# Can We Always Build a Field out of an Integral Domain?

Link to Hungerford's Text

Let $R$ be an integral domain, and $F$ its quotient field (or field of fractions). Assuming that $\phi: R \rightarrow F$ is isomorphic, $R[x]$ is isomorphic with $F[x]$ with $g: R[x] \rightarrow F[x]$ given by:

$$g\big( \sum {\alpha_k}{x^k} \big)=\sum {\phi(\alpha_k)}{x^k}$$

Hungerford (Algebra, p. 145) proved that if $E$ is a field that contains $R$, then $F$ is isomorphic to $\iota(R)$ under the inclusion map $\iota: R \rightarrow E$. Then, $R \cong \iota(R) \cong F$, which is my assumption above. Well, how do I get the field, $E$, containing $R$?

-- Additional Info --

I was asking this question while learning Gauss's lemma: Suppose that $R$ is a UFD, and $F$ its field of fractions. Then if $p(x) \in R[x]$ is reducible in $F[x]$, then $p(x)$ is reducible in $R[x]$ too. I consulted a couple of sources online and they all begin the same way....

Let $p(x) = a(x)b(x)$ be a nontrivial factorization in $F[x]$. Multiply both sides by the appropriate $r, s \in R$ such that $rsp(x) = ra(x)sb(x)$ is a factorization in $R[x]$ .... This is akin to clearing the denominators in $\mathbb Q$.

1. The product $a(x)b(x)$ is in $F[x]$, so how does it equal $p(x) \in R[x]$?
2. Similarly, how do you multiple $a(x)b(x)$ by $r, s \in R$ and get back something in $R[x]$?
3. I know that I can have $r/1 \in F$, but that is still not $r \in R$. (I can map $r$ to $r/1$.)

This is why I am trying to find an isomorphism from $R$ to $F$ or $R[x]$ to $F[x]$.

• I think that you are making a lot of confusion. A domain is isomorphic to its fraction field if and only if it is a field. By the way, your question does not make much sense. – Crostul Jun 15 '15 at 19:46
• Surely you mean $F$ is isomorphic to the field of fractions of $\iota(R)$? – GPerez Jun 15 '15 at 19:50
• I made an error. Suppose that $f: F \rightarrow E$ is the injective homomorphism produced by the universal quotient property, then $F \cong f(F)$, and $f(F)$ contains $R$. – Andy Tam Jun 16 '15 at 16:04

## 2 Answers

This is the basic idea: suppose we have an integral domain, $R$. We can build a (usually bigger) integral domain $F$, which is "nicer", because it's a field, by taking the field of fractions.

Now we can take the polynomial ring $R[x]$, and create a (usually bigger) ring $F[x]$. We can think of $R[x]$ "sitting inside" $F[x]$ in the same way that we think of $R$ sitting inside $F$. So if we have the injective ring-homomorphism:

$r \mapsto r/1$, we can send:

$a_0 + a_1x + \cdots + a_nx^n \in R[x]$ to:

$(a_0/1) + (a_1/1)x + \cdots + (a_n/1)x^n \in F[x]$.

If $R$ is "an even nicer ring" than just a mere integral domain-say, a UFD, we can "go both ways", turning a factorization (of something in $R[x]$) in $F[x]$ into one in $R[x]$.

For example, consider: $7x^2 + 20x + 12 \in \Bbb Z[x]$. This factors in $\Bbb Q[x]$as:

$21\left(\dfrac{2x}{3} + \dfrac{4}{7}\right)\left(\dfrac{x}{2} + 1\right)$.

Let's re-write this as:

$21\left(\dfrac{14x}{21} + \dfrac{12}{21}\right)\left(\dfrac{x}{2} + 1\right)$

$= (14x + 12)\left(\dfrac{x}{2} + 1\right) = 2\left(\dfrac{x}{2} + 1\right)(7x + 6)$

$= (x + 2)(7x + 6)$, which is a factorization over $\Bbb Z[x]$.

At a more "abstract level", enlarging a UFD $R$, to its field of fractions $F$, doesn't help with factoring stuff in $R[x]$. This is somewhat surprising, because enlarging $F[x]$ to polynomials over a splitting field $E$, definitely does help with factoring, for example:

$x^2 - 2$ does not factor over $\Bbb Q$, but it does factor over any subfield of the complex numbers that contains $\sqrt{2}$.

• Ok, I see it now. Set $S = \{ r/1 : r \in R \} \subset F$. Then the mapping $\sum {\alpha_k}{x^k} \mapsto \sum \frac {\alpha_k}{1}{x^k}$ is an isomorphism between $R[x]$ and $S[x]$. TY! – Andy Tam Jun 16 '15 at 15:36

Let's use an example. $\Bbb Z$ is an integral domain; we can build the field of fractions $\Bbb Q$ out of it. If an integral domain $D$ is already a field, then its fields of fractions is isomorphic to $D$.

Exercise: What is the general element of the field of fractions of a polynomial ring $F[x]$?

A general element is of the form $$\dfrac{a_nx^n + a_{n-1}x^{n-1} + \dotsb + a_0}{b_nx^n + b_{n-1}x^{n-1} + \dotsb + b_0}$$ where $a_0, b_0, a_1, b_1 \dotsc \in F$. i.e. A polynomial divided by a polynomial.