How does $s'$ define from the theorem?

57 Views Asked by At

I encounter this theorem from my introductory algebra textbook. I'm quite not good at defining some of the definitions. Here is the theorem:

Theorem (3.2)

Let $m_{1}$, $m_{2}$ be positive integers. Let $d$ be a positive generator for the ideal generated by $m_{1}$ and $m_{2}$. Then $d$ is the greatest common divisor of $m_{1}$ and $m_{2}$.

Proof: Since $m_{1}$ lies in the ideal generated by $m_{1}$ and $m_{2}$ (because $m_{1}=1m_{1}+0m_{2}$), there exists an integer $q_{1}$ such that: $m_{1}=q_{1}d$, whence $d$ divides $m_{1}$, Similarly, $d$ divides $m_{2}$. Let $e$ be a non-zero integer dividing both $m_{1}$ and $m_{2}$, say $m_{1}=h_{1}e$ and $m_{2}=h_{2}e$ with integers $h_{1}$, $h_{2}$. Since $d$ is in ideal generated by $m_{1}$ and $m_{2}$, there are integers $s_{1}$, $s_{2}$ such that $d=s_{1}m_{1}+s_{2}m_{2}$.

What do the $s_i$ define for $d$? I know this question is kinda foolish to ask here. I just want some clarification for this! Thanks.

1

There are 1 best solutions below

1
On BEST ANSWER

A basic theorem on the ring of integers $\mathbb{Z}$ is that every ideal thereof has a unique nonnegative generator. More precisely, if $I$ is an ideal, then there exists a unique $d\ge0$ such that $I=d\mathbb{Z}$.

Now it depends on how you define the greatest common divisor. Since your statement insists on the numbers being positive, I guess the definition is “the largest positive number $d$ such that $d$ divides both $m_1$ and $m_2$”.

The ideal $I$ generated by $m_1$ and $m_2$ consists of all integers of the form $x_1m_1+x_2m_2$, with $x_1,x_2$ arbitrary integers, because this is an ideal (prove it) and it is the smallest one containing both $m_1$ and $m_2$ (prove it).

By the basic theorem, there exists $d\ge0$ such that $I=d\mathbb{Z}$ and, since $I\ne\{0\}$, we get that $d>0$.

In particular $m_1,m_2\in d\mathbb{Z}$, so $d$ divides both. Also, since $d\in I$, we can find integers $s_1,s_2$ such that $d=s_1m_1+s_2m_2$ by the description of $I$ given above.

Suppose $e>0$ divides both $m_1$ and $m_2$. Then $m_1=h_1e$ and $m_2=h_2e$, for some integers $h_1,h_2$. Then $$ d=s_1m_1+s_2m_2=s_1h_1e+s_2h_2e=(s_1h_1+s_2h_2)e $$ which implies $e$ is a divisor of $d$. Since both are positive, this forces $e\le d$.

Therefore $d$ is the greatest common divisor.