# Gilmer-Parker Theorem

## Theorem

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

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

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

## Proof

Let $K$ be the field of quotients of $R$.

Let $R \xrightarrow \varphi R \sqbrk x \xrightarrow \psi K \sqbrk x$

where $\varphi, \psi$ - embedding homomorphisms.

Let $\map {\operatorname{Prim} } {R \sqbrk x}$ be set of primitive polynomials of $R \sqbrk x$.

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

0. $p_i = r_j * f_l \implies r_j \sim 1$

Let $\cont {f_i}$ be content of $f_i$.

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

As soon as polynomial domain over fraction field is Euclidean domain, it is GCD domain.

Let $\map {\operatorname{lcd} } {k_i}$ be the lowest common denominator of cofficients of $k_i$.

$k_0 = \gcd \set {p_1, p_2} \in K \sqbrk x$

$t_0 = \map {\operatorname {lcd} } {k_0} * k_0 \ \xrightarrow {\psi^{-1} } R \sqbrk x$

$t = \dfrac t {\cont {t_0} } \xrightarrow {\psi^{-1} } \map {\operatorname{Prim} } {R \sqbrk x}$

$\map {\operatorname{lcd} } {k_0}, \cont {t_0} \in K \sqbrk x^* \implies t \sim k_0$

$t \sim \gcd \set {p_1, p_2}$

$d = \gcd \set {\cont {f_1}, \cont {f_2 } } \in R$

l1. $\gcd \set {p_i, r_j} = 1 \in \R \sqbrk x$

$1 \divides p_i, \ r_j$

$x \divides r_j \implies x \in R$ (by in ID $\map \deg {f * g} = \map \deg f + \map \deg g)$

By 0:

$\paren {x \divides p_i \land x \in R} \implies x \sim 1$

So, any common divisor is associated with $1$

l2. $\gcd \set {a, b} = 1 \implies \paren {a \divides b * c \implies a \divides c}$

By $\gcd \set {a, b} * \lcm \set {a, b} = a b$:

$\gcd \set {a, b} = 1 \implies \lcm \set {a, b} = a b$

$\paren {a \divides b * c \land b \divides b * c} \implies \lcm \set {a, b} \divides b * c \implies a * b \divides b * c \implies a \divides c$

l3. $\begin{cases} t \sim \gcd \set {p_1, p_2} \in K \sqbrk x \\ t \xrightarrow {\psi^{-1} } \map {\operatorname{Prim} } {R \sqbrk x} \end{cases} \implies t \ \sim \gcd \set {p_1, p_2} \in R \sqbrk x$

3.1 $t \divides p_i \in K \sqbrk x \implies t \divides p_i \in R \sqbrk x$

$t \divides p_i \in K \sqbrk x \iff p_i = t * k_i$

$k_i = \dfrac {g_i} {\map {\operatorname{lcd} } {k_i} } = g_i * \map {\operatorname{lcd} } {k_i}^{-1} \implies$

$p_i = t * g_i * \map {\operatorname {lcd} } {k_i}^{-1}$

$p_i * \map {\operatorname{lcd} } {k_i} = t * g_i \implies$

$\begin{cases} t \divides p_i * \map {\operatorname{lcd} } {k_i} \\ \gcd \set {t, \map {\operatorname {lcd} } {k_i} } = 1 \ \text{(by l1)} \end{cases} \implies t \divides p_i \in R \sqbrk x$ (by l2)

3.2 $g \in R \sqbrk x \ g \divides p_1, p_2 \implies g \divides t \in R \sqbrk x$

$g \divides p_1, p_2 \in R \sqbrk x \implies \ (by \ \psi )$

$g \divides p_1, p_2 \in K \sqbrk x \implies \ (by \ t - \gcd \set {p_1, p_2} )$

$g \divides t \in K \sqbrk x \implies \ (by \ \psi^{-1} )$

$g \divides t \in R \sqbrk x$

I. $d * t \ | f_1, f_2$

4. $d \divides \cont {f_i} \in R \implies d \divides \cont {f_i} \in R \sqbrk x \ (by \ \varphi)$

5.

$\begin{cases} d \divides \cont {f_i} \\ t \divides p_i \end{cases} \in R \sqbrk x \implies$

$\begin{cases} d * t \divides \cont {f_i} * t \\ \cont {f_i} * t \divides \cont {f_i} * p_i \end{cases} \implies d * t \divides f_i$

II. $\forall h \in R \sqbrk x (h \divides f_1, f_2 \implies h \divides d * t)$

6. Let $h \divides f_1, f_2$

$h = \cont h * p_3$

$\cont h, p_3 \divides h \divides f_i$

$\begin{cases} \cont h, p_3 \divides \cont {f_i} * p_i \\ \gcd \set {p_i, \cont h} = 1 \ (by \ l1) \\ \gcd \set {p_3, \cont {f_i} } = 1 \end{cases} \implies$ (by l2)

$\begin{cases} p_3 \divides p_i \\ \cont h \divides \cont {f_i} \end{cases}$

7. $\cont h \divides \cont {f_i}, \cont {f_2} \implies$

$\cont h \divides \gcd \set {\cont {f_i}, \cont {f_2} } \ (by \ \varphi \ gcd \ is \ same \in R \text{ and } R \sqbrk x)$

$\cont h \divides d$

$\cont h * p_3 \divides d * p_3$

$h \divides d * p_3$

8. $p_3 \divides p_1, p_2$

$p_3 \divides t$ (by l3)

$d * p_3 \divides d * t \implies$ (by 7)

$h \divides d * t$

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

$\blacksquare$

## Source of Name

This entry was named for Robert William Gilmer and Thomas G. Parker.