Gilmer-Parker Theorem

Theorem
Let $\struct {\mathbf R, +, *}$ be a GCD Domain.

Let $\mathbf R \sqbrk x$ be a polynomial ring over $\mathbf R$.

Then $\mathbf R \sqbrk x$ is also a GCD Domain.

Proof
Let $\mathbf K$ be the field of quotients of $R$.

Let $\mathbf R \xrightarrow{\varphi} \mathbf R \sqbrk x \xrightarrow{\psi} \mathbf K \sqbrk x$

where $\varphi, \psi$ - embedding homomorphisms

Let $\forall i \in \N \set {r_i \in \mathbf R, \ f_i, g_i \in \mathbf R \sqbrk x, \ k_i \in \mathbf K \sqbrk x, \ p_i \in \mathbf{Prim(\mathbf R \sqbrk x)} }$

0. Let $\mathbf{Prim(\mathbf R \sqbrk x)}$ be set of primitive parts of $\mathbf R \sqbrk x \ \Leftrightarrow \ p_i \in \mathbf{Prim(\mathbf R \sqbrk x)} \Leftrightarrow p_i = r_j * f_l \Rightarrow r_j \sim 1$

Let $\map c {f_i} = content(f_i)$ - content of $f_i$

Let $f_1 = \map c {f_1} * p_1$, $f_2 = \map c {f_2} * p_2$

As soon as polynomial domain over fraction field is Euclidean domain, it is gdc-domain. $lcd(k_i)$ - lowest common denominator of cofficients of $k_i$.

$ \\ k_0 = \gcd \set {p_1, p_2} \in \ \mathbf K \sqbrk x \\ t_0 = lcd(k_0) * k_0 \ \xrightarrow{\psi^{-1}} \mathbf R \sqbrk x \\ t = \dfrac t {\map c {t_0} } \xrightarrow{\psi^{-1}} Prim(\mathbf R \sqbrk x) \\ lcd(k_0), \map c {t_0} \in \mathbf K \sqbrk x^* \Rightarrow t \sim k_0 \\ t \sim \gcd \set {p_1, p_2} \\ \ \\ d = \gcd \set {\map c {f_1}, \map c {f_2 } } \in \mathbf R \\ \ \\ l1. \gcd \set {p_i, r_j} = 1 \in \mathbf R \sqbrk x \\ 1 \divides p_i, \ r_j \\ x \divides r_j \Rightarrow x \in \mathbf R \ \text{(by in ID $\map deg {f, g} = \map \deg f + \map \deg g)$} \\ \begin{cases} x \divides p_i \\ x \in \mathbf R \end{cases} \Rightarrow x \sim 1 \text{(by 0.)} \\ \text{So, any common deviser is associated with 1} \\ \ \\ \ \\ l2. \ \gcd \set {a, b} = 1 \Rightarrow ( a \divides b*c \Rightarrow a \divides c ) \\ \gcd \set {a, b} = 1 \Rightarrow lcm(a,b) = ab \ \text{(by \gcd \set {a, b} * lcm(a,b) = ab)} \\ \begin{cases} a \divides b*c \\ b \divides b*c \end{cases} \Rightarrow \lcm \set {a, b} \divides b*c \ \Rightarrow a*b \divides b*c \ \Rightarrow a \divides c \\ \ \\ \ \\ l3. \begin{cases} t \sim \gcd \set {p_1, p_2} \ in \ \mathbf K \sqbrk x \\ t \xrightarrow{\psi^{-1}} Prim(\mathbf R \sqbrk x) \end{cases} \Rightarrow \ t \ \sim \gcd \set {p_1, p_2} \in \ \mathbf R \sqbrk x \\ \ \\ \ \\ 3.1 \ t \divides p_i \in \ \mathbf K \sqbrk x \ \Rightarrow \ t \divides p_i \ in \ \mathbf R \sqbrk x \\ \ \\ t \divides p_i \in \ \mathbf K \sqbrk x \ \Leftrightarrow \ p_i = t*k_i \\ k_i = \dfrac {g_i} {lcd(k_i)} = g_i * lcd(k_i)^{-1} \ \Rightarrow \\ p_i = t * g_i * lcd(k_i)^{-1} \\ p_i*lcd(k_i) = t*g_i \Rightarrow \\ \begin{cases} t \divides p_i*lcd(k_i) \\ gcd(t,lcd(k_i)) = 1 \ \text{(by l1)} \end{cases} \ \Rightarrow \ t \divides p_i \in \ \mathbf R \sqbrk x \ \text{(by l2)} \\ \ \\ \ \\ 3.2 \ g \in \mathbf R \sqbrk x \ g \divides p_1, p_2 \ \Rightarrow \ g \divides t \in \ \mathbf R \sqbrk x \\ g \divides p_1, p_2 \in \ \mathbf R \sqbrk x \ \Rightarrow \ (by \ \psi ) \\ g \divides p_1, p_2 \in \ \mathbf K \sqbrk x \ \Rightarrow \ (by \ t - \gcd \set {p_1, p_2} ) \\ g \divides t \in \ \mathbf K \sqbrk x \ \Rightarrow \ (by \ \psi^{-1} ) \\ g \divides t \in \ \mathbf R \sqbrk x \\ \ \\ \mathbf{I}. \ d*t \ | f_1, f_2 \\ \ \\ 4. \ d \divides c(f_i) \ in \ \mathbf R \ \Rightarrow \ d \divides c(f_i) \in \ \mathbf R \sqbrk x \ (by \ \varphi) \\ \ \\ 5. \\ \begin{cases} d \divides \map c {f_i} \\ t \divides p_i \end{cases} in \ \mathbf R \sqbrk x \ \Rightarrow \\ \ \\ \ \\ \begin{cases} d*t \divides \map c {f_i} * t \\ \map c {f_i} * t \divides \map c {f_i} * p_i \end{cases} \Rightarrow \ d*t \divides f_i \\ \ \\ \ \\ \mathbf{II}. \ \forall h \in \mathbf R \sqbrk x (h \divides f_1, f_2 \ \Rightarrow \ h \divides d*t) \\ \ \\ 6. \ let \ h \divides f_1, f_2 \\ h = \map c h * p_3 \\ \ \\ \map c h, p_3 \divides h \divides f_i \\ \begin{cases} \map c h, p_3 \divides \map c {f_i} * p_i \\ \gcd \set {p_i, \map c h} = 1 \ (by \ l1) \\ \gcd \set {p_3, \map c {f_i} } = 1 \end{cases} \Rightarrow (by \ l2) \begin{cases} p_3 \divides p_i \\ \map c h \divides \map c {f_i} \end{cases} \\ \ \\ \ \\ 7. \ \map c h \divides \map c {f_i}, \map c {f_2} \Rightarrow \\ \map c h \divides \gcd \set {\map c {f_i}, \map c {f_2} } \ (by \ \varphi \ gcd \ is \ same \ in \ \mathbf R \ and \ \mathbf R \sqbrk x) \\ \map c h \divides d \\ \map c h * p_3 \divides d * p_3 \\ h \divides d * p_3 \\ \ \\ \ \\ 8.\ p_3 \divides p_1, p_2 \\ p_3 \divides t \ (by \ l3) \\ d*p_3 \divides d*t \ \Rightarrow \ (by \ 7) \\ h \divides d*t $

So, for any $f_1, f_2 \in \mathbf R \sqbrk x$, we have that $\gcd \set {f_1, f_2} = d*t$