Are Principal Ideals of Associates Equal?

[Bleys]

I'm wondering something about principal ideals which I'm using to prove something.
K-field, let f,h be non-constant in K[X]. If f and h are associates, does it follow (f) = (h) ?
I tried to just prove it myself but I'm not sure if it's correct.
f, h associates means f=ch for some unit c. Then (f) = {fg : g in K[X]} = {hcg : g in K[X]}. Now I want to put " = (h) " but I'm not sure if that's correct. I think it is, because g runs over all of K[X], and cK[X] = K[X], so {hcg : g in K[X]} = {hq : q in K[X]}. So is my statement true?

 

[lavinia]

(f) = {fg : g in K[X]} = {hcg : g in K[X]} shows that any element in the ideal generated by f is in the ideal generated by g. Since c is a unit the converse is also true.

 

[Bleys]

(f) = {fg : g in K[X]} = {hcg : g in K[X]} shows that any element in the ideal generated by f is in the ideal generated by g. Since c is a unit the converse is also true.

Did you mean h?

Also, is the equality {hcg : g in K[X]} = {hq : q in K[X]} incorrect?

 

[lavinia]

yes I meant h

 

[dmatador]

a = bu, where u is a unit. also from this you get b = au^-1. obviously a is in (b), which means that (a) is a subset of (b), and b is in (a), so (b) is a subset of (a). by double inclusion, they are equal.