0

Let $K$ be a field and let $A$ be a finitely generated commutative $K$-algebra which is also a local ring. Must $A$ be Artinian?

Geometrically, I belive that this should be true: A local ring is "essentially a point" and an Artinian ring is "essentially a finite union of points".

At Atiyah-Macdonald, Exercise 8.3: A finitely generated $k$-algebra is Artinian iff it is a finite $k$-algebra. , it is shown that for a finitely generated $K$-algebra $A$, the following are equivalent:

  1. $A$ is Artinian

  2. $A$ is finite-dimensional as a $K$-vector space

This means that my question reduces to showing that if $A$ is a finitely generated $K$-algebra which is also a local ring, then $A$ is a finite-dimensional $K$-vector space.

Unfortunately, I don't know how to show this statement.

I have the following criteria from my lecture which imply that a ring $R$ is Artinian:

  1. "Every finite-dimensional $K$-algebra is Artinian." This means that my question would be solved if I could show the following: If $A$ is a finitely generated $K$-algebra which is also a local ring, then $A$ is a finite-dimensional $K$-algebra.

  2. "A ring $R$ is Artinian if and only if it is Noetherian of Krull dimension zero." Additionally, I know that every finitely generated $K$-algebra is Noetherian. This means that my question would be solved if I could show the following: If $A$ is a finitely generated $K$-algebra which is also a local ring, then every prime ideal $\mathfrak{p} \subseteq A$ is also maximal.

  3. "A ring $R$ is Artinian if and only if it has finite length as an $R$-module." This means that my question would be solved if I could show the following: If $A$ is a finitely generated $K$-algebra which is also a local ring, then $A$ has finite length as an $A$-module.

  4. "Let $(R, \mathfrak{m})$ be a Noetherian local ring. Exactly one of the following statements are true: (a) $\mathfrak{m}^n \neq \mathfrak{m}^{n + 1}$ for all $n \in \mathbb{N}$ (b) $\mathfrak{m}^n = 0$ for some $n$, in which case $R$ is Artinian." This means that my question would be solved if I could show the following: If $A$ is a finitely generated $K$-algebra which is also a local ring with maximal ideal $\mathfrak{m}$, then there is $n$ with $\mathfrak{m}^n = 0$.

Unfortunately, I don't know how to show any of the statements in (1.), (2.), (3.) or (4.).

Let's look at some examples. For example, we have the local finitely generated $K$-algebra $\newcommand\quotient[2]{{^{\Large #1}}/{_{ \Large #2}}} \quotient{K[x]}{\langle x^n \rangle}$. This is a finite-dimensional $K$-vector space, hence Artinian. We also have $\quotient{K[x, \ y]}{\langle x^2, \ xy, \ y^2 \rangle}$. This is also a finite-dimensional $K$-vector space, hence Artinian. There is also $K[[x]]$, which is a local $K$-algebra which is not Artinian, but this is not finitely generated.

Martin Brandenburg
  • 181,922
Smiley1000
  • 4,219

1 Answers1

1

Let $P$ be a prime ideal. Since $A$ is Jacobson, it follows that $P$ is the intersection of the maximal ideals that contain it. But there is just one maximal ideal, since $A$ is local. It follows that $P$ is the maximal ideal.

This shows that the dimension of $A$ is at most* $0$, which shows that $A$ is Artinian.

*Notice that the zero ring has dimension $-\infty$ (see MO). In most books and most contexts, "Krull dimension $0$" is not correct and needs to be replaced with "Krull dimension $\leq 0$".

Martin Brandenburg
  • 181,922