This question was asked in my mid term exam of Module Theory(now over) and I need help in proving this result. I couldn't solve it in exam and tried again back home.
Let $(p_i)_{i\in \mathbb{N}}$ be a sequence of prime ideals with $p_i\nsubseteq p_j$ for all i,j $\in \mathbb{N}$ with i<j and put $a_n=\cap_{i\leq n|}p_i$, $n\in \mathbb{N}$ form a strictly descending chain of ideals in A. Deduce that if A is artinian(Take A to be commutative ring):
(a) Spec A= Spm A is a finite set
Attempt: 1st proving that Spec A is finite:A is artinian implies that there exists an n such that $a_n=a_{n+i}$ for all $i \geq 1$.
So, $ P_1 \cap P_2 … \cap P_n = P_1 \cap P_2\cap ….P_n \cap… P_{n+i}$ which implies that $P_{n+1} \cap …\cap P_{n+i} \subseteq P_1 \cap P_2 … \cap P_n$. Now , I tried to use the fact that $p_i\nsubseteq p_j$ for all i,j $\in \mathbb{N}$ with i<j to prove that $P_{n+1} \cap …\cap P_{n+i}=\phi$. But I am unable to form a rigorious argument and I think I need help.
Assuming that it is true , I tried to prove that SpecA =Spm A: By Krull's Theorem as A is non-zero so $Spm A \subseteq Spec A$. For converse: I tried by taking an ideal I in Spec A , ie ideal I is a prime ideal. On the contrary assume that it is not maximal ie there exists an ideal I' such that $I\subset I'\subset A$, but I am unable to find any contradiction.
(b) $nil A= m_A$
If (a) holds then (b) is clear.
(c) $A/ m_A$ is a finite product of fields.
$m_A= \cap_{i=1,…n } M_i$ , where $M_i$ i=1 to n is set of maximal ideals of A. If I factor a ring by maximal ideal I will get a field but I am not sure how exactly I will get product of fields. Can you please give hint on this?
I shall be really thankful of any help that I recieve.
Best Answer
The fact that prime ideals are maximal in an Artinian ring is a very frequently asked duplicate.
Assuming all the $p_i$ are prime (as the notation suggests) and hence maximal, they satisfy your setup. Think of what it would mean for $a_n\subseteq p_j$ for some $j>n$. It would mean that $\prod_{i\leq n} p_i\subseteq a_n\subseteq p_j$, in which case $p_i\subseteq p_j$ for some $i\leq n$. This has been ruled out by hypothesis though.
So in fact, $a_n$ is not contained in any such $p_j$, therefore $a_n\cap p_j$ is strictly contained in $a_n$. This amounts to saying that if there were infinitely many such $p_i$, they would necessarily produce a strictly descending chain with members $a_n$. Since that is not possible, the family of primes can only be finite.
(There are certainly duplicates for the question itself but I could not find any using the strategy requested. They could still be out there.)
b) as you said is obvious given a)
c) Assuming $m_A$ is the jacobson radical, equal to the intersection of all prime/maximal ideals in this scenario, you have immediately by the Chinese Remainder theorem that $A/m_A\cong \prod_P A/P$ where $P$ ranges over the prime ideals. Since each prime $P$ is maximal, these are all fields, and we've established that there are only finitely many.