Frac$(R)=F$, quotient field of the integral domain $R$, then Frac$(R[x]) \cong F(x)$ and also Frac$(R[x]) \cong F(x_{1},x_{2},....,x_{n})$.

Question

Let $R$ be an integral domain, and consider $\operatorname{Frac}(R)=F$ the quotient field of $D$. Then $\operatorname{Frac}(R[x]) \cong F(x)$ and also $\operatorname{Frac}(R[x]) \cong F(x_{1},x_{2},....,x_{n})$.

For the first part of the statement, as $\operatorname{Frac}(R)= \lbrace ab^{-1} / a,b \in R\rbrace$, so more than an isomorphism, I think this problem as an equality, so I need to see that if $p(x) \in \operatorname{Frac}(R[x])$ if and only if $p(x)= \frac{q(x)}{r(x)}$ with $q(x), r(x) \in R[x]$ and also show that $p(x_{1},x_{2},...,x_{n}) \in \operatorname{Frac}(R[x_{1},...x_{n}])$ if and only if $p(x_{1},x_{2},...,x_{n})= \frac{q(x_{1},x_{2},...,x_{n})}{r(x_{1},x_{2},...,x_{n})}$ with $q(x_{1},x_{2},...,x_{n}), r(x_{1},x_{2},...,x_{n}) \in F(x_{1},...,x_{n})$? If this way is right then Im supposed to use that every coefficient of $p(x) \in F(x)$ is also in $\operatorname{Frac}(R)$? Thanks

Answer 1

Hint for the first part:

Consider the ring homomorphisms:

$$f: R[X] \rightarrow F(X), p \mapsto \frac{p}{1}$$

$$i: R[X] \rightarrow Frac(R[X]), p \mapsto p$$

$$f': Frac(R[X]) \rightarrow F(X), pq^{-1} \mapsto \frac{p}{q}$$

Note that $f = f' \circ i$.

You need to show that those are indeed ring homomorphisms and that $f'$ is well-defined, surjective an injective. Remember that since $R$ is an integral domain, so is $R[X]$, and thus $Frac(R[X])$ is a field, and ring homomorphisms between fields are injective.

You should know that $i$ is a ring homomorphism from the universal property of localization. $i$ is also injective onto $Frac(R[X])$, since $R[X]$ is an intergral domain and does not contain zero divisors.

Also if $f$ and $f'$ are ring homomorphisms, then then $f'$ is unique.

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For the second part we already know that $Frac(R[X])≅F(X)$ and by the thread posted in the comments you know that $R[X_1, X_2, ..., X_n] \cong S[X]$, where $S=R[X_1, X_2, ..., X_{n-1}]$.

We also know that if $R$ was an integral domain $R[X]$ is an integral domain. So $S$ will be an integral domain too, and we can use our base argument.

So $Frac(R[X_1, ... X_n]) \cong Frac(S[X]) \cong F_S(X) $, where $F_S=Frac(S)$ by the base case. Can you show that this is isomorphic to $F(X_1, X_2, ..., X_n)$?

Note that $F_S(X)$ are fractions of functions dependant on $X$ with coefficients in $S=R[X_1, ..., X_{n-1}]$ and $F(X_1, ..., X_n)$ are fractions of functions in $n$ variables. Why are they the same?