Let a = a₀ + a₁X + ··· + a_mX^m and b = b₀ + b₁X + ··· + b_mX^m be two polynomials that are congruent modulo n (according to the convention in Chapter 3 we may assume the highest power of a monomial in both a and b to be equal to  m). Then a_i and b_i differ by a multiple of n for i = 0, 1, ..., m. This implies that a_i = b_i mod n for i = 0, 1, ..., m. So our definition does not depend on the representative a or b that we may have chosen.

If I(a₀ + a₁X + ··· + a_mX^m + (n)) = I(b₀ + b₁X + ··· + b_mX^m + (n)), then a_i = b_i mod n for i = 0, ..., m. This implies that n | (a_i - b_i) for each i and hence a₀ + a₁X + ··· + a_mX^m and   b₀ + b₁X + ··· + b_mX^m are congruent modulo n.

If a₀ + a₁X + ··· + a_mX^m  represents a polynomial in (Z/nZ)[X], then  I(a₀ + a₁X + · · · + a_mX^m + (n)) = a₀ + a₁X + ··· + a_mX^m.

Let a₀ + a₁X + ··· + a_mX^m be a representative of a and b₀ + b₁X + ··· + b_mX^m a representative of b. Then:

I(a + b) = a₀ + b₀ + (a₁ + b₁)X + ··· + (a_m + b_m)X^m,

(first add a and b according to the definition and then apply the map I), while

I(a) + I(b) = (a₀ + a₁X + ··· + a_mX^m) + (b₀ + b₁X + ··· + b_mX^m).

Since the right-hand sides of the expressions for I(a + b) and I(a) + I(b) coincide, the required equality follows.

Since I respects addition, we have, for any a,

I(a) = I(a + 0+(n)) = I(a) + I(0+(n)),

and so I(0 + (n)) = 0.

The proof is similar to the one for addition. It is left to the reader.

The proof is similar to the one for the zeros. It is left to the reader.