Proof that an integral domain that is a finite-dimensional $F$-vector space is in fact a field

Weintraub's 15-line proof is correct but clumsy. Here is a 2-line proof:

Given $0\neq r\in R$ the $F$-linear map $R\to R:x\mapsto rx$ is injective ($R$ is an integral domain!), hence surjective ($R$ is finite-dimensional!). So $1$ is the image of some $s\in R$, i.e. $sr=1$ and so $s=r^{-1}$ belongs to $R$.

Another easy solution. I will explicitly construct the inverse.

Since $R$ is finite dimensional over $F$, we have $\{1,r,r^{2},...,r^{n}\}$ is a linear dependent set for some finite $n$ over $F$. In particular, if $r \neq 0$ and $r \in R$, then $a_{n}r^{n}+a_{n-1}r^{n-1}+\cdots+a_{0} =0$ has a nontrivial solution where each $a_{i} \in F$. If $a_{0}=0$ then $$a_{n}r^{n}+a_{n-1}r^{n-1}+\cdots+a_{1}r=0 \implies r(a_{n}r^{n-1}+a_{n-1}r^{n-2}+\cdots+a_{1})=0 \implies a_{n}r^{n-1}+a_{n-1}r^{n-2}+\cdots+a_{1}=0$$ since $R$ is an integral domain. If $a_{1}=0$ repeat the previous step. Clearly this process will terminate once we get to some nonzero $a_{i}$. Therefore we may assume WLOG that $a_{0} \neq 0$. But then $$a_{n}r^{n}+a_{n-1}r^{n-1}+\cdots+a_{0} =0 \implies a_{n}r^{n}+a_{n-1}r^{n-1}+\cdots+a_{1}r=-a_{0} \implies b_{n}r^{n}+b_{n-1}r^{n-1}+\cdots+b_{1}r=r(b_{n}r^{n-1}+b_{n-1}r^{n-2}+\cdots+b_{1}) =1$$ where $b_{i}=-a^{-1}_{0}a_{i}$, showing that $r$ has an inverse in $R$.

$\{1,2,4,8,\ldots\}$ is certainly $\mathbb{Q}$-linearly dependent in $\mathbb{Q}[\sqrt{2}]$; in fact, it is linearly dependent in $\mathbb{Q}$ already! $0 = 2(1) -1(2)$, with the elements in parentheses being the vectors. So this is a nontrivial linear combination of the vectors in the set which is equal to $0$.

For $\sqrt{2}$, the set is $\{1,\sqrt{2},2,2\sqrt{2},4,\ldots\}$. Again, this is $\mathbb{Q}$-linearly dependent, since $0 = 2(1) + 0(\sqrt{2}) -1(2)$. Again, this is a nontrivial linear combination of the vectors in the set which is equal to $0$.

What is it that makes it look "not dependent" to you?