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The rational function field k(t) is not finite over k[t]

Statement

Let $k$ be a field. Then the rational function field $k (t)$ is not finitely generated as a $k [t]$ -module.

Facts & Assumptions

Given: A field $k$ , the polynomial ring $k [t]$ , and its fraction field $k (t)$ .

[L1]

The rational function field is the fraction field of $k [t]$ (For a field $F$ , $F (t) = Frac (F [t])$ is its rational function field; in particular $R (t) = Frac (R [t])$ ).

[L2]

The polynomial ring over a field is a unique factorisation domain (For every field $F$ , $F [x]$ is a unique factorisation domain).

Proof

technique · direct

1.1

Suppose that $k (t)$ is generated as a $k [t]$ -module by finitely many fractions $f_{1} / g_{1}, \dots, f_{m} / g_{m}$ with $f_{i}, g_{i} \in k [t]$ and $g_{i} \neq 0$ . Let $g = g_{1} \dots g_{m}$ . Then every $k [t]$ -linear combination of the generators has denominator dividing $g$ , so it can be written as $h / g$ for some $h \in k [t]$ .

givenL1

2.1

If $g$ is constant, then $k [t]$ itself would equal $k (t)$ , which is false because $1 / t \notin k [t]$ . So $g$ is nonconstant. By [L2], the nonunit $g + 1 \in k [t]$ has an irreducible factor $q$ . Since $q$ divides $g + 1$ , it does not divide $g$ .

L2step 1.1algebra

3.1

The fraction $1 / q$ lies in $k (t)$ by [L1]. If it belonged to the $k [t]$ -module generated by the chosen fractions, step 1.1 would give $1 / q = h / g$ for some $h \in k [t]$ , hence $g = h q$ . But then $q$ would divide $g$ , contrary to step 2.1. Therefore the assumed finite generating set cannot exist, so $k (t)$ is not finite over $k [t]$ .

L1step 1.1step 2.1algebra∎

Depends on

Used by

Dependency tree · two levels

9 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

J. S. Milne, A Primer of Commutative Algebra, v4.03, Theorem 13.1 (standard reference, not scraped)
Allen B. Altman and Steven L. Kleiman, A Term of Commutative Algebra, 13th ed., Theorem (15.4) (standard reference, not scraped)