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?

  • $\begingroup$ 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. $\endgroup$ Mar 1 '12 at 8:58
  • $\begingroup$ 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. $\endgroup$ Mar 1 '12 at 9:23
  • $\begingroup$ @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. $\endgroup$ Mar 1 '12 at 10:34
  • $\begingroup$ For those who would be interested, here are some details about $m/m^2$... $\endgroup$
    – Watson
    Jan 24 '17 at 12:26

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)


The claim also seems to follow from the Noether normalization lemma:

Let $B:=k[x_1,..,x_n]$ with $k$ any field and let $I \subseteq B$ be any ideal. Since $A$ is a finitely generated $k$-algebra you may let $A:=B/I$. By the Noether normalization lemma it follows there is a finite set of elements $y_1,..,y_d \in A$ with $d=dim(X)$ and the property that the sub-ring $k[y_1,..,y_d] \subseteq A$ generated by the elements $y_i$ is a polynomial ring. The ring extension $k[y_1,..,y_d] \subseteq A$ is an integral extension of rings. If $d=0$ it follows from the same lemma the ring extension $k \subseteq A$ is integral and since $A$ is finitely generated as $k$-algebra by the elements $\overline{x_i}$ and since each element $\overline{x_i}$ is integral over $k$, it follows $dim_k(A) < \infty$.

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

Answer: It seems from the argument above you can use the Noether normalization lemma to give another proof of your implication, different from the proofs given above. Hence now you have two proofs of your result.


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