Let $R$ be an integral ring (has no zero divisors) and let $F$ be a field contained in $R$. Supposse that $R$ is a vector space over $F$. Show that $R$ is a field.
This promem seems very easy, since I just have to found the multiplicative inverse, but I can't figure out who is it. Thanks.
This is clearly false: consider a field $F$ and the ring of polynomials $R=F[x]$.
You're missing the most important hypothesis: that $R$, under the natural structure, is a finite dimensional vector space over $F$.
Let $r\in R\setminus F$; then there exists $n>0$ such that $\{1,r,\dots,r^n\}$ is linearly dependent. Thus there exists a monic polynomial $f(x)=a_0+a_1x+\dots+a_{n-1}x^{n-1}+x^n$, with $a_0\ne0$, such that $f(r)=0$ (work out the details). Then, with a simple transformation, $$ 1=b_1r+b_2r^2+\dots+b_nr^n $$ and so $$ r^{-1}=b_1+b_2r+\dots+b_nr^{n-1}\in R $$
Here is another proof, assuming that $R$ is a finite dimensional vector space over $F$:
Take $r \in R$, $r\ne 0$ and consider $\mu: R \to R$ given by $x \mapsto xr$.
Then $\mu$ is a linear transformation which is injective because $R$ has no zero divisors. Therefore, because $R$ is finite dimensional over $F$, $\mu$ is surjective and so there is $s \in R$ such that $sr=1$.
Thus every non zero element of $R$ has an inverse and so $R$ is a field.
