0
$\begingroup$

I am reading an article. There is a step in which I suspect that they use a "result" that "Let $A$ be a local artinian $k$-algebra with residue field $k$. If $A$ is regular then $A$ is nothing but $k$."

I do not know whether it is true or not . If it is true, can anyone give me a source including it .

Thank you in advance .

Improve this question
$\endgroup$

1 Answer 1

Reset to default
3
$\begingroup$

A local ring $(A,\mathfrak{m})$ is regular iff the minimal number of generators of $\mathfrak{m}$ equals the Krull dimension of $A$. But Artinian rings have Krull dimension $0$, i.e. $\mathfrak{m}=0$, hence $A=k$. The latter fact can for example be found in Atiyah and MacDonald's book on Commutative Algebra.

Improve this answer
$\endgroup$
3
  • $\begingroup$ @AnKhuongDoan: you are most welcome! $\endgroup$
    – M.G.
    Commented Sep 12, 2018 at 18:31
  • 1
    $\begingroup$ Note that it's even true that an artin local ring $k$ which is reduced is equal to its residue field. This is because the unique prime ideal of an artin local ring is its nilradical (the nilradical is always the intersection of all prime ideals). It's also equivalent to the other statement since reduced = regular in codimension $0$. $\endgroup$
    – dorebell
    Commented Sep 13, 2018 at 1:59
  • $\begingroup$ @dorebell: good point! Thanks for mentioning this! $\endgroup$
    – M.G.
    Commented Sep 13, 2018 at 2:37

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .