Every set with more than $k$ elements is dependent

Question

If $F$ is a free $R-$module, $R$ a domain, with rank=$k$ and $X\subset F$ show that $X$ is linear independent $\Rightarrow |X|\leq k$

There is a hint saying "consider the embedding of $R$ in it's field of fractions $R'$"

Attempt

Let $\phi:R\to R'$ be this embedding and $X\subset F$ with $|X|>k,X=\{x_1,...,x_m\}$

We may consider $F$ as a $R'-$module ($*$) so $R'-$ vector space and then since $dim_{R'}F=k$, $X$ is linear dependent over $R'$. Then I showed that $X$ is linear dependent over $R$.

So my question is :

How can I justify $(*)$ (if it is true)?? Is there a quicker way to answer this exercise??

Thanks!

Answer 1

A priori you can only view F as an R-module. However, the span of the image under a linear extension of the embedding can be viewed as a free module over the field of fractions. I.e. as a vector space over the field of fractions.

The nice thing is, that the embedding preserves the property of being linearly independent over R. Now the only thing you need to prove is that linear independence over R implies linear independence over R’ when R is a domain (multiply by every denominator in the R’ linear combination).

I hope this helps.

Answer 2

Alternatively, you could consider the tensor product $$F':=F\otimes_R R',$$ which is a vector space over $R'$.

A linear $R$-combination in $F$ $$r_1a_1+\cdots + r_na_n=0,$$ $a_i\in F, r_i\in R$ with some $r_j\neq0$, lifts to a linear $R$-combination in $F'$ $$r_1(a_1\otimes 1)+\cdots + r_n(a_n\otimes 1)=0,$$ which is clearly also a linear $R'$-combination with some $r_j\neq0$.

Reciprocally, any linear $R'$-combination in $F'$ of the form $$q_1(a_1\otimes 1)+\cdots + q_n(a_n\otimes 1)=0,$$ $a_i\in F, q_i\in R'$, $q_i=r_i/s_i$ with some $q_j\neq0$, gives a linear combination $$r_1(a_1\otimes 1)+\cdots + r_n(a_n\otimes 1)=0,$$ where $r_i=sq_i\in R$, with $s=s_1\cdots s_n$, so with $r_j\neq0$, which in turn projects to a linear combination $$r_1a_1+\cdots + r_na_n=0$$ in $R$ with some $r_j\neq0$.

These two facts together prove that a basis for $F$ gives a basis for $F'$ of the same cardinal, that all bases for $F$ have the same cardinal (since $F'$ is a vector space), that the rank of the $F$-module is indeed well defined, and that any set of more than rank$(F)$ vectors in $F$ must be linearly dependent (otherwise, it would generate an independent set in $F'$ with more vectors than $\dim_{R'} F'$, a contradiction).