Proof

The proof is divided into three steps.

S is a vector space.

The vectors 1, X, ..., X^{n - 1} of S span S.

The vectors 1, X, ..., X^{n - 1} of S are independent.

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First we specify the zero vector and the opposite of a vector:

• The zero vector is the class of the zero polynomial.
• The opposite of a vector coincides with the opposite of that element in the ring S.

The arithmetical rules for the ring S imply that all the axioms of a vector space over R are satisfied. For example, the `scalar' r R and the `vectors' f, g S satisfy

r(f + g) = r · f + r · g.

This is straightforward, as each residue class contains an element of degree at most n - 1.

Suppose that

₀·1 + ₁·X + ··· + _{n - 1}·X^{n - 1} = 0,

with ₀ , ..., _{n - 1} R, is a linear combination of the system whose sum is 0. The equality implies that ₀ + ₁ X + ··· + _{n - 1}X^{n - 1} is a representative of the class containing 0. Now every class contains exactly one representative of degree less than n. We therefore have ₀ + ₁ X + ··· + _{n - 1}X^{n - 1} = 0 in R[X] and hence

₀ = ₁ = ... = _n-1 = 0.