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Let $R$ be a domain and let $Q$ be its field of fractions. Show that the field of fractions of $R[X]$ is isomorphic to $Q(X)$.

By the way, I don't know exactly what $Q(X)$ is. It means $Q[X]$? Or $Q$ times the ideal generated by $X$ in $R[X]$?

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Generally, $A[\alpha]$ means "the ring generated by A and $\alpha$", and $A(\alpha)$ means "the field generated by A and $\alpha$". I have only ever seen the latter when $A$ itself is a field.

In the case of an indeterminate variable $X$, $A[X]$ would be the ring of polynomials (with coefficients in $A$), and $A(X)$ would be the fraction field of $A[X]$: the field of rational functions (with coefficients in $A$).

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$Q(X)$ denotes the field of fractions of $Q[X]$, which is the smallest field generated by adjoining $X$ to $Q$. Essentially, you are being asked to prove that taking the field of fractions commutes with adjoining an indeterminant.

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$(X+1)^{-1} \notin Q[X, X^{-1}]$ – Hurkyl Mar 19 '12 at 7:05
$Q[X,X^{-1}]$ usually denotes the ring of Laurent polynomials, which is much smaller than the quotient field. – Mariano Suárez-Alvarez Mar 19 '12 at 7:06
@Hurkyl Quite right, fixed. – Alex Becker Mar 19 '12 at 7:08
@Alex: Having the tendency to make that mistake myself a few times each year, I feel better to have seen someone else do it too. :) – Hurkyl Mar 19 '12 at 7:09
Am I being asked to prove the field of fractions of R[X] is isomorphic to the field of fractions of Q[X]? – Yangzhe Lau Mar 19 '12 at 7:37

Hint $\:$ $Q(X)$ denotes the fraction field of $Q[X]$, i.e. rational functions in $X$. By the universal property of fraction fields, the natural injection $R[X] \to Q(X)$ lifts to the fraction field of $R[X]$, and the lift is surjective, since every element of $Q(X)$ is equivalent to a fraction over $R[X]$, by scaling both numerator and denominator by a common denominator $\in R$ of all the coefficients.

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