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Exercise 2218Algebra › Rings and fieldsM1Added 3 Nov 2026

Localization

#localization#fraction field#ring

Statement

Let RR be an integral domain. Show that RR embeds in its fraction field Frac⁡(R)\operatorname{Frac}(R), and compute Frac⁡(Z)=Q\operatorname{Frac}(\mathbb{Z})=\mathbb{Q}.

Hint

The fraction field consists of equivalence classes of pairs (a,b)(a,b) with a∈Ra\in R, b∈R∖{0}b\in R\setminus\{0\}, under (a,b)∼(c,d)  ⟺  ad=bc(a,b)\sim(c,d)\iff ad=bc.

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Solution

Equivalence. Define (a,b)∼(c,d)  ⟺  ad=bc(a,b)\sim(c,d)\iff ad=bc. This is an equivalence relation (reflexive, symmetric, transitive, using that RR is an integral domain for cancellation).

Field structure. Addition: ab+cd=ad+bcbd\frac ab+\frac cd=\frac{ad+bc}{bd}. Multiplication: ab⋅cd=acbd\frac ab\cdot\frac cd=\frac{ac}{bd}. These are well-defined. The zero is 01\frac01, the unit is 11\frac11. Every nonzero ab\frac ab has inverse ba\frac ba.

Embedding. R↪Frac⁡(R)R\hookrightarrow\operatorname{Frac}(R) by a↦a1a\mapsto\frac a1.

Example. Frac⁡(Z)=Q\operatorname{Frac}(\mathbb{Z})=\mathbb{Q}: every pq\frac pq with p,q∈Zp,q\in\mathbb{Z}, q≠0q\neq0, is a rational number, and every rational is of this form.

Conclusion. Every integral domain embeds in a field.