Principal Ideals: Homework Solution

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In summary: By the way, what is the difference between x and x^2 in this context? Since we are working in a commutative ring, I'm not sure what the difference is between the two. Could you help me with this doubt?In summary, to prove part a), it suffices to show that <x^3 + x> is contained in J. To do this, we take any polynomial f in J and show that (x^3 + x) divides f. This can be done by noting that since f^2 is in I, (x^4 + x^2) divides f^2, which implies that (x^2 + 1) divides f^2. Since x is irreducible,
  • #1
playa007
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Homework Statement


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

Homework Equations





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.
 
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  • #2
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
Just think of what membership in each ideal implies about what irreducible factors the polynomials must have.
 
  • #4
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
If x|f*g then x|f or x|g. Doesn't that follow from the irreducibility of x?
 

1. What is a principal ideal?

A principal ideal is an ideal in a ring that is generated by a single element. In other words, every element in the ideal can be written as a multiple of a single element in the ring.

2. How do I find the generator of a principal ideal?

To find the generator of a principal ideal, you can use the definition of a principal ideal and determine which element in the ring can generate the ideal. This can involve checking if the ideal is closed under multiplication and addition, and if it contains all multiples of the generator.

3. Can a principal ideal have more than one generator?

No, a principal ideal can only have one generator. This is because the definition of a principal ideal states that it is generated by a single element.

4. Are all ideals in a ring principal ideals?

No, not all ideals in a ring are principal ideals. There are many examples of rings where not all ideals are generated by a single element. However, all principal ideals are ideals in a ring.

5. How are principal ideals related to principal ideal domains?

A principal ideal domain is a special type of ring where every ideal is a principal ideal. This means that every ideal in a principal ideal domain can be generated by a single element. Therefore, all principal ideal domains have the property that every ideal is a principal ideal.

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