# Field of Fractions

Throughout this article, A denotes an integral domain (which may not be a UFD).

Definition.

The field of fractions of A is an embedding of A into a field K,

$i : A\hookrightarrow K$

such that every element of K can be expressed as $\frac a b$, where $a\in A$ and $b\in A-\{0\}$.

Exercise A

Prove that the field of fractions of A is unique up to isomorphism.

Examples

1. $\mathbb Q$ is a field of fractions of $\mathbb Z$.

2. $\mathbb Q[\sqrt{-1}]$ is a field of fractions of $\mathbb Z[\sqrt{-1}]$.

3. A field of fractions of $k[X]$ (k = field) is given by the set of all $\frac{f(X)}{g(X)}$ where $f(X), g(X)\in k[X]$ and $g(X) \ne 0$.

The third example strongly hints at a general construction.

Proposition 1.

Every domain A has a field of fractions.

The trick is to “formally” invert elements. Although the proof is long, most of it is tedious mechanical work and not particularly illuminating.

Proof

We take the set of all pairs $(a,b) \in A\times (A - \{0\})$, together with the relation:

$(a,b) \sim (a',b') \iff ab' = a'b.$

This is an equivalence relation: to check transitivity, if $ab' = a'b$ and $a'b'' = a''b'$ where $b, b', b'' \ne 0$, then $(ab'' - a''b)b' = (ab')b'' - (a''b')b = (a'b)b'' - (a'b'')b = 0$ and since $b'\ne 0$ we have $ab'' = a''b$.

Write $\frac a b$ for the equivalence class of $(a,b)$ and let K be the set of equivalence classes. We define addition and product on K as follows.

$\frac a b + \frac {c}{d} = \frac{ad + bc}{bd}, \qquad \frac a b \times \frac {c}{d} = \frac{ac}{bd}.$

Now show that these operations are well-defined, e.g. for addition if $\frac {c}{d} = \frac{c'}{d'}$ then $\frac{ad + bc}{bd} = \frac{ad' + bc'}{bd'}$, which follows from

$(ad+bc)(bd') = abdd' + b^2(cd') = abdd' + b^2(c'd) = (ad' + bc')(bd).$

We leave it to the reader to verify that K, with the above operations, forms a commutative ring with 1. It is a field since if $\frac a b \ne 0$, then $a\ne 0$ so $\frac a b \times \frac b a = 1$. ♦

# Field of Fractions of UFD

Let A be a UFD and K its field of fractions. We say that $x,y\in K^*$ are associates if $\frac x y$ is a unit in A.

Now fix a set of primes $P \subset A$ modulo associates. For example, in ℤ, we can take $P = \{2, -3, -5, 7, 11, 13, \ldots\}$.

Proposition 2.

Any $x\in K^*$ can be uniquely written as

$x = u\pi_1^{e_1} \pi_2^{e_2} \ldots \pi_n^{e_n},$

where $u\in U(A)$, $\pi_i \in P$ and $e_i \in \mathbb Z - \{0\}$. We call this the prime factorization of $x\in K^*$.

Proof

Write $x = \frac a b$ where $a,b\in A-\{0\}$. Each of a and b can be written as $u \pi_1^{e_1} \ldots \pi_k^{e_k}$ where u is a unit, $\pi_i \in P$ and $e_i \in \mathbb Z_{\ge 0}$. Expanding $x = \frac a b$ and removing terms with exponent 0 gives us an expression as above.

For uniqueness, suppose $u \pi_1^{e_1} \ldots\pi_k^{e_k} = u' \pi_1^{f_1} \ldots \pi_k^{f_k}$ where $e_i, f_i \in \mathbb Z$. Without loss of generality assume $e_1 \ge f_1$. If $e_1 > f_1$, then $u \pi_1^{e_1 - f_1} \pi_2^{e_2} \ldots \pi_k^{e_k} = \pi_2^{f_2} \ldots \pi_k^{f_k}$. The LHS is a multiple of $\pi_1$ in A but the RHS is not; this gives a contradiction. Hence $e_1 = f_1$. This holds for the remaining terms as well. ♦

Now we can extend the following notions for $x,y \in K^*$. Let

$x = u\pi_1^{e_1} \ldots \pi_k^{e_k}, \ \ y = u'\pi_1^{f_1} \ldots \pi_k^{f_k}, \quad u, u' \in U(A), \pi_i \in P, e_i, f_i \in \mathbb Z$

be their prime factorizations.

