# Thread: every finite field is noetherian

1. ## every finite field is noetherian

i really need some help in proofing that Every Finite Field is Noetherian .
if anyone could help me i would be greatfull

2. Originally Posted by cig
i really need some help in proofing that Every Finite Field is Noetherian .
if anyone could help me i would be greatfull
Every field has only two ideals the trivial and the non-proper. Is this really what you meant?

3. Originally Posted by Drexel28
Every field has only two ideals the trivial and the non-proper. Is this really what you meant?
Actually,the question is : "Is it true that every finite ring is noetherian ? " I have to JUSTIFY my answer.. Nothing else is specified . Thank you in advance .

4. Originally Posted by cig
Actually,the question is : "Is it true that every finite ring is noetherian ? " I have to JUSTIFY my answer.. Nothing else is specified . Thank you in advance .
What would the existence of a non-terminating AC imply about the cardinality? Think about it..

5. Originally Posted by Drexel28
What would the existence of a non-terminating AC imply about the cardinality? Think about it..

I haven't understand the AC condition. In a finite ring it means that i have specific number of Ideals and that somehow it ends the series of ideals that are subsets. so by this condition it means that it is true ? I am really confused !

6. The problem is that you don't know what Noetherian means. Take care of that, and the answer to this problem will become obvious.

7. Originally Posted by Tinyboss
The problem is that you don't know what Noetherian means. Take care of that, and the answer to this problem will become obvious.

I know, as i said i am confused. Can you give me a hint or help me somehow ??

8. Originally Posted by cig
I know, as i said i am confused. Can you give me a hint or help me somehow ??
Suppose that $\#(R)=n$ then what if for any $\left\{A_1,\cdots,A_{n+1}\right\}\subseteq 2^R$ one has $A_1\subsetneq\cdots\subsetneq A_{n+1}$?

9. Originally Posted by Drexel28
Suppose that $\#(R)=n$ then what if for any $\left\{A_1,\cdots,A_{n+1}\right\}\subseteq 2^R$ one has $A_1\subsetneq\cdots\subsetneq A_{n+1}$?

This gives us a finite strincly increasing chain,contradicting the ascending chain condition that stabilises for Io=In for n>n0 ?

10. Originally Posted by cig
This gives us a finite strincly increasing chain,contradicting the ascending chain condition that stabilises for Io=In for n>n0 ?
You're thinking about this way too hard. What you want to prove is that any ascending chain terminates, right? But for any (non-empty) subsets $A_1,\cdots,A_{n+1}$ of your ring $R$ with $n$ elements if $A_1\subsetneq\cdots\subsetneq A_{n+1}$ then this implies that $\#(A_1)<\cdots\#(A_n)$ and so evidently $\#(A_{n+1})\geqslant n+\#(A_1)\geqslant n+1$...but this isn't possible since $\#(A_{n+1})\leqslant \#(R)=n$. Get the point?

11. Originally Posted by Drexel28
You're thinking about this way too hard. What you want to prove is that any ascending chain terminates, right? But for any (non-empty) subsets $A_1,\cdots,A_{n+1}$ of your ring $R$ with $n$ elements if $A_1\subsetneq\cdots\subsetneq A_{n+1}$ then this implies that $\#(A_1)<\cdots\#(A_n)$ and so evidently $\#(A_{n+1})\geqslant n+\#(A_1)\geqslant n+1$...but this isn't possible since $\#(A_{n+1})\leqslant \#(R)=n$. Get the point?
So since that this isnt possible for the cardinality of our #R=n ,then all A's are subsets and satisfies the AC condition and thus it is true that every finite ring is noetherian ?