# Thread: UFD that is not a PID

1. ## UFD that is not a PID

I have to prove that $\mathbb{Z}[x]$ is a unique factorization domain that is not a principal ideal domain.

I´m supposed to notice that $<2,x>$ cannot be generated by an ideal of one element.

Can anyone explain me why is that?

Thank you!

2. Suppose $<2,x>$ can be generated by a single element. That means that there exist $\alpha$ such that $\alpha|2$ and $\alpha|x$ but since $(2,x) = 1$ so it is impossible.

3. Originally Posted by vincisonfire
Suppose $<2,x>$ can be generated by a single element. That means that there exist $\alpha$ such that $\alpha|2$ and $\alpha|x$ but since $(2,x) = 1$ so it is impossible.
you also need to show that: $<2,x> \neq \mathbb{Z}[x]$: since the constant term of every polynomial in $<2,x>$ is a multiple of 2, we have $1 \notin <2,x>$ and thus $<2,x>$ is a proper ideal of $\mathbb{Z}[x].$

similarly you can easily prove a general result which is quite useful: if $R$ is a non-zero ring, then the polynomial ring $R[x]$ is a PID if and only if $R$ is a field.

4. NonCommAlg:
I´m trying to prove the last thing you said, but I can´t get it.
Can you give me any hints?
Thank you!

5. Originally Posted by Inti
NonCommAlg:
I´m trying to prove the last thing you said, but I can´t get it.
Can you give me any hints?
Thank you!
if $R$ is a field, we know that $R[x]$ will be a PID. that's a well-known fact. conversely, suppose $R[x]$ is a PID. then since $R$ is a subring of $R[x],$ it has to be an integral domain. we'll prove that

every non-zero element of $R$ is a unit, which means $R$ is a field. so let $0 \neq r \in R[x].$ let $I=.$ since $R[x]$ is a PID, there exists $f(x) \in R[x]$ such that $I=.$ hence $f(x)$ must

be a unit element of $R$ because $f(x) \mid r, \ f(x) \mid x.$ so $1 \in I,$ i.e. there exists $g(x), h(x) \in R[x]$ such that $rg(x) + xh(x)=1.$ now put $x=0$ to get $rg(0)=1,$ i.e. $r$ is a unit element of $R. \ \Box$