Field of Fractions and free A-module( Proof verification)

Question

The following question was asked in my assignment of commutative algebra and I am not able to make considerable progress on this question.

Question: Let $A$ be an integral domain and let $K$ be it's field of fractions. Assume that $K\neq A$. Then show that $K$ is not a free A-module.

Proof: Let on the contrary $K$ is a free A-module. So, there exists an basis $k_1,...,k_r$ such that every $k\in K$ can be represented as $k= a_1 k_1+... + a_k k_r$. But we are given that $K\neq A$. So, there exists $a \in A/K$ such that $a$ cannot be written in terms of basis. But this is a contradiction as A is embedded in K (so there exists an inclusion map $i : A\to K.$ If I am not wrong inclusion / embedding means that A becomes a subset of K and using the above argument as $K$ is a free A-module and $A\neq K$).

Can you please let me know if the proof is correct? If not what is the correct way to prove this?

Answer 1

In your proof, "there exists $a\in K\setminus A$(?) such that..." is not reasonable, since you already have assumed "every $k\in K$ can be represented...".

One proof of the proposition goes as follows. Suppose $K$ was free. Let $x=a/b, y=c/d$ be arbitrary in $K$, then $bcx-ady=0$, which means no two elements are linearly independent over $A$. Therefore, $K$ must be of rank 1. But then, if $e=m/n$ is a basis, the element $m/n^2\in K$ can be written as $m/n^2=ae=am/n,$ then follows that $an=1$ since $A$ is a domain, so $e=am/an=am\in A$ and $K=Ae\subset A$, a contradiction.