CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-13

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
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For a field $F$ , $F (t) = Frac (F [t])$ is its rational function field; in particular $R (t) = Frac (R [t])$

Statement

For every field $F$ , the polynomial ring $F [t]$ is an integral domain, and $F (t) : = Frac (F [t]) = {\frac{f (t)}{g (t)} : f, g \in F [t], g \neq 0}$ is a field containing an embedded copy of $F [t]$ . It is called the rational function field over $F$ . In particular, $R (t) = Frac (R [t])$ .

Facts & Assumptions

Given: A field $F$ .

[F1]

Every field is an integral domain (Every field is a commutative ring with $1 \neq 0$ ; it is an integral domain, and it is a commutative division ring).

[F2]

If $R$ is an integral domain, then $R [t]$ is an integral domain (A polynomial ring over an integral domain is an integral domain).

[F3]

The field of fractions of a domain consists of fractions with nonzero denominator and is a field containing the domain injectively (The field of fractions $Frac (D) = (D ∖ {0})^{- 1} D$ of an integral domain, $Frac (D)$ is a field and $d \mapsto d / 1$ embeds the integral domain $D$ ).

Proof

technique · direct

1.1F1F2F3

By [F1] and [F2], $F [t]$ is an integral domain. Applying [F3] gives the displayed set of fractions, its field structure, and the embedding of $F [t]$ .

2.1step 1.1∎

Taking $F = R$ gives the final assertion.

Depends on

Dependency tree · 2 levels

Sources

The CRing Project, Chapter 13: Fields and Extensions

For a field FF, F(t)=Frac⁡(F[t])F(t)=Frac(F[t]) is its rational function field; in particular R(t)=Frac⁡(R[t])R(t)=Frac(R[t])