# Thread: Principal ideal and PID

1. ## Principal ideal and PID

(i) Show that (2) +(x) ={ f2 + gx: f, g in Z[x]} is not a principal ideal. So Z[x] is not a PID even though Z is a PID

(ii) Let r in R* where R is an integral domain. Show that (x) + (r) is a principal ideal <==> r in U(R). Hence show that R[x] is a PID <==> R is a field.

This is my attempt for part (i)

Let (a) = (2) + (x) for some a belongs to Z[x]
Clearly, (a) is the least principal ideal which contains both (2) and (x). Hence (2,x) exists and (2,x) = (a)
Since (a) = (2)+(x) we have (2) is the subset of (a) and (x) is a subset of (a). So, a|2 and a|x.
But also, a is in (a) = (2) +(x) and so a = 2f + gx.
Up to here, I dont know how to show it leads to contradiction so that (a) is not a principal ideal

Thank you in advanced for helping me

2. Originally Posted by knguyen2005
(i) Show that (2) +(x) ={ f2 + gx: f, g in Z[x]} is not a principal ideal. So Z[x] is not a PID even though Z is a PID

(ii) Let r in R* where R is an integral domain. Show that (x) + (r) is a principal ideal <==> r in U(R). Hence show that R[x] is a PID <==> R is a field.

This is my attempt for part (i)

Let (a) = (2) + (x) for some a belongs to Z[x]
Clearly, (a) is the least principal ideal which contains both (2) and (x). Hence (2,x) exists and (2,x) = (a)
Since (a) = (2)+(x) we have (2) is the subset of (a) and (x) is a subset of (a). So, a|2 and a|x.
But also, a is in (a) = (2) +(x) and so a = 2f + gx.
Up to here, I dont know how to show it leads to contradiction so that (a) is not a principal ideal

Thank you in advanced for helping me

Hint: The ideal (2)+(x) is the set of all the pol's in $\displaystyle \mathbb{Z}[x]$ with even free coefficient.

Tonio