# Atiyah-Macdonald, Exercise 8.3: Artinian iff finite k-algebra.

Atiyah Macdonald, Exercise 8.3. Let $k$ be a field and $A$ a finitely generated $k$-algebra. Prove that the following are equivalent:
(1) $A$ is Artinian.
(2) $A$ is a finite $k$-algebra.

I have a question in the proof of (1$\Rightarrow$2): By using the structure theorem, we may assume that $(A,m)$ is an Artin local ring. Then $A/m$ is a finite algebraic extension of $k$ by Zariski lemma. Since $A$ is Artinian, $m$ is the nilradical of $A$ and thus $m^n=0$ for some $n$. Thus we have a chain $A \supseteq m \supseteq m^2 \supseteq \cdots \supseteq m^n=0$. Since $A$ is Noetherian, $m$ is finitely generated and hence each $m^i/m^{i+1}$ is a finite dimensional $A/m$-vector space, hence a finite dimensional $k$-vector space.

But now how can I deduce that $A$ is a finite dimensional $k$-vector space?

• I have a question: What if we assume that if R is a field then an $R$-module $V$ is artininan, then does this imply that $V$ is finite dimensional? I know it does but how would one show this? Thank you. – Hugo Martin Mar 1 '12 at 8:58
• Dear Hugo: Suggestion: Try to prove this: Let $X$ be a set and $Y$ the set of subsets of $X$. Then $Y$ is an artinian poset iff $X$ is finite. – Pierre-Yves Gaillard Mar 1 '12 at 9:23
• @Hugo: You need to ask new questions separately, by going here. You shouldn't post a question as an answer on another question, even it is related. I have now converted your post below into a comment here on the question. – Zev Chonoles Mar 1 '12 at 10:34
• For those who would be interested, here are some details about $m/m^2$... – Watson Jan 24 '17 at 12:26

## 2 Answers

Prove the following result from general linear algebra:

If $V$ is a vector space over a field $k$ and if $\{0\}=V_0\subseteq V_1\subseteq\cdots\subseteq V_{n-1}\subseteq V_n=V$ where $V_{i}/V_{i-1}$ is a finite dimensional $k$-vector space for all $1\leq i\leq n$, then $V$ is a finite dimensional $k$-vector space.

Hint: The following steps lead to a solution. Please try to prove this on your own referring to each subsequent step as you need it. For example, first try to prove the result on your own; if you are truly stuck, read step (1). Then try to prove the result with step (1) as a hint and only if you are truly stuck should you read step (2).

(1) We work by induction on $n$. The result is clear if $n=1$. In general, let us assume that we have proven the result for $n-1$ and wish to prove the result for $n$. Show that it suffices to establish the following result:

If we have a tower $\{0\}\subseteq W\subseteq V$ where $W$ and $V/W$ are finite dimensional $k$-vector spaces, then $V$ is a finite dimensional $k$-vector space.

(2) Let us prove the result directly above. Since $V/W$ is finite dimensional, we may choose a tuple $(v_1+W,\dots,v_n+W)$ that spans $V/W$ where $v_i\in V$ for all $1\leq i\leq n$. Similarly, since $W$ is finite dimensional, we may choose a tuple $(w_1,\dots,w_m)$ that spans $W$ where $w_j\in W$ for all $1\leq j\leq m$. Prove that the tuple $(v_1,\dots,v_n,w_1,\dots,w_m)$ spans $V$.

(3) Conclude that $V$ is finite dimensional by induction.

I hope this helps!

Alternatively, the chain can be refined to a composition series which will be of finite length since each quotient is of finite length and for k-spaces this is equivalent to finite dimension. (Prop. 6.10 in A-M)