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TheoremStatement: 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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C=R[x]/(x2+1) is a field, every element is uniquely a+bi, and every nonzero element has inverse (a−bi)/(a2+b2)

Statement

C=R[x]/(x2+1) is a field containing the embedded copy of R. Every complex number has a unique form a+bi with a,b∈R, and (a+bi)+(u+vi)=(a+u)+(b+v)i, (a+bi)(u+vi)=(au−bv)+(av+bu)i. If a+bi≠0, then (a+bi)−1=a−bia2+b2.

Facts & Assumptions

Given: The quotient construction C=R[x]/(x2+1).

[F1]

The polynomial x2+1 is irreducible over R (x2+1 is irreducible over R).

[F2]

For a monic irreducible polynomial p of degree n, F[x]/(p) is a field extension of F and every class has a unique representative of degree below n (F[x]/(p) for monic irreducible p is a field extension containing the root x+(p) with unique reduced representatives).

[F3]

The real numbers form an ordered field (The reals form a totally ordered field).

[F4]

In an ordered field every nonzero square is positive (Squares of nonzero elements are positive).

[F5]

C is the quotient R[x]/(x2+1) and i=x+(x2+1) (The complex numbers as R[x]/(x2+1), with the real embedding and imaginary unit i).

Proof

technique · direct
1.1

Apply [F2] to [F1]. With the construction in [F5], this proves that C is a field, that the constant-class map embeds R, and that every class has a unique representative a+bx, hence a unique form a+bi.

F1F2F5
2.1

Quotient addition and multiplication, followed by i2=−1, give the two displayed coordinate formulas. All field axioms are inherited from the field in step 1.1.

F5step 1.1algebra
2.2

If a+bi≠0, uniqueness in step 1.1 gives a≠0 or b≠0. The corresponding square is positive by [F4], the other square is nonnegative, and therefore a2+b2>0 in the ordered field [F3].

F3F4step 1.1
3.1

Direct multiplication using step 2.1 gives (a+bi)(a−bi)=a2+b2. Since the real denominator is nonzero by step 2.2, division proves the inverse formula.

step 2.1step 2.2algebra∎

Depends on

Used by

…and 24 more results.

Cited to discharge well-definedness by The complex numbers as ℝ[x]/(x²+1), with the real embedding and imaginary unit i.

Dependency tree · two levels

20 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