# Thread: UFD that is not a PID

1. ## UFD that is not a PID

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

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

Can anyone explain me why is that?

Thank you!

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

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

similarly you can easily prove a general result which is quite useful: if $\displaystyle R$ is a non-zero ring, then the polynomial ring $\displaystyle R[x]$ is a PID if and only if $\displaystyle R$ 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 $\displaystyle R$ is a field, we know that $\displaystyle R[x]$ will be a PID. that's a well-known fact. conversely, suppose $\displaystyle R[x]$ is a PID. then since $\displaystyle R$ is a subring of $\displaystyle R[x],$ it has to be an integral domain. we'll prove that

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

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