# Principal Ideals

1. Dec 3, 2008

### playa007

1. The problem statement, all variables and given/known data
Let I be an ideal of the commutative ring R, and let J = {y in R such that y^2 in I}
a) If R is the polynomial ring Q[x] and I is the principal ideal of R generated by x^4 + x^2, show that J is the principal ideal of R generated by x^3 + x
b) If R is a Principal Ideal Domain, show that J is an ideal of R

2. Relevant equations

3. The attempt at a solution
To solve for part a), I tried to write I = (x^4 + x^2) and (x^3 + x) explicitly and it turned out be long messy polynomial equations and i couldn't really show it rigorously that J = (x^3 + x) is true. For part b), one can take R to be Q[x] since Q[x] is a P.I.D. (since Q is a field), and by a), J must then be an ideal of R.

Any help on part a) particularly and b) would be very much appreciated.

2. Dec 3, 2008

### Office_Shredder

Staff Emeritus
I used < > isntead of ( ) for denoting ideal generated by since I thought it might be confusing when I'm referring to a polynomial vs. the ideal generated by it

If I = <x4 + x2>... that IS I... what kind of form does "explicitly" take? A generic polynomial is going to be one of the form f(x)(x4 + x2) where f is a polynomial.

A good place to start would be to show <x3 + x> is a subset of J... show x3 + x is in J, then for any polynomial f(x), f(x)*(x3 + x) is contained in J also.

Your answer to (b) is wrong. You can't assume R is a specific principal ideal domain, you have to show that the result holds for any principal ideal domain (for example, what if the domain was Z?)

3. Dec 3, 2008

### Dick

Just think of what membership in each ideal implies about what irreducible factors the polynomials must have.

4. Dec 5, 2008

### playa007

i've solved part b) and one direction of a); precisely the part that <x^3 + x> is contained in J but I have some difficulty proving the other inclusion. This is what I have so far: take any polynomial f in J, so f^2 is in I => so (x^4 + x^2) divides f^2 => x^2 divides f^2 (or even x divides f^2) and (x^2 + 1) divides f^2 but I can't show how to deduce that (x^3 + x) divides f => J is contained in (x^3 + x)

5. Dec 5, 2008

### Dick

If x|f*g then x|f or x|g. Doesn't that follow from the irreducibility of x?