Birkhoff prime ideal theorem
If is prime, then we are done. Let . Then . Order by inclusion. This turns into a poset. Let be a chain in . Let . If , then and for some ideals . Since is a chain, we may assume that , so that as well. This means . Next, assume and . Then for some ideal , so that also. This shows that is an ideal. If , then for some , contradicting the definition of . So and also. This shows that every chain in has an upper bound. We can now appeal to Zorn’s lemma, and conclude that has a maximal element, say .
We now want to show that is the candidate that we are seeking: is a prime ideal in and . Since , is an ideal such that . So the only thing left to prove is that is prime. This amounts to showing that if , then or . Suppose not: . Let be the ideal generated by elements of and , and the ideal generated by and . Since and properly contain , and . Write and , where . Then and . Take the meet of these two expressions, and we obtain . Since is distributive, on the left hand side, we get . On the right hand side, we have . As the left hand side is less than or equal to the right hand side, we get that . Since , , a contradiction. Therefore, is prime and the proof is complete. ∎
In the proof, we use the fact that, an element belongs to the ideal generated by ideals iff is less than or equal to a finite join of elements, each of which belongs to some .
The theorem can be generalized: if we use a subset instead of an element , there is a prime ideal containing but excluding .
|Title||Birkhoff prime ideal theorem|
|Date of creation||2013-03-22 17:02:18|
|Last modified on||2013-03-22 17:02:18|
|Last modified by||CWoo (3771)|