• Divisibility : we write $x|y$ if $\frac y x \in A$, equivalently $e_i \le f_i$ for each $1\le i \le k$.
• Gcd : write $\gcd(x,y) = \pi_1^{\min(e_1, f_1)} \ldots \pi_k^{\min(e_k, f_k)}$.
• Lcm : write $\mathrm{lcm}(x,y) = \pi_1^{\max(e_1, f_1)} \ldots \pi_k^{\max(e_k, f_k)}$.

As before if $\gcd(x_1, \ldots, x_n) = g$, then $y|x_i$ for each i if and only if $y|g$. Similarly, if $\mathrm{lcm}(x_1, \ldots, x_n) = h$, then $x_i|y$ for each i if and only if $h|y$.

Example

In $\mathbb Q$ we have

$\gcd(\frac{20}3, \frac{50}9, \frac{8}5) = \frac{2}{45}, \quad \mathrm{lcm}(\frac{20}3, \frac{50}9, \frac{8}5) = 200.$

# Gauss’s Lemma

Here is our theorem of the day.

Theorem (Gauss).

If $A$ is a UFD, so is its ring of polynomials $A[X]$.

Proof

We say $f \in A[X] - \{0\}$ is primitive if its coefficients have gcd 1.

Step 0. Any prime $\pi \in A$ is also prime in $A[X]$.

Because $A[X]/(\pi) \cong (A/\pi)[X]$ is an integral domain.

Step 1. The product of two primitive polynomials is primitive.

Let $f = a_0 + a_1 X + \ldots \in A[X]$ and $g = b_0 + b_1 X + \ldots \in A[X]$ be primitive.

• Let $\pi \in A$ be any prime; it suffices to show there is a coefficient of fg not divisible by $\pi$.
• Since f is primitive there is a maximum d such that $a_d \not\in (\pi)$; similarly, there is a maximum e such that $b_e \not\in (\pi)$.
• The coefficient of $X^{d+e}$ in fg is not a multiple of $\pi$ since it is:

$a_0 \overbrace{b_{d+e}}^{\in (\pi)} + a_1 \overbrace{b_{d+e-1}}^{\in (\pi)} + \ldots + \overbrace{a_d b_e}^{\not\in (\pi)} + \ldots + \overbrace{a_{d+e}}^{\in (\pi)} b_0.$

Step 2. Extend A into a field.

Let K be the field of fractions of A. For any $g \in K[X] - \{0\}$, let $c(g)$ be the gcd of the coefficients of $g$ and $p(g) := c(g)^{-1} g$. Note that $p(g) \in A[X]$ since $c(g)\in K^*$ divides every coefficient of $g$. Also, the gcd of all coefficients of $p(g)$ is 1 so we have:

$g = c(g)p(g), \quad c(g)\in K^*, p(g) \in A[X] \text{ primitive.}$

This expression is clearly unique: if $g = c'p$ where $c'\in K^*$ and $p\in A[X]$ is primitive, we have $c' = c(g)$ and $p = p(g)$.

Step 3. Prove that c and p are multiplicative.

To show that $c(fg) = c(f)c(g)$ and $p(fg) = p(f)p(g)$, write

$f = c(f)p(f), \ g = c(g)p(g), \quad c(f), c(g)\in K^*, p(f), p(g) \in A[X] \text{ primitive.}$

Then $c(fg) p(fg) = fg = c(f)c(g)p(f)p(g)$. By step 1, $p(f)p(g)$ is primitive; hence $c(fg) = c(f)c(g)$ and $p(fg) = p(f)p(g)$.

