Let $A$ be an unitary integral domain , and let $Fr (A)$ its fractionary field; this field is determined (to isomorphism field) by the following universal property:
a) the ring $A$ is injected by a ring homomorphism $i$ in $Fr (A)$,
b) for any monomorphism ring $j$ from $A$ in a field $k$, there exists a unique morphism field $f$ from $Fr (A)$ to $k$, such that $fi = j$.
Let $A [X]$ the ring of polynomials with coefficients in a unitary domain $A$; If $A = k$ is a field then the field $Fr (k [X])$ is noted $k (X)$. The question is to demonstrate that $Fr (A [X]) = Fr (A) (X)$.
I proceeds by saying that $A [X]$ is injected into $Fr (A) (X)$ by $j:$, so by universal property of $Fr (A [X])$ there is a field morphism from $Fr (A [X])$ to $Fr (A) (X)$ which of course is injective, but how to show the surjection? or if we can proceed otherwise and thank you in advance for any participation.
As $A \hookrightarrow A[X]$, $Fr(A) \hookrightarrow Fr\biggr(A[X]\biggl)$. Now $Fr(A)[X] \hookrightarrow Fr\biggr(A[X]\biggl)$, and thus $Fr(A)(X) \hookrightarrow Fr\biggr(A[X]\biggl)$ by the definition of fraction fields. Now you have already have a map $j$ from $Fr\biggr(A[X]\biggl)$ to $Fr(A)(X)$, check that $j$ actually is the inverse of the map constructed above.