Prove it is not a PID.

• Mar 29th 2011, 01:53 PM
joestevens
Prove it is not a PID.
Fix a field $k$ and consider the subring $k[x^2,x^3] \subseteq k[x]$ (the former ring is all polynomials with zero coefficient in front of x). Prove that $k[x^2,x^3]$ is not a principal ideal domain.

I just need to find an ideal $I$ that is not principal, right?

$I=(x^2,x^3)= \{$polynomials in $x$ s.t. $a_0=a_1=0 \}$

Is this actually true? $(x^2,x^3)= \{$polynomials in $x$ s.t. $a_0=a_1=0 \}$

Is $I$ principal?
• Mar 29th 2011, 03:31 PM
Beaky
Suppose it is. Then $\exists r \in k[x^2,x^3]$ such that $I = (r)$.

$r = a_{0} + a_{1}x + ... + a_{n}x^n$

If $a_{n} \neq 0$ for any n > 2 then clearly (r) won't generate I since it will only contain polynomials of degree > 2. So $r = a_{0} + a_{1}x + a_{2}x^2$.

If $a_{2} = 0$ then show I is a strict subset of (r). Then $a_{2} \neq 0$ and show (r) contains no polynomials of degree 3 and so it can't be I.

I think that's fine anyways.
• Mar 29th 2011, 04:00 PM
hatsoff
Sorry, correction..

Let $I:=(x^2,x^3)=\{\sum_n a_ix^i:a_0=a_1=0,n\in\mathbb{N}\}$

Suppose towards a contradiction that $I=(p(x))$ for some nonzero $p\in(x^2,x^3)$. Then $x^2=p(x)q_1(x)$ and $x^3=p(x)q_2(x)$ for some $q\in k[x^2,x^3]$. So $\deg p+\deg q_1=2$. We know that $\deg p\geq 2$ since $p\neq 0$, so $0=\deg q_1$. This gives us $\deg p=2$ and therefore $2+\deg q_2=\deg p+\deg q_2=3$. Then $\deg q_2=1$, a contradiction.