Step 4. Every irreducible non-constant f in A[X] is irreducible in K[X].

Let $f \in A[X]$ be non-constant and irreducible. Note that it must be primitive since a prime element of $A$ is also prime in $A[X]$ (step 0). If $f$ is reducible in $K[X]$ we have $f = gh$ where $g, h\in K[X]$ are not constant. Then $f = p(f) = p(g)p(h)$ gives a factorization in $A[X]$ as well, which is a contradiction.

Step 5. Every irreducible f in A[X] is prime.

First suppose $f\in A[X]$ is non-constant and irreducible.

Suppose $gh = fh'$ where $g, h, h' \in A[X]$. Then $p(g)p(h) = p(f)p(h') = f\cdot p(h')$ since $f$ is primitive. Hence we may assume $g, h, h'$ are primitive. Since $K[X]$ is a PID, one of $g, h$ is a multiple of $f$ in $K[X]$ (by step 4). Without loss of generality write $g = fg'$ for some $g' \in K[X]$. But now $g = p(g) = p(f)p(g') = f\cdot p(g')$ so in fact $g$ is a multiple of $f$ in $A[X]$.

Finally if $f\in A[X] - \{0\}$ is constant and irreducible, it is irreducible in A. ♦

### Exercise B

Complete the proof by showing that $A[X]$ satisfies a.c.c. on principal ideals.

# Consequences

While proving the above theorem, we also showed

Proposition 3.

Let A be a UFD and K its field of fractions. Then $f\in A[X] - \{0\}$ is irreducible in $A[X]$ if and only if either

• $f$ is a prime element of A, or
• $\deg f > 0$, $f$ is primitive, and irreducible in $K[X]$.

Thus $k[X_1, \ldots, X_n]$ and $\mathbb Z[X_1, \ldots, X_n]$ are UFDs for any field k. Looking at the spectra of these rings, we see a striking difference between PIDs and general UFDs.

Proposition 4.

Every non-zero prime ideal of a PID $A$ is maximal.

Proof

Suppose we have ideals $\mathfrak p \subseteq \mathfrak a$ of A with $\mathfrak p$ a non-zero prime. Since A is a PID we write $\mathfrak p = (\pi)$ and $\mathfrak a = (\alpha)$ where $\pi \ne 0$. Then $\pi = \alpha\beta$ for some $\beta\in A$. Since $\pi$ is irreducible either $\alpha$ or $\beta$ is a unit. In the former case, $\mathfrak a = (1)$; in the latter case $\mathfrak a = \mathfrak p$. Hence $\mathfrak p$ is maximal. ♦

In particular, if A is a PID and not a field, its Krull dimension is 1.

In contrast, each $k[X_1, \ldots, X_n]$ is a UFD of Krull dimension at least n. Thus UFDs are structurally much more complicated than PIDs. They can be of arbitrarily high dimension.

### Exercise C

Prove that $\mathbb C[X, Y, Z]/(Z^2 - X^2 - Y^2)$ is not a UFD.

[Hint: define some norm function to an easier ring. ]

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### 6 Responses to Commutative Algebra 17

1. Vanya says:

In the lemma of the section on consequences, it is mentioned that $\text{deg} f >0$ is primitive. I guess you mean $f$ is primitive and degree f >0.

• limsup says:

Thanks. Corrected. 🙂

2. Vanya says:

Can you please post the solution to Exercise C?

• limsup says:

Here’s a hint. Use the fact that $\mathbb C[X, Y]$ is a UFD. In the extension $\mathbb C[X, Y] \hookrightarrow \mathbb C[X, Y, Z]/(Z^2 - X^2 - Y^2)$, define a norm map, similar to the case of $\mathbb Z \hookrightarrow \mathbb Z[\sqrt{-5}]$.

3. Vanya says:

Do we need A to be a PID in proposition 3? I suppose it is sufficient that A is a UFD.

• limsup says:

Yup I think you’re right.