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Let $R^2 = R \times R$ be a free module of rank $2$ over a principal ideal domain.

I am trying to prove that for every non-zero element $a$ of $R^2$ there is a basis such that $a$ belongs to one of the basis axes.

Is this correct?

My attempt to prove it:

Two elements $(a, b)$ and $(c,d)$ of $R^2$ form a basis if and only if there are scalars $x$, $y$, $v$, $w$ such that:

$x(a,b) + y(c,d) = (1,0)$
$v(a,b) + w(c,d) = (0,1)$

or

$xa + yc = 1$
$xb + yd = 0$
$va + wc = 0$
$vb + wd = 1$

The equations hold if

$ad - bc = 1$

then

$x = d$
$y = -b$
$v = -c$
$w = a$

Now let's take an arbitrary non-zero element $(a',b')$ from $R^2$.

In every PID the Bezout’s identity holds:

for any scalars $a'$, $b'$ of a PID there are scalars $c$, $d$ such that

$a'd - b'c = gcd(a',b')$

If $gcd(a',b') \ne 0$ (which is true if $a' \ne 0$ or $b' \ne 0$), then there are also scalars $a$, $b$

such that

$a' = gcd(a',b')a$
$b' = gcd(a',b')b$

and

$ad - bc = 1$

The scalars $a$, $b$, $c$, $d$ satisfy the necessary and sufficient equations, so elements $(a,b)$ and $(c,d)$ form a basis of $R^2$.

The element $(a',b')$ belongs to the axis generated by $(a,b)$ since $(a',b') = g(a,b)$, where $g = gcd(a',b')$.

Alex C
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  • Can it be shown that $ad-bc=1$ is a consequence of the four formulas after "or"? That is, you did show that if $ad-bc=1$ then the equations hold, but didn't show the converse. – coffeemath Feb 07 '19 at 10:32
  • @coffeemath: Do I need it? $ab - dc = 1$ may not be true for another basis. I showed that there is always an axis that contains an arbitrary element, but it may not be unique. – Alex C Feb 07 '19 at 10:42
  • Since you use $ad-bc=1$ during the development, it should be proved. I don't really follow what you're trying to do, but if it is to show any $a$ can be one of the components of what you term a basis, then given arbitrary $a$ one should show there are $b,c,d$ for which $ad-bc=1.$ That may be easy to do in a PID, I don't know... – coffeemath Feb 07 '19 at 19:02
  • @coffeemath: I am not trying to show that any scalar or element can be a part of a basis. I trying to show that one can draw a basis line through any element. There is a basis element that defines the line, but it is not arbitrary, and may not be unique, but at least one of them can be calculated based on the coordinates of the selected element. And the solution always exists since there are always $a,b,c,d$ in PID such that $ad - bc = 1$. – Alex C Feb 07 '19 at 19:19
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    @coffeemath: I added the detail steps into the original post. Please check if it is clear now. – Alex C Feb 08 '19 at 08:50
  • Yes, clear now. – coffeemath Feb 10 '19 at 11:12

1 Answers1

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Your intuition is correct. It's quite unusual to write $M$ for the regular module (free module of rank $1$), but I'll stick to it.

Suppose $(r,s)\in M^2$. Then let $d$ be a greatest common divisor of $r$ and $s$. If $r=da$ and $s=db$, you know that $1$ is a greatest common divisor of $a$ and $b$, so by Bézout's identity, there exist $c$ and $d$ such that $ad-bc=1$.

Then $\{(a,b),(c,d)\}$ is a basis of $M^2$ and $(r,s)=d(a,b)$.

egreg
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