# Prove it is not a PID.

• Mar 29th 2011, 12:53 PM
joestevens
Prove it is not a PID.
Fix a field $\displaystyle k$ and consider the subring $\displaystyle k[x^2,x^3] \subseteq k[x]$ (the former ring is all polynomials with zero coefficient in front of x). Prove that $\displaystyle k[x^2,x^3]$ is not a principal ideal domain.

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

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

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

Is $\displaystyle I$ principal?
• Mar 29th 2011, 02:31 PM
Beaky
Suppose it is. Then $\displaystyle \exists r \in k[x^2,x^3]$ such that $\displaystyle I = (r)$.

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

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

If $\displaystyle a_{2} = 0$ then show I is a strict subset of (r). Then $\displaystyle a_{2} \neq 0$ 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, 03:00 PM
hatsoff
Sorry, correction..

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

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