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Let $R^n$ be a finitely generated free module of rank $n > 0$ over a principal ideal domain.

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

Is this correct? Is it a known fact? Is it true for an infinitely generated module?

My attempt to prove it:

Using mathematical induction:

  1. Base case

The statement is true for $n = 1$ and $n = 2$ (Axes of a free module over a PID).

  1. Induction step

Let's assume it is true for some $n \ge 2$.

Let's prove that it is also true for $n + 1$.

A non-zero element belongs an axis $A_k$ if and only if the coordinate of the element on $A_k$ is not $0$, and the coordinates of the element on other axes are $0$.

$R^{n+1} = R^n \oplus R$

For $R^n$ the statement is true, so for an arbitrary element $a$ there is an axis $A_k$ such that the coordinate of $a$ is not $0$ on $A_k$, and $0$ on all other other axes $A_{i \ne k}$:

$R^n = \sum_{i \ne k}A_i \oplus A_k$

then

$R^{n+1} = \sum_{i \ne k}A_i \oplus A_k \oplus R$

The coordinates of $a$ are $0$ on all $A_{i \ne k}$, not $0$ on $A_k$, and may or may not be $0$ on $R$.

But $A_k \oplus R$ is a module of rank $2$, and the property is true for it as well:

$A_k \oplus R = X \oplus Y$

where the coordinate of $a$ is $0$ on $X$, and not $0$ on $Y$.

Thus $R^{n+1} = \sum_{i \ne k}A_i \oplus X \oplus Y$

where the coordinate of $a$ is $0$ on all axes except $Y$.

Alex C
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    Hint. Write $a=(a_1,\dots,a_n)$ and use that $a\sim(d,0,\dots,0)$, where $d=\gcd(a_1,\dots,a_n)$. – user26857 Feb 09 '19 at 21:26
  • @user26857: Thank you! I was able to prove the $gcd$ formula for rank $2$, but still struggling to find a simple way for higher ranks. Can I read it somewhere? – Alex C Feb 10 '19 at 08:56
  • Found the $gcd$ formula in here: https://math.stackexchange.com/questions/391040/when-is-an-element-of-a-free-module-over-a-principal-ideal-domain-contained-in-a?rq=1. – Alex C Feb 10 '19 at 10:21

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