$1.$ Let $F$ be a field and let $R$ be the integral domain in $F[x]$ generated by $x^2$ and $x^3.$ Show that $R$ is not a unique factorization domain
Attempt: $R = \langle x^3,x^2 \rangle = \{p(x)~ x^3 + q(x)~x^2~~|~~p(x),q(x) \in F[x]\}$
I think we need to find an element in $R$ with more than one factorization. However, at the moment, I am not able to think of one. Please tell me how do I find one.
$2.$ In $\mathbb Z[\sqrt{-7]}$, show that $6+2 \sqrt{-7}$ and $1+3 \sqrt{-7}$ are not associated.
Attempt: Suppose $6+2 \sqrt{-7} = u~(1+3 \sqrt{-7})$ . We need to show that $u$ is not a unit. But $N(1+3 \sqrt{-7})=N(6+2 \sqrt{-7}) =64 \implies N(u)=1$ which is possible if and only if $u$ is a unit $\implies$ the two quantities must be associates.
Where could I be making a mistake?
$3.$ Prove or disprove that a sub domain of a Euclidean domain is a Euclidean Domain.
Attempt: Every Euclidean Domain is a principal ideal domain which means every Euclidean domain is a unique factorization domain as well.
So, if we prove that there exists a sub domain of a Euclidean domain which is not a unique factorization domain, then we are sure that a sub domain of a Euclidean domain is not a Euclidean domain.
Or we prove that every every sub ring of an Euclidean domain is a Euclidean domain, then we can prove it. Somewhere, I feel that this could be true.
If $d$ is the measure , A Euclidean domain $D$ satisfies : $(i) d(a) \leq d(ab)~\forall~a,b \in D$
$(ii)$ If $a =bq+r$, then $d(b) > d(r)$.
I think every sub ring of a Euclidean domain satisfies this? Hence, every sub domain of a Euclidean domain must be Euclidean
Am I correct?
Thank you for your help.
Best Answer
(1) $\langle x^2,x^3\rangle$ in $F[x]$ is not an integral domain because it's missing an identity. The ring we're talking about is $R=F[x^2,x^3]$ (or $K[x^2,x^3]$ where $K$ is the prime subfield, it doesn't really matter). Have you considered something that's $=$ to a product of $x^2$s or $x^3$s?
(2) You're assuming $u$ is an element of $\Bbb Z[\sqrt{-7}]$. Simplify $\frac{6+2\sqrt{-7}}{1+3\sqrt{-7}}$ to find it's not in $\Bbb Z[\sqrt{-7}]$ because every element of $\Bbb Q[\sqrt{-7}]$ has a unique representation as $a+b\sqrt{-7}$ with $a,b\in\Bbb Q$.
(3) The issue is that if $a,b$ are in a subdomain of an ED, the quotient $q$ and remainder $r$ in any division expression $a=bq+r$ need not be in the subdomain as well. An ED would be a UFD as you note, so it'd suffice to find a subdomain of an ED which is not a UFD, and you've already done this (see (1)).