Let ∅_a : Z₅[x] → Z₅ be an evaluation homomorphism as in Theorem 22.4. Use Fermat's theorem to evaluate  ∅₃(x²³¹ + 3x¹¹⁷ - 2x⁵³ + 1).

Data from Theorem 22.4

(The Evaluation Homomorphisms for Field Theory) Let F be a subfield of a field E. let a be any element of E, and let x be an indeterminate. The map ∅_α : F [x] → E defined by ∅_α(a₀ + a₁x + a_nxⁿ) = a₀ + a₁α + ..... +a_nαⁿ

for (a₀ + a₁x + • • • + a_n xⁿ) ∈ F[x] is a homomorphism of F[x] into E. Also,∅_α(x) = α, and ∅_α maps F isomorphically by the identity map; that is, ∅_α(a) = a for a ∈ F. The homomorphism ∅_α is evaluation at α.

Proof The subfield and mapping diagram in Fig. 22.5 may help us to visualize this situation. The dashed lines indicate an element of the set. The theorem is really an immediate consequence of our definitions of addition and multiplication in F [ x]. The map ∅_α is well defined, that is, independent of our representation off (x) ∈ F [ x] as a finite sum a₀ + a₁x + • • • + a_n xⁿ) since such a finite sum representing f (x) can be changed only by insertion or deletion of terms 0xⁱ, which does not affect the value of ∅_αU(x)). 

If f(x)=a₀+a₁x+··•+a_nxⁿ,g(x)=b₀+b₁x+···+b_mx^m, and h(x)= f(x) + g(x) =c₀+ c₁x + • · · + c_rx^r, then ∅_α(f(x) + g(x)) = ∅_α(h(x)) =c₀+ C₁α + · · · + c_rα^r, while  ∅_α(f(x))+ ∅_α (g(x)) = (a₀ + a₁α ..... + a_nαⁿ) + (b₀ + b₁α + ..... b_mα^m).

Since by definition of polynomial addition we have c_i = a_i + b_i, we see that ∅_α(f(x) + g(x)) = ∅_α(f(x)) + ∅_α(g(x)). 

Turning to multiplication, we see that if f(x)g(x) =d₀ +d₁x + ...... + d_sx^s, then ∅_α(f(x)g(x) = d₀ + d₁α + ..... +d_sα^s, while [∅_α (f(x))] [∅_α (g(x))] = (a₀ + a₁α + ..... + α_nαⁿ) (b₀ + b₁α + .... +b_mα^m).

Since by definition of polynomial multiplication d_j = ∑_i^j=0 a_ib_j-i, we see that  ∅_α(f(x)g(x)) = [∅_αf(x))][∅_α(g(x))]. 

Thus ∅_α is a homomorphism. 

The very definition of ∅_α applied to a constant polynomial a ∈ F[x], where a ∈ F. gives ∅_α(a) = a, so ∅_α maps F isomorphically by the identity map. Again by definition of ∅_α, we have ∅_α(X) = ∅_α(1x) = 1α= α.