If $R$ is an integral domain, then $Frac(R[x_1,…,x_n])= Frac(K[x_1,…,x_n])$ where $K=Frac{(R)}$ I’m trying to show that if $R$ is an integral domain, then $Frac(R[x_1,…,x_n])= Frac(K[x_1,…,x_n])$ where $K=Frac{(R)}$ (here $Frac{(R)}$ denotes “field of fractions of $R$”).
I think I can use the universal property for the polynomial ring $R[x_1,\dots,x_n]$ applied to the map $R  \hookrightarrow K \hookrightarrow K[x_1,\dots ,x_n] \hookrightarrow Frac(K[x_1,\dots,x_n])$ to show that there exists a unique morphism of $R$-algebras $\phi : R[x_1,\dots,x_n] \rightarrow Frac(K[x_1,\dots,x_n]) $ such that $\phi(x_i)=x_i$ for all $i=1,\dots ,n$.
I guess the idea now would be to show that we can obtain an isomorphism $Frac(R[x_1,\dots, x_n)= Frac(K[x_1,\dots, x_n)$ but I’m not sure how to reach that conclusion. Any help is appreciated.
 A: Any polynomial with coefficients in $R$ can be regarded as a polynomial with coefficients in $K$, so there is a map $\phi:Frac(R[x_1,\ldots,x_n])\to Frac(K[x_1,\ldots,x_n])$ that regards rational functions with coefficients in $R$ as rational functions with coefficients in $K$ (a coefficient $r\in R$ maps to a coefficient $\frac{r}{1}\in K$). You should check that this is a well-defined ring map (appealing to universal properties will make this easier). Clearly $\phi$ is not the zero map and hence must be injective since the domain is a field. It remains to see that $\phi$ is surjective. Given a rational function with coefficients in $K$, one can multiply the top and bottom by suitable elements of $K$ so as to "clear denominators", thus showing that any element of $Frac(K[x_1,\ldots,x_n])$ "looks like" a rational function with coefficients in $R$, so the map $\phi$ is surjective and hence an isomorphism.
Example: take $n=1$, $R=\mathbb{Z}$ and $K=\mathbb{Q}$.
Then $\frac{2x}{5x^2}$ maps to $\frac{\frac{2}{1}x}{\frac{5}{1}x^2}$ under $\phi$.
Given $\frac{\frac{1}{2}x}{\frac{7}{3}x^5}$ we can "clear fractions" by multiplying the top and bottom by $\frac{2}{1}\cdot\frac{3}{1}$ to get $\frac{\frac{3}{1}x}{\frac{14}{1}x^5}$ which gets mapped onto by $\frac{3x}{14x^5}$ via $\phi$.
