Bezout Subring of Q[x] (9.3.5)

Dummit and Foote Abstract Algebra, section 9.3, exercise 5:

Let $R=\mathbb{Z}+x\mathbb{Q}[x] \subset \mathbb{Q}[x]$.

(a) Let $f(x),g(x)∈\mathbb{Q}[x]$ with $x^r$ the maximal power of $x$ dividing both $f(x)$ and $g(x)$, and let $f_r$ and $g_r$ be the coefficients of $x^r$ in $f(x)$ and $g(x)$ respectively. Then $\mathbb{Z}f_r+\mathbb{Z}g_r=\mathbb{Z}d_r$ for some nonzero $d_r\mathbb{Q}$. Prove there is a polynomial $d(x)∈\mathbb{Q}[x]$ such that $d(x)$ is a gcd of $f(x)$ and $g(x)$, and its minimal term is $d_rx^r$.
(b) Prove $f(x)=d(x)q_1(x)$ and $g(x)=d(x)q_2(x)$ for some $q_1(x),q_2(x)∈R$.
(c) Prove $d(x)=f(x)a(x)+g(x)b(x)$ for some $a(x),b(x)∈R$.
(d) Conclude $Rf(x)+Rg(x)=Rd(x)$ in $\mathbb{Q}[x]$, implying in part that $R$ is a Bezout domain.
(e) Show that there must be ideals of $R$ which are not finitely generated, and in fact $x\mathbb{Q}[x]$ is an ideal of $R$ that is not finitely generated.

Proof: (a) We can compute the gcd of $f(x)$ and $g(x)$ in the Euclidean domain $\mathbb{Q}[x]$. By the Euclidean algorithm we have $d(x)=v_1(x)f(x)+v_2g(x)$ for some polynomials $v_1(x),v_2(x)∈\mathbb{Q}$, so that $x^r \mid d(x)$. Furthermore, letting $f_r$ be the one of its pair that is guaranteed nonzero, by $d(x) \mid f(x)$ we must have the coefficient of $x^r$ in $d(x)$ is nonzero. Since this gcd is unique up to multiplication by a nonzero rational number, we can arrange for a gcd $d(x)$ with minimal term $d_rx^r$.

(b) Since $\mathbb{Z}f_r+\mathbb{Z}g_r=\mathbb{Z}d_r$ we have $d_rq_1=f_r$ and $d_rq_2=g_r$ for some $q_1,q_2∈\mathbb{Z}$. By definition of a gcd we have $f(x)=d(x)q_1(x)$ and $g(x)=d(x)q_2(x)$ for some $q_1(x),q_2(x)∈\mathbb{Q}[x]$, and by observation we see the constant terms of $q_1(x)$ and $q_2(x)$ are $q_1$ and $q_2$ respectively, putting these polynomials in $R$.

(c) We can observe $d(x)=f(x)a(x)+g(x)b(x)$ for some $a(x),b(x)∈\mathbb{Q}[x]$, and by 9.2.11 we have a full solution set for $a(x),b(x)$ given by $$a'(x)=a(x)+m(x) \dfrac{g(x)}{d(x)}$$ $$b'(x)=b(x)-m(x) \dfrac{f(x)}{d(x)}$$ as $m(x)$ ranges over $\mathbb{Q}[x]$.

Assume $f_r$ and $g_r$ are nonzero. First, observe that $d_r=f_r\alpha+g_r\beta$ for some $\alpha,\beta∈\mathbb{Z}$, so that $1=\dfrac{f_r}{d_r}\alpha+\dfrac{g_r}{d_r}\beta$, implying that $v_f=\dfrac{f_r}{d_r}$ is relatively prime to $v_g=\dfrac{g_r}{d_r}$ (which are, foremost, seen to be integers).

Letting $a$ and $b$ be the constant terms of $a(x)$ and $b(x)$ respectively, we can see by our mentioned equation that $(f_rx^r)a+(g_rx^r)b=d_rx^r$, so that $f_ra+g_rb=d_r$ implying $v_fa+v_gb=1$. As such, we can see $v_fz_1+v_gz_2=v_fa+v_gb$ for some $z_1,z_2∈\mathbb{Z}$, so that $v_fz_1-v_fa=-v_gz_2+v_gb$ and we can set $$m=\dfrac{z_1-a}{v_g}=\dfrac{-z_2+b}{v_f}$$ so that with some manipulation and replacement we have $$z_1=a+m\dfrac{g_r}{d_r}$$ $$z_2=b-m\dfrac{f_r}{d_r}$$ Now, by letting $m(x)=m$ and taking the solutions $a'(x)$ and $b'(x)$ as above, by noting that the constant terms of $\dfrac{f(x)}{d(x)}$ and $\dfrac{g(x)}{d(x)}$ must be $\dfrac{f_r}{d_r}$ and $\dfrac{g_r}{d_r}$ respectively we see that the equations above actually describe the constant terms of $a'(x)$ and $b'(x)$, which implies $a'(x),b'(x)∈R$.

Assume one of $f_r$ or $g_r$ is zero, letting $f_r$ be the nonzero. Since now $d_r \mid f_r$ and $f_r \mid d_r$ we have $f_r=\pm d_r$, so that in the equation $d(x)=f(x)a(x)+g(x)b(x)$ we can see the constant term of $a(x)$ must be $\pm 1$ as the latter summand doesn't affect the sum coefficient of $x^r$. Continuing the terminology of the last procession we still have the same complete set of solutions, although observing the constant terms $z_1$ and $z_2$ of $a'(x)$ and $b'(x)$ simplifying to $$z_1=a+m\dfrac{g_r}{d_r}=a=\pm 1$$ $$z_2=b-m\dfrac{f_r}{d_r}=b-m$$ So choose $m(x)=b$ so that $a'(x),b'(x)∈R$ are valid solutions.

(d) This is evident from parts (a) and (c) as greatest common divisors in $R$ both exist and are expressible as $R$-linear combinations of elements from $R$, so that $(~f(x),g(x)~)=(~d(x)~)$. This implies every finitely generated ideal of $R$ is principle, i.e. $R$ is a Bezout domain.

(e) Since $R$ has been shown in the previous exercise to not be a UFD, it cannot be a PID, so that $R$ must contain infinitely generated nonprinciple ideals. Now, assume $x\mathbb{Q}[x]=(a_1,...,a_n)$ is finitely generated; then $x\mathbb{Q} \subseteq x\mathbb{Q}[x]=a_1R+...+a_nR$. By observing the first degree terms, we have $\mathbb{Q}=a_1\mathbb{Z}+...+a_n\mathbb{Z}$, which translates to saying that $\mathbb{Q}$ as an additive group is finitely generated by the $a_i$, a contradiction by 2.4.14(d).$~\square$