ideals contained in a union of ideals
Assume that is a commutative ring.
Lemma. Let , , be ideals in such that . Then or .
Proof. Assume that this is not true. Then there are such that , and , . Obviously and without loss of generality we may assume that . Then . Contradiction.
Proposition. Let , be ideals in such that each is prime. If , then there exists such that .
Proof. We will use the induction on . For our lemma applies. Let . Assume that . For define
Now for any define . Then we have
and thus there is such that . Since for any , then we have that
But is prime, so there is such that . Contradiction.
Counterexample. We will show, that if ’s are not prime, then the thesis no longer hold, even when . Consider the ring of polynomials in two variables over a simple field of order , i.e. . Let . For we shall write . Then it is easy to see, that
It can be easily checked, that are all ideals and but obviously for any .
|Title||ideals contained in a union of ideals|
|Date of creation||2013-03-22 19:03:55|
|Last modified on||2013-03-22 19:03:55|
|Last modified by||joking (16130)|