$ be an artinian local ring with residue field $k$.

  1. Suppose $k$ is algebraically closed. Is $A$ necessarily a $k$-algebra? If not what are some simple counterexamples?
  2. Suppose $k$ has characteristic $0$ (not necessarily algebraically closed). Is $A$ necessarily a $\mathbb{Q}$-algebra? If not what are some simple counterexamples?

EDIT: Later I realized this question is very much related to this one. I think it's still slightly simpler and might get an answer so I will not delete it yet.

user26857
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Saal Hardali
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1 Answers1

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For 2. this is alright. Clearly we have $\mathbb{Z}\subset A$ and all non-zero elements of $\mathbb{Z}$ are units in $A$, no artinian assumption is necessary. Thus, for any local ring with $k=A/\mathfrak{m}$, $\mathbb{Q}\subset A$, with no artinian assumptions.

If $k=A/\mathfrak{m}$ is a field of characteristic zero, then, Hensel's lemma would ensure that $k\subset A$.

Finally, in the case of $k$ as above is characteristic $p>0$, then the theory of Witt vectors says that $A$ may not be a $k$-algebra.

user26857
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Mohan
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