Problem in finding gcd of $a$ and $b$ in a UFD.

Question

Let $R$ be a UFD.Let $a,b \in R$ where $a = {p_{1}}^{r_{1}}.{p_{2}}^{r_{2}}...{p_{n}}^{r_{n}}$ and $b = {p_{1}}^{s_{1}}{p_{2}}^{s_{2}}...{p_{n}}^{s_{n}}$, where $r_{i},s_{i} \in \mathbb N_{0}$ ,$p_i$'s are irreducible in $R$ and $p_{i}$ is not an associate of $p_{j}$ if and only if $i \neq j$.Then show that $d = {p_{1}}^{t_{1}}{p_{2}}^{t_{2}}...{p_{n}}^{t_{n}}$, is the gcd of $a$ and $b$, where $t_{i} = \min \{r_{i},s_{i} \}$.

My attempt :

First of all it can be easily observed that $d$ divides both $a$ and $b$.

Let $x$ be a common divisor of $a$ and $b$.

Let $k_i$ be the highest power of $p_i$ such that ${p_{i}}^{k_{i}}|x$.If $k_i > r_i$ then I think ${p_{i}}^{k_{i}}$ does not divide $a$.Though I fail to prove it.If I assume it then it is quite simple to prove that $x|d$ which proves the result.Please help me in showing ${p_{i}}^{k_{i}}$ does not divide $a$.Then I can do the remaining part.

Thank you in advance.

Answer 1

$\renewcommand{\epsilon}{\varepsilon}$$\renewcommand{\phi}{\varphi}$$\newcommand{\Set}[1]{\left\{ #1 \right\}}$I think you forgot to mention that $t_{i} = \min\Set{r_{i}, s_{i}}$.

You will need the following. In your notation,

$\epsilon = {p_{1}}^{e_{1}}{p_{2}}^{e_{2}} \cdots {p_{n}}^{e_{n}}$ divides $\phi = {p_{1}}^{f_{1}}{p_{2}}^{f_{2}}\cdots{p_{n}}^{f_{n}}$ iff for each $i$ we have $e_{i} \le f_{i}$.

Proof. $\epsilon \mid \phi$ means there is $\gamma = {p_{1}}^{g_{1}}{p_{2}}^{g_{2}}\cdots {p_{n}}^{g_{n}}$ such that $\phi = \epsilon \gamma$. Therefore $$ \phi = {p_{1}}^{f_{1}}{p_{2}}^{f_{2}}\cdots{p_{n}}^{f_{n}} = \epsilon \gamma = {p_{1}}^{e_{1} + g_{1}}{p_{2}}^{e_{2}+g_{2}} \cdots {p_{n}}^{e_{n}+g_{n}}. $$ Since $R$ is a UFD, it follows that for each $i$ we have $f_{i} = e_{i} + g_{i} \ge e_{i}$. Conversely, if for each $i$ we have $f_{i} \ge e_{i}$, take $g_{i} = f_{i} - e_{i}$ and build $\gamma$ as above to get $\epsilon \mid \phi$.

Then $c = {p_{1}}^{e_{1}}{p_{2}}^{e_{2}}\cdots{p_{n}}^{e_{n}}$ is a common divisor of $a$ and $b$ iff for each $i$ you have $e_{i} \le r_{i}$ and $e_{i} \le s_{i}$ iff for each $i$ you have $e_{i} \le \min\Set{r_{i}, s_{i}}$. Therefore $c$ is the greatest common divisor of $a, b$ when you have $e_{i} = \min\Set{r_{i}, s_{i}}$.