- #1

- 26

- 0

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter mathmajor2013
- Start date

- #1

- 26

- 0

- #2

- 26

- 0

- #3

disregardthat

Science Advisor

- 1,866

- 34

If neither I nor J is contained in P, some element a in I and b in J are neither in P, as you say. But this is a contradiction, since ab is in P and P is prime, hence one of the ideals is contained in P.

Follow-up: Show this for an arbitrary finite number of ideals.

Follow-up: Show this for an arbitrary finite number of ideals.

Last edited:

- #4

Deveno

Science Advisor

- 908

- 6

but since I is an ideal, ab is in I. since J is an ideal ab is in J. therefore ab is in I∩J, and thus in P. since P is prime, either a is in P, contradicting our choice of a, or b is in P, contradicting our choice of b.

the only conclusion is that we cannot pick such a in I AND b in J outside of P, either I or J must lie within P.

- #5

disregardthat

Science Advisor

- 1,866

- 34

Share: