Is the Transitivity of Ideals Always Guaranteed in Rings?

[esisk]

Hi All,
If I is an ideal of J and J is an ideal of R (the ring)... Is it possible that I is not an ideal of R. When are we guaranteed to have this transitivity, if at all? Thank you

 

[mparks]

I'm still trying to work out the details, but would this be an example?

Let Z be the ring of integers, then 2Z is an ideal of Z and 4Z is an ideal of 2Z. But 4Z is also an ideal of Z. I would guess that is you have the situation you described, then I would always be an ideal of R.

Let R be a ring and let J be an ideal of R. We must first ensure that J is a ring in order for it to have any ideals. So as long as its closed under subtraction and multiplication, its a ring, which we can confirm. Next let I be an ideal of J. Then all elements in I must be in R, since I is an ideal of J and J is one of R. So, as long as we can take any element in R, multiply by an element in I and still stay in I, then I would say that I is an ideal of R. Give that a try and see what happens.

 

[Jose27]

Hi All,
If I is an ideal of J and J is an ideal of R (the ring)... Is it possible that I is not an ideal of R. When are we guaranteed to have this transitivity, if at all? Thank you

In general this is not true: Consider I= \langle x^2 \rangle = \{ fx^2 : f\in J= \langle x \rangle \} with both I and J considered as subsets of R= \mathbb{R} [x]. By definition I ideal of J and J ideal of R, but I is not an ideal of R since ax^2 \notin I for all a\neq 0,1 such that a\in \mathbb{R} . As for conditions that guarantee transitivity I can't think of any but that J=R.

 

[Spartan Math]

Well, in all honesty, it's a bit silly to talk about "an ideal of an ideal" since we define (particularly in commutative algebra) ideals to be subsets of rings and proper ideals are not rings (ie. they are not subrings).

 

[Hurkyl]

Some authors use the word "ring" to refer to rngs, so ideals of ideals makes sense (because any ideal of a rng is a rng).

In what I've read I've never seen that usage -- although most authors will make an explicit statement they are not adopting that convention.

I'm going to assume the opening poster is talking about rngs.

 

[Hurkyl]

Jose, there is an error in your example: because J doesn't have a unit, the ideal of J generated by x² is not equal to x²J.

 

[Jose27]

Jose, there is an error in your example: because J doesn't have a unit, the ideal of J generated by x² is not equal to x²J.

You're right, but couldn't we "close" it under addition and then it still won't have any, for example, irrational miltiples of x^2?

 

[Hurkyl]

I believe that is true.